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CONTENTS A
Definitions
8
A1
Important Properties of Engineering Materials
8
A2
Simple Stresses and Strains
9
A3
General Definitions
10
B
Conversion Tables
26
B1
Introduction
26
B2
List of Units and Abbreviations
26
B3
Conversion factors Grouped by Category
28
B4
Miscellaneous Data
33
C
Section Shear Center
34
C1
Location of Shear Center – Open Section
35
C2
Shear Center – Curved Web
36
C3
Shear Center Beam with Constant Shear Flow Between Booms
38
C4
Shear Center – Single Cell Closed Section
39
C5
Shear Center – Single Cell Closed Section with Booms
40
C6
Shear Center for Standard Section (Tables)
41
D
Warping Constant
46
D1
Warping Constant – Typical Sections (Tables)
48
E
Mechanical Properties
53
E1
DATA Basis
53
E2
Stress Strain Data
54
E3
Definitions of Terms
54
E4
Temperature Effects
57
E5
Fatigue Properties
57
E6
Generalised Stress - Strain Equation
59
E6.1
Problems
61
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1
Prismatic Bar In Tension
63
1.1
Limit Stress Analysis
63
1.2
Ultimate Stress Analysis
63
1.3
Problem
64
2.
Prismatic Bar In Compression
65
2.1
Compression Stress Analysis
65
2.1.2
Problem
65
2.2
Euler Buckling
67
2.3
For Other Buckling Cases
67
2.3.1
Buckling For Eccentric Loading
67
2.3.2
Beam-Column
68
2.3.2.1
Notations And Conventions
68
2.3.2.2
Terms Used
69
2.3.2.3
Calculation Of Bending Moment
69
2.3.2.4
Allowable Stress
74
2.3.3
Local Buckling (Thin Walled Structures)
75
2.3.3.1
Flanges
75
2.3.3.2
Thin Webs
75
2.3.3.3
Plates And Shells
76
2.3.3.4
Analysis Of Thin Walled Structures
80
2.4
Problems
80
2.4.1
Problem For Euler Buckling
80
2.4.2
Problem For Beam – Column
83
2.4.3
Problem For Local Buckling
86
2.4.3.1
Problem For Combined Bending And Shear
87
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3
Lug Analysis
88
3.1
Failure In Tension
88
3.2
Failure By Shear Tear Out
89
3.3
Failure By Bearing
90
3.4
Lug Strength Analysis Under Transverse Loading
90
3.5
Lug Strength Analysis Under Oblique Loading
91
3.6
Lug Analysis (Avro Method)
92
3.7
Problems
93
4.
Member In Bending
97
4.1
Introduction To Bending
98
4.2
Stress Analysis Of Beam
100
4.3
Bending Stresses
101
4.3.1
Limit Stress Analysis
101
4.3.2
Ultimate Stress Analysis
101
4.3.3
Rectangular Moment Of Inertia and Product of Inertia
102
4.3.4
Parallel Axis Theorem
103
4.4
Elastic Bending Of Homogeneous Beams
114
4.5
Elastic Bending Of Non Homogeneous Beams
128
4.6
Inelastic Bending of Homogeneous Beams
133
4.7
Elastic Bending Of Curved Beams
146
4.8
Miscellaneous Problems
148
4.8.1
Bending Stress In Symmetrical Section
151
4.8.2
Bending Stress In Unsymmetrical Section
153
4.8.3
Beams Of Uniform Strength
159
4.8.4
Beams Of Composite Section
160
4.8.5
Bending Of Unsymmetrical Section About Principal Axis
163
4.8.6
Bending Of Unsymmetrical Section About Arbitrary Axis
164
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5
Member In Torsion
172
5.1
Torsion of Circular Bars
173
5.1.1
Elastic and Homogeneous
176
5.1.2
Elastic and Non-homogeneous
178
5.1.3
Inelastic and Homogeneous
180
5.1.4
Inelastic and Non-homogeneous
182
5.1.5
Residual Stress Distribution
184
5.1.6
Power Transmission
185
5.2
Torsion of Non-Circular Bars
189
5.3
Elastic Membrane Analogy
190
5.4
Torsion of Thin-Wall Open Sections
194
5.5
Torsion of Solid Non-Circular Shapes
198
5.6
Torsion of Thin-Wall Closed Sections
202
5.7
Torsion-Shear Flow Relations in Multiple-Cell Closed Sections
208
5.8
Shear Stress Distribution and Angle of Twist for Two-Cell Thin-Wall Closed Section
5.9
Shear Stress Distribution and θ for Multiple-Cell Thin-Wall Closed Section
214
5.10
Torsion of Stiffened Thin-Wall Closed Sections
215
5.11
Effect of End Restraint
217
5.12
Miscellaneous problems
222
5.12.1
Problem For Torsion On Circular Section
222
5.12.2
Problem For Torsion In Non Circular Section
225
5.12.3
Problem For Torsion In Open Section Composed Of Thin Plates
226
5.12.4
Problem For Torsion On Thin Walled Closed Section
228
5.12.5
Problem For Torsion In Thin Walled Unsymmetrical Section
233
6
Transverse Shear Loading Of Beams With Solid Or Open Sections 236
6.1
Introduction
237
6.2
Shear Center
238
6.3
Flexural Shear Stress And Shear Flow
239
6.4
Shear Flow Analysis For Symmetric Beams
261
209
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6.5
Shear Flow Analysis For Unsymmetric Beams
286
6.6
Analysis Of Beams With Constant Shear-Flow Webs (I.E., Skin-Stringer Type Sections)
298
7.
Transverse Shear loading Of Beams With Closed C.S
310
7.1
Single-Cell Unstiffened Box Beam: Symmetrical About One Axis
311
7.2
Statically Determinate Box Beams With Constant-Shear-Flow Webs
325
7.3
Single-Cell Multiple-Flange Box Beams. Symmetric And Unsymmetric Cross Sections
331
7.4
Multiple-Cell Multiple-Flange Box Beams. Symmetric And Unsymmetric Cross Sections
7.5
The Determination Of The Flexural Shear Flow Distribution By
338
Considering The Changes In Flange Loads (The Delta P Method) 346 7.6
Shear Flow In Tapered Sheet Panels
357
8.
Combined Transverse Shear, Bending, And Torsion Loading
360
9
Internal Pressure
360
9.1
Membrane Equations Of Equilibrium
360
9.2
Special Problems In Pressurized Cabin Stress Analysis
368
10
Analysis Of Joints
373
10.1
Analysis Of Riveted Joints
373
10.1.1
Shear Failure Of Rivets
375
10.1.2
Tensile Failure Of Plate Along Rivet Line
375
10.1.2.1
Single Row And Rivets Having Uniform Diameter And Equal Pitch375
10.1.2.2
More Than One Row And Rivets With Equal Spacing
376
10.1.2.3
Rivet Line With Varying Number Of Rivets
377
10.1.2.4
Rivets Of Varying Diameter, Sheet Thickness And Pitch Distance 378
10.1.3
Failure By Double Shear Of Plate
378
10.1.4
Rivets Subjected To Tensile Loads
379
10.1.5
Rivets Subjected To Eccentric Loads
380
10.1.6
Inter Rivet Buckling
381
10.1.7
Calculation Of Fir Using ESDU Data Sheets
382
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10.1.8
Problems
382
10.2
Analysis Of Bolted Joints
397
10.2.1
Failure By Bolt Shear
397
10.2.2
Failure By Bolt Bending
398
10.3
Analysis Of Welded Joints
399
10.3.1
Shear Failure In Weld
399
10.3.2
Tensile Failure Of Weld
400
11
Combined Stress Theories For Yield And Ultimate Failure
401
11.1
Determination Of Yield Strength Of Structural Member
401
11.1.1
Maximum Shearing Stress Theory
401
11.1.2
Maximum Energy Of Distortion Theory
402
11.1.3
Maximum Strain Energy Theory
402
11.1.4
Octahedral Shear Stress Theory
403
11.2
Determination Of Ultimate Strength Of A Structural Member
404
11.2.1
Maximum Principal Stress Theory
404
11.3
Problem
404
12
Cutouts In Plane Panels
409
12.1
Framing Members Around The Cutouts
409
12.2
Framing Cutouts With Doublers And Bends
412
12.3
Ring Or Donut Doublers For Round Holes
414
12.4
Webs With Lightning Holes Having Flanges
416
12.5
Structures With Non Circular And Non Rectangular Cutouts
417
12.6
Cutouts In Fuselage
418
12.6.1
Load Distribution Due To Fuselage Skin Shear
419
12.6.2
Load Distribution Due To Fuselage Bending
421
12.6.3
Load Distribution Due To Fuselage Cabin Pressurization
422
12.6.3.1
Panels Above And Below The Cutout
423
12.6.3.2
Panels At Sides Of The Cutout
424
12.6.3.3
Corner Panels Of The Cutout
424
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12.7
Problem
425
13
Tension Clips
430
13.1
Single Angle
433
13.2
Double Angle
433
13.3
Problem
435
14
Pre - Tensioned Bolts
438
14.1
Optimum Preload
439
14.2
Total Bolt Load
440
14.2.1
Mating Surfaces Not In Initial Contact
440
14.2.2
Mating Surfaces In Initial Contact
440
14.3
Preload For A Given Wrench Torque
441
14.4
Bolt Strength Requirements
441
14.5
Ultimate Stress Analysis
442
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A. DEFINITIONS: A1. Import Properties of Engineering Material: Its necessary to know from engineering point of view the following import properties of solids. a. Elasticity: It is property by virtue of which the body regains its original shape and size after the removal of the external force acting on the body. b. Isotropic Material: It is the material, which is equally elastic in all directions. c. Ductility: It is the property of the material by virtue of which it can be drawn out into wires of smaller dimension e.g., copper, Aluminium. d. Plasticity: It is the property of the material wherein the strain does not disappear even after the load or stress is not acting on the material. The material becomes plastic in nature and behaves like a viscous liquid under the influence of large forces. e. Malleability: It is the property possessed by the material enabling the material to extend uniformly in a direction without rupture. As such the material can be hot rolled, forged or drop stamped. A malleable material is highly plastic. f. Brittleness: A material is said to be brittle when it cannot be drawn out into thon wires. This due to lack of ductility and the material breaks into pieces under loading. g. Toughness: It is property of the material due to which it is capable of absorbing energy without fracture. h. Hardness: It is the property of the material by virtue of which the is capable of resisting abrasion or indentation.
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A2. Simple Stresses and Strains. Allowable Stress (Working Stress): If a member is so designed that the maximum stress as calculated for the expected condition of service is less than some certain value, the member will have a proper margin of security against damage or failure. This certain value is the allowable stress of the kind and for the material and condition of service in question. The allowable stress is less than the damaging stress because of uncertainty as to the conditions of service, non uniformity of material, and inaccuracy of stress analysis. The margin between the allowable stress and the damaging stress may be reduced in proportion to the certainty with which the condition of service are known, the intrinsic reliability of material, the accuracy with which the stress produced by the loading can be calculated and the degree to which failure is unaffected by danger or loss. (Compare Damaging stress, Factor of Safety; Factor of utilization; Margin of safety) Apparent Elastic Limit: The stress at which the rate of change of strain with respect to stress is 50 percent greater than at zero stress. It is more definitely determine from the stress strain diagram than is the proportional limit, and is useful for comparing material s of the same general class(Compare Elastic limit, Proportional limit, Yield point; Yield strength). Apparent Stress: The stress corresponding to a given unit strain on the assumption of uniaxial elastic stress is called apparent stress. It is calculated by multiplying the unit strain by the modulus of elasticity and may differ from the true stress because the effect of transverse stresses is not taken into account.
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A3. General Definitions: Bending Moment: The bending moment at any section of the beam is the moment of all forces that act on the beam to the left of that section, taken about the horizontal axis of the section. The bending moment is positive when clockwise and negative when counter clock wise; a positive bending moment therefore bends it so that it is concave downward. The moment equation is an expression for the bending moment at any section in terms of x, the distance to that section measured from a chosen origin, usually taken at the left end of the beam. Boundary Conditions: As used in strength of materials, the term usually refers to the condition of stress, displacement, or slope at the ends or edges of a member, where these conditions are apparent from the circumstances of the problem. Thus for a beam with fixed ends the zero slope at each end is a boundary condition are apparent from the circumstances of the problem. Thus for beam with fixed ends, the zero slope at each is a boundary condition: for a pierced circular plate with freely supported edges, the zero radial stress at each edge is a boundary condition. Brittle Fracture: The tensile failure with negligible plastic deformation of an ordinarily ductile metal is called brittle fracture. Bulk Modulus of Elasticity: The ratio of a tensile or compressive stress, triaxial and equal in all direction(e.g., hydrostatic pressure), to the relative change it produces in volume. Central Axis (Centroidal Axis): A central axis of an area is one that passes through the centroid; it is understood to lie in the plane of the area unless contrary is stated. When taken normal to the plane of the area, it is called the central polar axis.
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Centroid of an Area: That point in the plane of the area about any axis through which the moment of the area is zero; it coincides with the center of gravity of the area materialized as an infinitely thin homogenous and uniform plate. Corrosion Fatigue: Fatigue aggravated by corrosion , as in parts repeatedly stressed while exposed to salt water. Creep: Continuous increase in deformation under constant or decreasing stress. The term is ordinarily used with reference to the behaviour of metals under tension at elevated temperatures. The similar yielding of a material under compressive stress is usually called plastic flow, or flow. Creep at atmospheric temperature due to sustained elastic stress is sometimes called drift, or elastic drift. Another manifestation creep, the diminution in stress when deformation is maintained constant, is called relaxation. Damaging Stress: The least unit stress of a given kind and for a given material and condition of service that will render a member unfit for service before the end of its normal life. It may do this by producing excessive set, by causing creep to occur at an excessive rate, or by causing fatigue cracking, excessive strain hardening, or rupture.. Damping Capacity: The amount of work dissipated into heat per unit of strain energy present at maximum strain for a complete cycle.
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Deformation (Strain): Change in the form or dimensions of a body produced by stress. Elongation is often used for tensile strain, compression or shortening for compressive strain, and detrusion for shear strain. Elastic deformation is such deformation as disappears on removal of stress; permanent deformation is such deformation as disappears on remains on removal of stress. Eccentricity: A load or component of a load normal to a given cross section of a member is eccentric with respect to that section if it does not act through the centroid The perpendicular distance from the line of action of the load to either principal central axis is the eccentricity with respect to that axis. Elastic: Capable of sustaining stress without permanent deformation; the term is also used to denote conformity to the law of stress strain proportionality. An elastic stress or strain within the elastic limit. Elastic Axis: The elastic axis of a beam is the line, lengthwise of the beam, along which transverse loads must be applied in order to produce bending only, with no torsion of the beam at any section. Strictly speaking, no such line exists except for a few conditions of loading. Usually the elastic axis is assumed to be the line that passes through the elastic center of every section. The term is most often used with reference to an airplane wing of either the shell or multiplespar type (Compare Torsional center; Flexural center; Elastic center). Elastic Center: The elastic center of given section of a beam is that point in the plane of the section lying midway between the flexural center and center of twist of that section. The three points may be identical and are usually assumed to be so. (Compare Flexural center; Torsional center; Elastic axis).
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Elastic Curve: The curve assumed by the axis of a normally straight beam or column when bent by loads that do not stress it beyond the proportional limit. Elastic Limit: The least stress that will cause permanent set (Compare Proportional limit; Apparent elastic limit; Yield point; Yield strength). Elastic Ratio: The ratio of the elastic limit to the ultimate strength Ellipsoid of Strain: An ellipsoid that represent the state of strain at any given point in a body; it has the form assumed under stress by a sphere centered at that point. Ellipsoid of Stress:. An ellipsoid that represents the state of stress at a given point in a body; its semi axes are vectors representing the principal stresses at that point, and any radius vector represents the resultant stress on a particular plane through the point. For a condition of plane stress (one principal stress zero) the ellipsoid becomes the ellipse becomes the ellipse of stress. Endurance Limit(Fatigue strength): The maximum stress that can be reversed an indefinitely large number of times without producing fracture of a material. Endurance Ratio: Ratio of the endurance limit to the ultimate static tensile strength is called endurance ratio.
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Endurance Strength: The highest stress that a material can withstand with repeated application or reversal without rupture for a given number of cycles is the endurance strength of that material for that number of cycles. Unless otherwise specified, reversed stressing is usually implied (Compare Endurance Limit) Energy of Rupture: The work done per unit volume in producing fracture is the energy of rupture. It is not practical to establish a definite energy of rupture value for a given material because the result obtained depends upon the form and proportion of the test specimen and manner of loading. As determined by similar tests specimen, the energy of rupture affords a criterion for comparing the toughness of different materials. Equivalent Bending Moment: A bending moment that, acting alone, would produce in a circular shaft a normal (tensile or compressive) stress of the same magnitude as a maximum normal stress produced by a given bending moment and a given twisting moment acting simultaneously. Equivalent Twisting Moment: A twisting moment that, acting alone would produce in a circular shaft a shear stress of the same magnitude as the shear produced by a given twisting and a given bending moment acting simultaneously. Factory of Safety: The ratio of the load that would cause failure of a member or structure to the load that is imposed upon it in service. The term usually has this meaning; it may also be used to represent the ratio of breaking to service value of speed, deflection, temperature variation, or other stress producing factor against possible increase in which the factor of safety is provide as a safeguard (Compare Allowable stress; Margin of safety).
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Factor of Strain: In the presence of stress raisers, localized peak strains are developed. The factor of strain concentration is the ratio of the localized maximum strain at a given cross section to the nominal average strain on that cross section. The nominal average strain behaviour of the material. In a situation where all stresses and strains are elastic, the factors of stress concentration and strain concentration are equal (Compare Factor of stress concentration). Factor of Stress Concentration: Irregularities of form such as holes, screw threads notches, and sharp shoulders, when present in a beam, shaft, or other member subject to loading may produce high localized stresses. This phenomenon is called stress concentration, and the form irregularities that causes it are called stress raisers. The ratio of the true maximum stress to the stress calculated by the ordinary formulas of mechanics (flexural formula, torsion formula etc.), using the net section but ignoring the changed distribution of stresses is the factor of stress concentration for the particulars type of stress raiser. Factor of Stress Concentration in Fatigue: At a specified number of loading cycles, the fatigue strength of a given geometry depends upon the stress concentration factor and upon material properties. The factor of stress concentration in fatigue is the ratio of the fatigue strength without a stress concentration. It may vary with the specified number of cycles as well as with material Factor of Utilization: The ratio of the allowable stress to the ultimate strength. For cases in which stress is proportional to load, the factor of utilization is the reciprocal to the factor of safety. Fatigue: Tendency of material to fracture under many repetitions of a stress considerably less than the ultimate static strength.
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Fibre Stress: A term used for convenience to denote the longitudinal tensile or compressive stress in a beam or other member subject to bending. It is sometimes used to denote this stress at the point or points most remote from the neutral axis, but the term stress in extreme fibre is preferable for this purpose. Also, for convenience, the longitudinal elements of filaments of which a beam may be imagined as composed are called fibres. Fixed (Clamped, built-in, encastre): A conditioned of support at the ends of a beam or column or at the edges of a plate or shell that prevents rotation and transverse displacements of the edge of the neutral surface but permits longitudinal displacements. Flexural Center (Shear Center): With reference to a beam, the flexural center of any section is that point in the plane of the section through which a transverse load, applied at that section, must act if bending deflection only is to be produced with no twist of the section (Compare Torsional center; Elastic center; Elastic axis). Form Factor: The terms pertains to a beam section of a given shape and means the ratio of the modulus of rupture of a beam having a section adopted as standard. The standard section is usually taken as rectangular or square; for wood it is a 2 by 2 in square, with edges horizontal and vertical. The term is also used to mean, for a given maximum fiber stress within the elastic limit, the ratio of the actual resisting moment of a wide-flanged beam to the resisting moment the beam would develop if the fiber stress were uniformly distributed across the entire width of the flanges. So used, the term expresses the strength –reducing effect of shear lag. Fretting Fatigue: Fatigue aggravated by surface rubbing, as in shaft with press fitted collars.
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Guided: A condition of support at the ends of a beam or column or at the edge of a plate or shell that prevents rotation of the edge of the neutral surface but permits longitudinal and transverse displacement (Compare Fixed; Guided; Supported). Held: A condition of support at the ends of a beam or columns or at the edge of a plate or shell that prevents longitudinal and transverse displacement of the edge of the neutral surface but permits rotation in the plane of bending (Compared Fixed; Guided Supported). Influence Line: Usually pertaining to a particular section of a beam, influence line is a curve drawn; so that its ordinate at any point represent the value of the reaction; vertical shear, bending moment, or deflection produced at the particular section by a unit load applied at the point where the ordinate is measured. An influence line may be used to show the effect of load position on any quantity dependent thereon, such as the stress in a given truss member, the deflection of a truss, or the twisting moment in a shaft. Isoclinic: A line (in a stressed body) at all points on which the corresponding principal stresses have the same directions. Isotropic: Having the same properties in all properties in all direction. In discussions pertaining to strengths of materials, isotropic usually means having the same strength and elastic properties (modulus of elasticity, modulus of rigidity, and poisson’s ratio) in all directions. Korn (Kornal): The korn is that area in the plane of the section through which the line of action of a force must pass if that force is to produce, at all points in the given section, the same kind of normal stress, i.e., tension throughout or compression throughout
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Margin of safety: As used in aeronautical design, margin of safety is the percentage by which the ultimate strength of a member exceeds the design load. The design load is the applied load, or maximum probable load, multiplied by a specified factor of safety. [The use of the term margin of safety and design load in this sense is practically restricted to aeronautical engineering] Mechanical Hysteresis: The dissipation of energy as heat during a stress cycle, which is revealed graphically by failure of the descending and ascending branches of the stress-strain diagram to coincide. Modulus of Elasticity (Young’s Modulus): The rate of change of unit tensile or compressive stress with respect to unit tensile or compressive strain for the condition of uniaxial stress within the proportional limit. For most, but not all, materials, the modulus of elasticity is the same for tension and compression. For nonisotropic material such as wood, it is necessary distinguish between the moduli of elasticity in different directions. Modulus of Resilience: The strain energy per unit volume absorbed up to the elastic limit under the condition of uniform uniaxial stress. Modulus of Rigidity (Modulus of Elasticity in Shear): The rates of change of unit shear stress with respect to unit shear strain for the condition of pure shear within the proportional limit. For non-isotropic materials such as wood, it is necessary to distinguish between the moduli of rigidity in different directions. Modulus of Rupture in Bending (Computed ultimate Twisting Strength): The fictitious tensile or compressive stress in the extreme fiber of a beam computed by the flexure equation σ =
Mc , where M is the bending moment that causes rupture. I
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Modulus of Rupture in Torsion (Computed Ultimate twisting moment): The fictitious shear stress at the surface of circular shaft computed by the tension formula τ =
Tr , where T is the Twisting moment that causes rupture. J
Moment of Area (First Moment of an Area, Statical Moment of an Area): With respect to an axis, the sum of the products obtained by multiplying each element of the area dA by its distance from the axis y; it is therefore the quantity ∫ dAy . An axis in the plane of the area is implied. Moment of Inertia of an area (Second Moment of an Area): The moment of inertia of an area with respect to axis is the sum of the products obtained by multiplying each element of the area dA by the square of its distance from the axis y; it is therefore the quantity
∫ dAy
2
. An axis in the plane of the area is implied, if the axis is
normal to that plane, the term polar moment of inertia. Neutral Axis: The line of zero fiber stress in any given section of a member subject to bending; it contains the neutral axis of every section. Notched – Sensitivity Ratio: Used to compare stress concentration factor kt and fatigue-strength reduction
factor kf, the notch sensitivity ratio is commonly defined as the ratio (k f − 1) / (k t − 1) .It varies
from 0, for some ductile materials, to 1, for some hard brittle materials. Neutral Surface:
The longitudinal surface of zero fibre stress in a member subject to bending; it contains the neutral axis of every section.
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Plasticity:
The property of sustaining appreciable (visible to the eye) permanent deformation without rupture. The term is also used to denote the property of yielding or flowing under steady load. Poisson’s Ratio:
The ratio of lateral unit strain to longitudinal unit strain under the condition of uniform and uniaxial longitudinal unit within the proportional limit. Principal Axes:
The principal axes of an area for a given point in its plane are the two mutually perpendicular axes, passing through the point and lying in the plane of the area, for one of which the moment of inertia is greater and for the other less than for any other coplanar axis passing through that point. If the point in question is the centroid of the area, these axes are called principal central axes. Principal Planes; Principal Stresses:
Through any point in stressed body there pass three mutually perpendicular planes, the stress on each of which is purely normal, tension, or compression; these are the principal planes for that point. The stresses on these planes are the principal stresses; one of them is the maximum stress at that point, and one of them is the minimum stress at that point. When one of the principal stresses is zero, the condition is one of uniaxial stress. Product of Inertia of an Area:
With respect to a pair of rectangular axes in its plane, the sum of the products obtained b multiplying each element of the area dA by its coordinates with respect to those axes x and y; it is therefore the quantity ∫ dAxy . Proof Stress:
Pertaining to acceptance tests of metals, a specified tensile stress that must be sustained without deformation in excess of a specified amount.
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Proportional Limit:
The greatest stress that a material can sustain without deviating from the law of stress-strain proportionality. (Compare Elastic limit; apparent elastic limit; Yield point; Yield strength) Radius of Gyration:
The radius of gyration of area with respect to a given axis is the square root of the quantity obtained by dividing the moment of inertia of the area with respect to that axis by the area. Reduction of Area:
The difference between the cross sectional area of a tension specimen at the section of rupture before loading and after rupture. Rupture Factor:
Used in reference to brittle materials, i.e. materials in which failure occurs through tensile rupture rather than through excessive deformation. For a member of given form, size, and material, loaded and supported in a given manner, the rupture factor is the ratio of the fictitious maximum tensile stress at failure, as calculated by the appropriate formula for elastic stress, to the ultimate tensile strength of the material, as determined by a conventional tension test. Section Modulus (Section Factor):
Pertaining to the cross section of a beam, the section modulus with respect to either principal central axis is the moment of inertia with respect to that axis divided by the distance from that axis to the most remote point of the section. The section modulus largely determines the ability of a beam to carry load with no plastic deformation Anywhere in the cross section (Compare plastic section modulus)
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Set (Permanent set, Permanent deformation, Plastic strain, Plastic deformation):
Strain remaining after removal of stress. Shakedown Load (Stabilizing Load):
The maximum load that can be applied to a beam or rigid frame and on removal leave such residual moments that subsequent application of the same or a smaller load will cause only elastic stresses. Shear Lag:
Because of shear strain, the longitudinal tensile or compressive bending stress in wide beam flanges diminishes with the distance from the web or webs, and this stress diminution is called shear lag. Singularity Functions:
A class of function that, when used with some caution, permit expressing in one equation what would normally be expressed in several separate equations, with boundary conditions being matched at the ends of the intervals over which the several separate expressions are valid. Ultimate Strength:
The ultimate strength of a material in tension, compression, or shear, respectively, is the maximum tensile, compressive, or shear stress that the material can sustain calculated on the basis of the ultimate load and the original or unstrained dimensions. It is implied that the condition of stress represents uniaxial tension, uniaxial compression, or pure shear, as the case may be. Unit Stress:
The amount of stress per unit of area. The unit stress (tensile, compressive, or shear) at any point on a plane is the limit, as ΔA approaches 0, ΔP / ΔA where ΔP is the total tension, compression, or shear on an area ΔA that lies in the plane and includes the point.
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Vertical Stress:
Refers to a beam, assumed for convenience to be horizontal and loaded and supported by forces that all lies in a vertical plane. The vertical shear at any section of the beam is the vertical component of all the forces that act on the beam to the left of the section. The shear equation is an expression for the vertical shear at any section in terms of x, the distance to that section measured from a chosen origin, usually taken at the left end of the beam. Yield Point:
The lowest stress at which strain increase in stress. For some purposes it is important to distinguish between the upper yield point, which is the stress at which is the stressstrain diagram first becomes horizontal, and lower yield point, which is the somewhat lower and almost constant stress under which the metal continues to deform. Only a few materials exhibit a true yield point; for other materials the term is sometimes used synonymously with yield strength (Compare yield strength; Elastic limit; Apparent elastic; proportional limit) Yield Strength:
The stress at which a material exhibits a specified permanent deformation or set. The set is usually determined by measuring the departure of the actual stress-strain diagram from an extension of the initial straight portion. The specified value is often taken as a unit strain of 0.002. Slenderness Ratio:
The ratio of length of a uniform column to the least radius of gyration of the cross section. Slip Lines:
Lines that appear on the polished surface of crystal or crystalline body that has been stressed beyond the elastic limit. They represent the intersection of the surface by planes on which shear stress has produced plastic slip or gliding.
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Strain:
Any forced change in the dimensions of a body. A stretch is tensile strain; a shortening is a compressive strain; and angular distortion is a shear strain. The word strain is commonly used to connote unit strain. Strain energy:
Other wise called as elastic energy, potential energy of deformation. Mechanical energy stored up in the stressed material. Stress with in the elastic limit is implied; therefore, the strain energy is equal to the work done by the external forces in producing the stress and is recoverable. Strain rosette:
At any point on the surface of a stressed body, strains measured on each of the intersecting gauge line make possible the calculation of the principal stresses. Such gauge lines and the corresponding strains are called strain rosettes. Stress:
Internal force exerted by either of two adjustant parts of a body upon the other across an imagined plane of separation. When the forces are parallel to the plane, stress is called shear stress; when the forces are normal to the plane, the stress is called normal stress; when the normal stress is directed towards the part on which it acts, it is called compressive stress; and when it is directed away from the part on which it acts, it is called tensile stress. Shear, compressive, and tensile stresses, respectively, resists the tendency of the parts to mutually slide, approach, or separate under the action of applied forces. Stress solid:
The solid figure formed by the surfaces bounding vectors drawn at all points of the cross section of a member and representing the unit normal stress at each such point. The solid stress gives a picture of the stress distribution on a section.
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Stress trajectory (isostatic):
A line in a stressed body tangent to the direction of one of the principal stresses at every point through which it passes id called stress trajectory. Torsional center:
Otherwise called as center of twist or center of torsion or center of shear. If a twisting couple is applied at a given section of a straight member, that section rotates about some point in its plane. This point which does not move when the member twists, is the torsional center of that section. True stress:
For an axially loaded bar, the load divided by corresponding actual cross section area. It differs from the stress as ordinarily defined because of the change in area due to loading. Twisting moment (torque):
At any section of the member, the moment of all forces that act on the member to the left of that section, taken about a polar axis through the flexure center of that section. For the sections that are symmetrical about each principal central axis, the flexural center coincides with the centroid. Ultimate elongation:
The percentage of permanent deformation remaining after tensile rupture measured over an arbitrary length including the section of rupture.
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B Conversion Tables: B1. Introduction:
The conversion factors in this section include the System International (S.I) Units as these are now widely accepted as the standard units of measurement. There are six basics units in the S.I. system. These are follows
The main advantage of the S.I. System is the ease with which mechanical units expressed in mass length and time can be related to electrical units. B2. List of Units & Abbreviations:
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Decimal Prefixes:.
The use of double prefixes should be avoided when single prefixes are available. Basic Units in Stress Analysis
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B3. Conversion Factors Grouped by Category:
The following pages give a list of various conversion factors used grouped by category Length:
Area:
Volume:
Force:
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Mass:
Density:
Velocity:
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Force Length Combination:
Moment of Inertia:
Mass Moment of Area:
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Energy:
Acceleration:
Electrical:
Angular:
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Power:
Stress or Pressure:
Temperature:
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Time:
Viscosity:
B4. Miscellaneous Data:
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C. Section Shear Centre:
The shear centre can be defined as that point in the plane of a cross-section through which an applied transverse load must pass for bending to occur unaccompanied by twisting. To obtain the shear centre a unit transverse load is applied to the section at that point. The shear stress distribution, as thus shear flow distribution, thus shear flow distribution is obtained for each element of the section. Equating the torque given by the shear distribution to the torque given by the transverse load will produce position of shear centre.
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C1. Location of Shear Centre-Open sections:
Section Constants
Zx = Zy =
Z xy =
I xx I xx I yy − I xy2 I yy I xx I yy − I xy2
I xy I xx I yy − I xy2
Shear flow in an element
[
]
q f = ∫ t t (xiV y + y iV x )Z xy − xiV x Z x − y iV y Z y ds + q o Where q0 is value of q f at s = 0 xi = horizontal distance from the element centroid to the section centroid yi = vertical distance from the element centroid to the section centroid Shear force in an element
F f = qi ds Torque/Moment given by element
Ti = Fi Ri Centroid chosen as datum point, but it could be any convenient point.
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Horizontal Location of shear Centre:
To Locate the horizontal position of the shear centre, we apply a vertical unit force through the unknown position of the shear centre x0,y0 ie Vy = 1, and Vx = 0. ∴ qi = ∫ (xi Z xy − y i Z y )t t ds + q 0
Shear forces and total torque ∑ Ti are calculated as defined previously. Vertical Location Of Shear Centre:
To locate the vertical position of the shear centre, we apply a horizontal unit through the unknown position of the shear centre x0,y0 ie Vy = 0, and Vx = 1 ∴ qi = ∫ ( y i Z xy − xi Z x )t t ds + q 0 0
C2. Shear Centre-curved Web:
Horizontal Location: θ ⎛ 1 ⎞ ⎟⎟tds qθ = ∫ − y⎜⎜ 0 ⎝ I xx ⎠
y = R cosθ
ds = Rdθ
qθ = qθ =
t I xx t I xx
∫
R cosθ ⋅ Rdθ
R 2 sin θ
Moment of force of element about 0
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= qds × R
= qRdθ × R R 4t sin θdθ Ix Total Moment
=
R 4t π sin θdθ I xx ∫0
R 4t [− cosθ ]π0 I xx Vertical component of force of element
qds sin θ ∴ Total Vertical Force =
R 3t sin 2 θdθ ∫ I xx
=
R 3 tπ 2 I xx
Taking Moment about 0 2R 4t R 3 tπ e= I xx 2 I xx Horizontal Location,
e=
4R
π
Alternatively for a solid section:
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with
q = web shear flow A = Enclosed area h = web depth Vertical Equilibrium
qh = Vy Moment Equilibrium
2 Aq = Vy × e . e = 2A / h
C3. Shear centre-Beams with constant shear flow between boom members:
I xx , I yy , I xy evaluated using boom areas only(skin shear carrying only) and Z x , Z y , Z xy using above values of I xx , I yy , I xy
Shear flow in each element
[
]
qi = ∑ (xV y + yV x )Z xy − xV x Z x − yV y Z y δA
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Horizontal location:
Firstly set Vy = 1 and Vx = 0
[
]
Giving qi = ∑ ( x )Z xy − ( y )Z y δA + q(i − 1) Where x and y are locations to element area δA Vertical location:
Firstly set Vy = 0 and Vx = 1
[
]
Giving qi = ∑ ( y )Z xy − ( x )Z x δA + q (i − 1) Where x and y are locations to element area δA
C4. Shear Centre – Single Cell Closed Section:
For Horizontal Location:
Cut element at 0 and assume q o = 0 Evaluate shear flow distribution around section. Take moment about 0to give the out of balance moment Mo. Balance Mo with balancing shear flow qb. Where qb =
M0 2A
With resultant shear flow distribution; take moments about convenient point to give moment MR. e=
MR vy
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For Vertical Location:
Cut element at a and assume q a = 0 Evaluate shear flow distribution around section. Take moment about 0to give the out of balance moment Mo. Balance Mo with balancing shear flow qb. Where qb =
M0 2A
With resultant shear flow distribution; take moments about convenient point to give moment MR. e=
MR vy
C5. Shear centre – single cell closed section with booms:
Initially the member is dealt with as a beam i.e. a cut is made at appropriate skin element thus reducing it from a single cell to an open beam section. Having established the open beam element shear floes, moments are taken about a convenient point and out of balance moment Mo established. Mo is balanced by shear flow qb =
Mo 2A
With resultant shear flow distribution; take moments about a convenient point to give MR. Shear centre horizontal location and vertical location: e=
M MR , e= R Vx Vy
MR calculated independently for unit Vy and Vx.
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C6. Shear Centre for Standard sections:
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D. Warping Constant:
When a torque, is applied to a non-circular beam, the beam will twist with each section of the beam rotating about torsional centre. The sections of the beam will not remain plane and will wrap out of plane. Depending on how the beam is constrained the effects of the applied torque, T will vary.
Consider a torque, T, applied to the section to the right. The diagrams below illustrate the effects of the torque with different constraints. The warping constant Γ is calculated by adding Γ1 and Γ2 . The constant Γ1 relates to the strain in the middle plane of the walls arising from variations of warping of the cross-section along length due to twist. The constant Γ2 relates to the strain across the thickness, t, of the walls. Therefore the warping constant is defined by Γ = (Γ1 + Γ2 )
Consider the following section
We can now show equations for the warping constants based on the generalised section shown above. The primary warping constant Γ1 is obtained from the formula
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1 A Γ1 = ∫ (W ) tds − ⎡ ∫ WdA⎤ ⎥⎦ 0 A ⎢⎣ 0 s
2
2
Where s
W = ∫ rt ds 0
The secondary warping constant Γ2 is obtained from the formula Γ2 =
1 A (rn ts )2 dA in6 ∫ 0 12
where A = Total cross-section area s = distance from any free end along middle of wall. rt = the perpendicular from the shear centre to the tangent of the point defined by s. rn = distance from the point of rotation to the normal of the tangent line of the middle of the wall. A = Total cross-section area rt = distance from the point of rotation to the tangent line of the centre line of the section wall at s s = distance from any free end along the middle of the wall. Constraint Warping about a Point It is sometimes necessary to know the warping constant (Γ1 )0 may be obtained by means of the formula
[Γ1 ]0 = Γ1 + x02 I xx + y 02 I yy − 2 x0 y 0 I xy The numerical method of calculating Γ for open section is shown,
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D1 Warping Constant – Tables for Typical Sections:
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E. Mechanical properties: E1. DATA Basis:
A-values: When applied loads are eventually distributed through a single member within an assembly, the failure of which would results in the loss of the structural integrity of the component involved, the guaranteed minimum design mechanical properties (“A” - values) must be met. At least 99% of the populations of values are expected to equal or exceed the A-value mechanical property allowable, with a confidence of 95% Examples of such items are: Single members such as drag struts, push-pull rods, etc. B-values: Redundant structures, in which the failure of individual elements would result in applied loads being safely distributed to other load-carrying members, may be designed on the basis of 90% probability. “B”-values are above which at least 90% of the population of values is expected to with a confidence of 95%. Examples of such items are: Sheet – stiffener combinations. Multi – rivet or multiple bolt connections. Thick skin type wing construction. S-values: The S-value is the minimum value specified by the governing industry specification of federal or military standards for the material. For certain products heat treated by the user the S- values may reflect a specified quality control requirement. Statistical assurance associated with this value is not known.
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E2. Stress-Strain Data:
The significant properties of material for structural design purposes are determined experimentally. Tension and Compression coupon tests provide the simplest fundamental information on material mechanical properties. Experimental data is recorded on stress – strain curves that record the measures units strain as a function of the applied force per unit area. Data of importance for structural analysis is derived from the stress –strain as shown above. E3. Definition of terms: Modulus of Elasticity:
Slope of initial straight-line part of the stress-strain curve, E (force per unit area) Secant Modulus:
Slope of secant line from the origin to any point (stress) on the stress-strain curve a function of the stress level applied, Es
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Tangent Modulus:
Slope of the line tangent to stress-strain curve at any point (stress) on the stressstrain curve, Et. The tangent modulus is the same as the modulus of elasticity in the range. Proportional Limit:
Stress at which the stress-strain curve deviates from linearity, indication of plastic or permanent set is called proportionality limit. Sometimes defined arbitrarily as stress at 0.01% offset, Fpl (psi Yield Strength or Stress:
At 0.2% offset, Fcy, Ftu (psi) Ultimate Strength or Stress:
Maximum force per unit area, Ftu (psi) Poisson’s Ratio:
Ratio of the lateral strain measured normal to the loading direction, to the normal strain measured in the direction of the load. Poisson ratio varies from the elastic value of around 0.3 to 0.5 in the plastic range. Secant stress: Stress at intercept of any given secant modulus line with the stress – strain curve is called secant stress. F0.7, F0.85 are secant lines of 0.7 and 0.85 % of young’s modulus. Elongation:
Permanent strain at fracture (tension) in the direction of loading e(%). Sometimes this property is considered a measure of ductility/brittleness. Fracture Strain:
Maximum strain at fracture of the material, ε e (in/in)
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Elastic Strain:
Strains equal to or less than the strain a t the proportional limit stress, ε e or the strain to the projected young’s modulus line of any stress level, Plastic Strain:
Increment of strain beyond the elastic strain at stresses above the proportional limit. Shear Properties:
The results of torsion tests on round tubes or round solid section s are sometimes plotted as torsion stress-strain diagrams. The modulus of elasticity in shear as determined from such a diagram is a basic shear property. Other properties, such as proportional limit and ultimate shearing stress, cannot be treated as basic properties because of the form factor. Bearing Properties:
Bearing strength are of value in the design of joints and lugs. Only yield and ultimate values are obtained from bearing tests. The bearing stress is obtained by dividing the load on a pin, which bears against the edge of the hole, by the bearing area, where the area is the product of the pin diameter and sheet thickness. The bearing test requires the use of special cleaning procedures as specified in ASTM E 238. In the “various Room Temperature Property Tables. When the indicated values are based on tests with clean pins, the values are footnoted as “dry pin” values. Designers should consider the use of a reduction factor in applying these values to structural analysis. In the definition of bearing values, t is sheet thickness, D is the hole diameter and e is the edge distance measured from the hole centre to the edge of the material in the direction of applied stress.
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E4. Temperature Effects:
Temperature below room temperature generally causes an increase in all strength properties of metals. However ductility usually decreases below room temperature. Temperatures above room temperature have the opposite effect which is dependent on many factors such as time of exposure and the characteristic of the material.
E5. Fatigue Properties:
Practically all materials will break under numerous repetitions of a stress is not as great the stress required to produce immediate rupture. This phenomena is known as fatigue. Fatigue crack initiation is normally associated with the endurance limit of a material below which fatigue does not occur is usually presented in the form of S-N curves, where the cycles stress amplitude is plotted versus the number of cycles to failure. Another major factor affecting the fatigue behaviour is the stress concentration caused by detail design of the structure. Metallurgical Instability:
In addition to the retention of load-carrying ability and ductility a structural material must also retain surface and internal stability. Surface stability refers to the resistance of the material to oxidizing of corrosive environments. Lack of internal stability is generally
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manifested by carbide precipitation, spheroidization, sigma-phase formation, temper embrittlement and internal or structural transformation, depending upon the material and the condition. Biaxial and Multiaxial Properties:
Many structural geometries, or load applications, are such that induced stresses are not uniaxial but are bi-or triaxial. Because of the difficulty in testing, few triaxial test data exist. However considerable biaxial testing has been conducted. If stresses are referred to as mutually perpendicular x, y, and z direction of the usual rectangular co-ordinates, a biaxial stress is a condition such that there are either a positive or negative stress in the x, y directions and the stresses in the z direction is usually zero. Fracture Strength:
The occurrence of flaws in a structural component is an unavoidable circumstance of material processing, fabrication, or service. These flaws may be cracks, metallurgical inclusion or voids, weld defects, design discontinuities, or combination thereof. If severe enough, these flaws can include structural failure at loads below those of normal design. The strength of a component containing a flaw is dependent on the flaw size, the component geometry, and a material property termed “fracture toughness”. The fracture toughness of a material is literally a measure of resistance to fracture. It also is considered a measure of its tolerance or lack of sensitivity to flaws. As with many other material properties, fracture toughness is dependent on processing variables, product form geometry, temperature, loading rate, and other environmental factors. Fatigue-Crack Propagation Behaviour:
Between the crack initiating phenomenon of fatigue and the critical instability of cracked structural elements as identified by fracture toughness and residual strength lies an important fact of material behaviour known as fatigue-crack propagation. In small size laboratory fatigue specimens, crack initiation and specimen failure may be nearly synonymous. However in larger structural components the existence of a crack does not
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necessarily indicate imminent failure of the component. The structural life during the cycle crack extension phase is the basis for damage tolerance certification of aircraft structure. Material Failures:
Fracture of a metal can be very complex, it can occur in either ductile or brittle state, in the same material depending on the state of stress and the environment. Fracture can occur after elongation of the metal over a relatively large uniform length, or after a concentrated elongation in a short length. Shear deformation will also vary dependent on metal and stress state, because of these variations in magnitude and mode of deformation the ductility of a metal can have a profound effect on the ability of a fabricated part to withstand applied loads. E.6 Generalised Stress-Strain Equations:
Introduction: This section describes a generalised method for representing the stress-strain curves of materials which exhibits a smooth curves. The method uses a reference stress and a characteristic index, in addition to the modulus of elasticity, and this section indicates how these two quantities may be determined Generalised Equations for Stress-Strain curves: A smooth continuous stress- strain curve can be represented by the equation
f 1⎡ f ⎤ = + ⎢ ⎥ fn fn m ⎣ fn ⎦
εE
m
in which f is the stress, fn is taken to be the stress at which the tangent modulus is one half the modulus of elasticity and the index m has been found to characterize the shape of the stressstrain curve and is therefore referred to as the material characteristic. The equation above only applies if f/fn is positive and this needs to be taken into account in cases where the chosen sign convention leads to either for ε being of opposite sign to fn This equation can then be rearranged and the value of Tangent Modulus obtained as Et (= df / dε )
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in which f is the stress, fn is taken to be the stress at which the tangent modulus is one half the modulus of elasticity and the index m has been found to characterize the shape of the stressstrain curve and is therefore referred to as the material characteristic. The equation above only applies if f/fn is positive and this needs to be taken into account in cases where the chosen sign convention leads to either for ε being of opposite sign to fn. This equation can then be rearranged and the value of Tangent Modulus obtained as Et (= df / dε ) ⎡ ⎡ f ⎤ m −1 ⎤ Et = E ⎢1 + ⎢ ⎥ ⎥ ⎢⎣ ⎣ f n ⎦ ⎥⎦
−1
This can be rearranged to give E f f ⎡ f ⎤ = +⎢ ⎥ f n Et fn ⎣ fn ⎦
m
The secant modulus Es, is the ratio of stress to total strain and may be determined directly from stress – strain curve equation. ⎡ 1 ⎡ f ⎤ m −1 ⎤ E s = E ⎢1 + ⎢ ⎥ ⎥ ⎢⎣ m ⎣ f n ⎦ ⎥⎦
−1
From the above set of equations the simple relationship between Es and Et follows ⎡E ⎤ 1⎡E ⎤ ⎢ − 1⎥ = ⎢ − 1⎥ ⎣ Es ⎦ m ⎣ Et ⎦
The value of Poisson’s ratio is according to the equation
v = v p − (E s / E )(v p − ve ) =
1⎡ f ⎤ ve + v p ⎢ ⎥ m ⎣ fn ⎦ 1⎡ f ⎤ 1+ ⎢ ⎥ m ⎣ fn ⎦
m −1
m −1
v e = Fully elastic value of Poisson’s ratio v p = Fully plastic value of Poisson’s ratio as the material yields Poisson’s ration will increase from its elastic value up to its fully plastic value
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The plastic value of Poisson’s ratio v p can be assumed to be 0.5 for all aluminium alloys. The elastic value for aluminium alloys can be assumed to be 0.3. Determination of m and fn The value of m, the material characteristic, can be found from standard fitting procedure if the full stress-strain curve is known. However, it may be estimated, provided that two points on the non-proportional region of the curve are known, from the equation. m=
( (
log ε R / ε R' log f R / f R'
) )
where fR and f’R are known stress values on the stress-strain curve of the actual material and ε R and ε R' are the strain in excess of the elastic strain corresponding to these stresses. For example , if fR and fR’ are the 0.1% and 0.2% tensile proof stresses of the material then ε R and ε R' are 0.001 and 0.002 respectively. For a known value of m the reference stress, fn, may be evaluated from the equation ⎡ mε E ⎤ fn = fR ⎢ R ⎥ ⎣ fR ⎦
−1 / ( m −1)
E.6.1. Problem:
Calculate the strain when a stress of 30000 lb.in2 is applied to sample 2024 T3 aluminium alloy sheet using B-basis in compression.(0.010
ε R = 0.002 fR = 3.9 × 10 4 lb/in2 m = 15 (material characteristic as defined in MIL-HDBK-5) We can calculate fn from ⎛ mε E ⎞ f n = f R ⎜⎜ R ⎟⎟ ⎝ fR ⎠
⎛ 1 ⎞ −⎜ ⎟ ⎝ m −1 ⎠
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Therefore fn = 33593.97 lbin2 To calculate the strain, ε , we must simplify the equation
f 1⎡ f ⎤ = + ⎢ ⎥ fn fn m ⎣ fn ⎦
εE
m
To give m ⎛ f ⎞ ⎜ + 1 ⎡ f ⎤ ⎟f ⎜ f n m ⎢⎣ f n ⎥⎦ ⎟ n ⎠ ε=⎝ E
Which gives us a strain of 0.003.
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1. PRISMATIC BAR IN TENSION:
For Ductile material yield stress is to be considered during analysis. If material is brittle or exhibits brittle fracture upon loading, ultimate stress is to be considered for analysis. Load considered for ultimate analysis is factor of safety times limit load. t2 values are the maximum permissible tensile stresses for the proof loading condition. ft values are given for the ultimate loading conditions. In case of a few materials, which have high proof to ultimate strength ratio, allowance may have to be made for the effects of stress concentration.
1.1 Limit Stress Analysis:
For limit stress analysis t2 is considered. It is then compared with limit stress or actual stress. Limit load = P Working stress σ =
P A
Allowable Proof stress (Or) 0.2% Proof stress = t 2 Reserve factor =
t2
σ
If reserve factor value is greater than 1 the design is considered to be safe. 1.2 Ultimate Stress Analysis:
For ultimate stress analysis ultimate load has to be considered. It has to be compared with UTS value of the material. Ultimate Working stress = (factor of safety† X actual proof stress)/A Allowable Ultimate stress (Or) UTS = ft. Reserve factor = ft / Actual Ultimate stress †
factor of safety is generally 1.5 according to MIL-A-8860
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1.3 Problem:
Material BSL 168T6511 t 2 = 370 Mpa
f t = 415Mpa Limit load P = 1500kg A = 10mm × 10mm Limit stress analysis: Applied proof stress σ = P
A
=
15000 = 150Mpa 100
Allowable proof stress t2 = 370Mpa Reserve factor = 370/150 = 2.47 Ultimate stress analysis: Actual ultimate stress = fos × actual proof stress = 1.5 × 150 = 225 Allowable ultimate stress = 415 Reserve factor = 415/225 = 1.84
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2. PRISMATIC BAR IN COMPRESSION:
For the stable members in pure compression an ultimate compressive stress equal to the 0.2% proof stress (c2) is recommended. Where the c2 value is not given tensile values t2 may be used. For thin sections or flanges, consideration should be given to the possibility of local or Torsional instability. For slender columns buckling load is to be considered. Where magnesium alloys are used in compression, it is recommended that special test made to be establish the compressive proof stress values, since they may be as low as 0.5 of the corresponding proof stresses.
2.1 Compression Stress Analysis: (Block Compression)
For Stress analysis c2 is considered. Actual Load = P Working Stress σ = P /A Allowable stress = c2. Reserve Factor = c2/ σ If the reserve factor value is greater than 1 design is safe. 2.1.2 Problem:
Material BSL 168T6511 Actual load = 1500 kg A = 10mm × 10mm
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Actual ultimate stress =
1.5 × 15000 = 225 N / mm 2 100
Allowable ultimate stress = 370 Reserve factor = 370/225 = 1.64
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2.2 Euler Buckling:
Buckling stress σ cr =
π 2 EI CL2 A
Pcr value depends upon end fixing condition C: Hinged - Hinged C = 1 Fixed - Free C = 4 Fixed - Fixed C = (1/4) Fixed - Hinged C = (1/2) 2.3 For other Buckling cases:
Section Modulus = Z 2.3.1 Buckling for Eccentric Loading:
In an actual structure it is not possible for a column to be perfectly straight or to be loaded exactly at the centroid of the area. Therefore apart from normal load there will be additional load due to moment caused by eccentricity.
σ cr
⎛L Pe sec⎜⎜ P ⎝2 = + A Z
P EI
⎞ ⎟ ⎟ ⎠
This is a Transcendental in P. The solution will be taken as Pall. Eccentricity of the load = e in mm. Moment of inertia = I in mm4.
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2.3.2 Beam - Columns:
A beam is called Beam-Column when it is submitted to both compression and bending loads. When a structural element similar to a beam is subjected simultaneously to a normal load and bending moment, superposition theorem cannot be used to determine the stresses. The lateral deflections appreciably change the moment arms of the compression forces and deflections are not proportional to the loads. The verification of Beam – Column will include two steps as following 1 .The study of column. 2. The study of Beam – column if axial load is lower than the critical load calculated in the previous step.
2.3.2.1 Notations and Conventions:
1. The compressive loads are taken to be positive. 2. A positive bending moment compresses the upper fibres of the beam. 3. A distributed lateral loads leads to the positive moment. 4. A positive lateral load compresses the upper fibre of the beam. 5. The compression stresses are positive.
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2.3.2.2 Terms used:
Compression load =P in N. Distributed lateral load = q in N/mm. Applied isolated load = F in N. Critical buckling stress = σ cr in N/mm2. Modulus of Elasticity in compression = Ec in N/mm2. Moment of inertia = I in mm4. Area of cross section = A in mm2. Clamping factor = K. Column effective buckling length = L in mm. Real column length = S in mm.
2.3.2.3 Calculation of Bending Moment:
Let us consider a beam where the ends are on single supports with compression loads and bending loads.
Beam with compression and bending loads: ⎡ M Z − M ZA The bending moment at X is M Z = M Z A + ⎢ B L ⎣
⎤ ⎥ x − (Py ) in N-mm. ⎦
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⎡ ⎛ L ⎞⎤ ⎢ M Z B − M Z A cos⎜⎜ ⎟⎟ ⎥ ⎝ j ⎠ ⎥ sin ⎛⎜ x ⎞⎟ + M cos⎛⎜ x ⎞⎟ in N-mm. Final bending moment is M Z = ⎢ ZA ⎜ j⎟ ⎢ ⎥ ⎜⎝ j ⎟⎠ ⎛ L⎞ ⎝ ⎠ sin ⎜⎜ ⎟⎟ ⎢ ⎥ ⎢⎣ ⎥⎦ ⎝ j⎠ ⎡ ⎛ L ⎞⎤ ⎢ M Z B − M Z A cos⎜⎜ ⎟⎟ ⎥ ⎝ j ⎠ ⎥ in N-mm. Maximum bending moment is at x = jArc tan ⎢ ⎢ ⎥ ⎛ L⎞ sin ⎜⎜ ⎟⎟ ⎢ ⎥ ⎝ j⎠ ⎣⎢ ⎦⎥
Common expression for Beam – Column bending moment ⎛ x⎞ ⎛ x⎞ M Z = C1 sin ⎜⎜ ⎟⎟ + C 2 cos⎜⎜ ⎟⎟ + j 2 f ( x ) . ⎝ j⎠ ⎝ j⎠
Where
j=
Et I and several cases can be combined as shown below. P
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2.3.2.4 Allowable Stress:
Two main cases must be considered when choosing allowable stress 1. Solid beams 2. Thin web beams Solid beams: No local buckling phenomenon occurs before general buckling, the allowable stress on tensional side is equal to σ R . Thin Web Beams: A local buckling phenomenon can occur before general buckling. This state does no give rise to an immediate collapse but can considerably decrease the allowable breaking stress. The allowable stress on compressive side is equal to σ all = MIN {σ local buckling ;σ R }
For further references refer, 1. Aircraft Structures P.J.Perry Page no 524 – 533. 2. Analysis and Design of flight Vehicle structures by E.F.Bruhn Supplement page no 2 3. Refer Aircraft Structures for Engineers THG Megson Page no 162-165
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2.3.3 Local buckling: (Thin Walled Structures):
If column is composed wholly or partially of thin material, local buckling may occur at a unit load less than the required load to cause failure of column as a whole. Thin walled sections are encountered in aircraft structures in the shape of longitudinal stiffeners. These elements fall in to two distinct categories – flanges which have a free unloaded edge and webs which are supported by adjustant plate elements in the column cross section.
2.3.3.1 Flanges:
For long flange having one end fixed and other edge is free. Modulus of Elasticity = E Elastic Poisson’s ratio = μ Thick ness of flange = t Breadth of flange = b Buckling stress f cr =
1.09 E t 1− μ 2 b
()
2
For flange having one edge simply supported and other edge is free. Buckling stress f c =
0.416 E t 1− μ 2 b
()
2
Also refer ESDU DATA sheet 01.01.08.
2.3.3.2 Thin webs:
Long thin web fixed along each edge Buckling stress f c =
5.73E t 2 1− μ 2 b
()
Simply supported along each edge Buckling stress f c =
3.92 E t 1− μ 2 b
()
2
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2.3.3.3Plates and Shells:
1) Rectangular plate under equal uniform compression on two opposite edges:
Sides of rectangular plate = a, b Compressive load = f Buckling stress f c = K
E t 2 1− μ b
()
2
Where K depend on a/b ratio and end fixing conditions Table for value of K for various end fixing condition: Manner of Support
Value of K For (a/b=1)
All edges simply supported
3.29
All edges clamped
7.7
Edges b simply supported, edges a clamped
6.32
edges b and edge a simply supported and one edge a is free
1.18
Refer Analysis and design of flight vehicle structures by E.F.Bruhn chapterC5 for K values for various a/b ratios. 2) Rectangular plate under uniform compression fx on b and fy on a:
Sides of rectangular plate = a, b Compressive loads = fx, fy. Poisson’s ratio = μ
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Buckling stress f c = K
E t 2 1− μ b
()
2
Where K depend on a/b ratio and end fixing conditions 3) Rectangular plate under linearly varying stress on edges b:
Sides of rectangular plate = a, b Compressive load = f0, fv. Buckling stress f c = K
E t 2 1− μ b
()
2
Where K depend on a/b ratio and α Where α =
σ0 σ 0 −σ v
Table for value of K for various α: α
0.5
0.75
1.00
1.25
1.50
∞
K(a/b=1)
21.1
9.1
6.4
5.4
4.8
3.29
Refer Analysis and design of flight vehicle structures by E.F.Bruhn chapterC5 for K values for various a/b ratios. 4) Rectangular plate under uniform shear on all edges:
Sides of rectangular plate = a, b Shear stress = fs. Poisson’s ratio = μ
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Buckling stress f sc = K
E t 2 1− μ b
()
2
Where K depend on a/b ratio and end fixing conditions Table for value of K for various end fixing condition: All edges simply supported
All edges clamped
a/b
1
2
∞
K
7.75
5.43
4.40
a/b
1
2
∞
K
12..7 9.5
7.38
Refer Analysis and design of flight vehicle structures by E.F.Bruhn chapterC5 for K values for various a/b ratios. 5) Rectangular plate under uniform shear on all edges, compression fx on b and fy on a:
Sides of rectangular plate = a, b Compressive loads = fx, fy. Shear load = fs. Poisson’s ratio = μ Buckling stress for all edges simply supported
(
f sc = C 2 2 1 − Where C =
fy C
+2−
0.823E t 1− μ 2 b
()
fx C
)(2 1 − Cf + 8 − Cf ) y
x
2
Buckling stress for all edges clamped
(
f sc = C 2 2.31 4 −
fy C
+
4 fx − 3 C
)(2.31 4 − Cf + 8 − Cf ) y
x
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6) Rectangular plate under uniform shear on all edges and bending stresses on b:
Sides of rectangular plate = a, b Bending loads = fb, Shear load = fs. Poisson’s ratio = μ Buckling stress for all edges simply supported fb = K
E t 2 1− μ b
()
2
Here K depends on fs/fsmax and on a/b Table for value of K for various end fixing condition: fs/fsmax
0
0.2
0.3
0.4
0.5
0.6 0.7 0.8
0.9 1.0
K(a/b=1)
21.1
20.4
19.6
18.5
17.7
16
8.2 0
14
11.9
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2.3.3.4 Analysis of Thin Walled Structures:
Rb = (Actual Bending Stress/Allowable Bending Stress) Rc = (Actual Compressive Stress/Allowable compressive Stress) Rs = (Actual Shear Stress/Allowable Shear Stress) Rl = (Actual Longitudinal Stress/Allowable Longitudinal Stress) For combined Bending and Longitudinal Compression: Reserve factor Rb1.75 + Rc For Combined Bending and Shear:
1
Reserve Factor
R + Rs2 2 b
For Combined Shear and Longitudinal direct Stress: Reserve Factor
2
Rl + Rl2 + 4 Rs2
2.4 Problems: 2.4.1 Problem for Euler Buckling:
Problem 1: Material BSL 168T6511 E = 70 × 10 3 Mpa t = 370
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ft = 415Mpa Actual load = 2500 kg Ultimate load = fos× 2500 = 1.5 × 2500 = 3750 end condition = hinged –hinged C=1 b = 10mm d = 10mm l = 100mm I=
bd 3 10 4 = = 833.33 12 12
Pcr =
π 2 EI l2
=
π 2 × 70 × 10 3 × 833.33
Reserve factor =
100 2
= 57572 N
Pcr 5757.2 = = 1.535 Pult 3750
Problem 2: Material BSL 168T6511 E = 70 × 10 3 Mpa t 2 = 370 Mpa
f t = 415Mpa l = 5000 mm P = 1000 kg
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Ultimate load = fos× P = 1.5 × 1000 = 1500kg
End condition = hinged –hinged = c = 1 x1 =
20 + 60 = 50mm 2
x2 =
20 = 10mm 2
x 3 = 20 +
60 = 50mm 2
y1 =
20 = 10mm 2
y2 =
100 = 50mm 2
y 3 = 80 +
20 = 90mm 2
A1 = 20 × 60 = 1200mm 2
A2 = 100 × 20 = 2000mm 2
A3 = 1200mm 2
X =
A1 x1 + A2 x 2 + A3 x3 1200 × 50 + 2000 × 10 + 1200 × 50 140000 = = 31.82mm = 1200 + 2000 + 1200 4400 A1 + A2 + A3
Y=
A1 y1 + A2 y 2 + A3 y 3 1200 × 10 + 2000 × 50 + 1200 × 90 220000 = = 50mm = 1200 + 2000 + 1200 4400 A1 + A2 + A3
IX X = =
b d3 b1 d13 b d3 + A1 h12 + 2 2 + A2 h22 + 3 3 + A3 h32 12 12 12 60 × 20 3 20 × 10 3 60 × 20 3 + 1200 × (18.18) 2 + + 2000 × 21.82 2 + + 1200 × 18.18 2 12 12 12
= 1780573.22mm4
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I YY = =
2 2 2 d b3 d1b13 d b3 + A1 h1 + 2 2 + A2 h 2 + 3 3 + A3 h 3 12 12 12
20 × 60 3 100 × 20 3 20 × 60 3 + 1200 × 40 2 + + + 1200 × 40 2 12 12 12
= 4626666.67mm4 Pcr =
π 2 EI Ll 2
=
π 2 × 70 × 10 3 × 1780573.22 1 × 8000 2
Reserve factor =
= 49205 N
4920.5 = 3.28 1500
Ultimate load = 18150N Limit load =
18150 18150 = = 12100 N fos 1.5
2.4.2 Problems for Beam – Column:
Problem 1: Considered a beam with both Transverse and Axial load with moment as shown in the diagram. The data for the set up is listed below calculate the maximum bending moment and reserve factor.
Data: S = 100 mm, a = 30 mm, c =10 mm. A = 100mm2, I = 833 mm4, F = 700 N, P = 11600 N. M1 = 12000 N – mm. M2 = 40000 N – mm. Material 7075 T 7651 E = 71711 N/mm2.
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Ec = 73800 N/mm2.
σ c 0.2 = 460 N/mm2. σ 0.2 = 460 N/mm2. σ R = 525 N/mm2. e% = 7. Check for beam does not buckle the action of P alone
σ cr =
π 2 Ec I 2
2
S K A
=
π 2 × 7380 × 833 100 × 1 × 100 2
2
= 607 N/mm2.
K = 1 for given clamping condition To apply plasticity correction
σ = ησ cr η=
Et Ec
Using Ram berg and Osgood model:
σ cr
⎛σ ⎞ ε= + 0.002⎜⎜ cr ⎟⎟ Ec ⎝ σ c 0.2 ⎠ Es =
nc
σ cr ε
1 − nc n 1 = c + Et E s Ec
σ = 411 N/mm2. Stress applied due to axial load σ =
P = 116 N/mm2. A
So, it can be deduced that column does not buckle. Maximum bending moment expression j =
Ec I = P
7380 × 833 = 728 1160
For x ≤ a , Using combination of cases 1 to 4 given in above table. M 2 − M 1 cos C1 =
S sin j
S j
jF sin +
S sin j
b j
= 81010 N-mm.
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C 2 = M 1 = 12000 N-mm.
M AC = 81010 sin
x x + 12000 cos j j
For x > a , Using combination of cases 1 and 5 given in above table. M 2 − M 1 cos C1 =
S sin j
S j
jF sin +
S j
S tan j
= 34310 N-mm.
C 2 = M 1 + jF sin
S = 32410 N-mm. j
M CB = 34310 sin
x x + 32410 cos j j
To determine the maximum bending moment At point A M A = 81010 sin 0 + 12000 cos 0 = 12000 Nmm
At point C M C = 81010 sin
30 30 + 12000 cos = 43440 Nmm 72.8 72.8
At point B M B = 34310 sin
100 100 + 32410 cos = 40000 Nmm 72.8 72.8
Between A and C dM AC 81010 x 12000 x = sin − cos =0 dx 728 728 728 728
The moment is maximum at abscissa point x = 728 Arc tan
81010 = 103.7mm But 103.7>XC 12000
Therefore MAC is maximum at XC MACmax = 43440 N-mm Between C and B dM CB 34310 x x 32410 = sin − cos =0 dx 728 728 728 728
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The moment is maximum at abscissa point x = 728 Arc tan
34310 = 59.3mm 32410
M CB max = 34310 sin
593 593 + 32410 cos = 47200 N − mm 728 728
Therefore, the maximum moment on beam AB is M AB max = MAX (M AC max ; M CB max ) = 47200 N − mm
Calculate the total stress and the reserve factor Stress due to the axial load
σc =
P = 116 N / mm 2 S
Stress due to the bending moment
σf =±
M c × = ±283N / mm 2 I 2
Maximum compression stress
σ MAX = σ c + σ f = 399 N / mm 2 Reserve factor R.F. =
525 = 1.32 399
2.4.3 Problem for Local Buckling:
Problem 1: Material BSL 168T6511 E = 70 × 10 3 ,a = 10,b = 10,t = 1
All edges are simply supported Compressed load f = 1000kg
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Actual ultimate load f ult = 1.5 × 100 = 150kg Actual stress =
15000 = 1500 N / mm 2 10 × 1 2
2
70 × 10 3 ⎛ 1 ⎞ E ⎛t⎞ Allowable stress = K ⎜ ⎟ = 2530.27N ⎜ ⎟ = 3.29 × 1 − 0.3 2 ⎝ 10 ⎠ 1− μ 2 ⎝ n ⎠ Reserve factor =
2530.27 = 1.69 1500
2.4.3.1 Problem for Combined Bending and Shear:
Problem 1: Material BSL 168T6511
E = 70 × 10 3 , a = 10, b = 10, t = 1 All edges are hinged. Bending load = 100kg, Shear load = 1000kg
My 500 × 0.5 × 12 = = 3000 N / mm 2 3 I 10 × 1
Actual bending stress = Actual shear stress =
T 10000 × 10 = 500 N / mm 2 = 2 At 2 × 10 × 10 × 1 2
K6E ⎛ t ⎞ 6 × 70 × 10 3 Buckling stress (bending) = (0.01) = 4615.3 N ⎜ ⎟ = 0.91 (1 − μ 2 ) ⎝ b ⎠ 2
KsE ⎛ t ⎞ 9.5 × 70 × 10 3 = (0.01) = 73.0769 Shearing buckling stress = ⎜ ⎟ 0.91 (1 − μ 2 ) ⎝ b ⎠
Rb =
3000 500 = 0.65 , R B = = 0.068 4615.3 7307.69
Reserve factor =
1 ( Rb ) + ( Rs ) 2
2
=
1 (0.65) + (0.068) 2 2
= 1.53
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3. LUG ANALYSIS:
The lug refers to that portion of the fitting that involves the hole for the single bolt that connects the male and female part of the fitting unit. The simplified assumptions regarding failing action and the resulting equations have been widely used for quick approximate check of the lug strength. The failure of the Lug may be due to Tension or shear tear out or due to bearing. Loading conditions considered are 1. Axial loading 2. Transverse loading 3. Oblique loading
3.1 Failure in Tension:
The above diagram indicates how a plate can be pull apart due to tension stresses on a section through the centre line of bolt hole. Both the male and female parts of the fitting must transfer the load past the centreline of the hole, thus both parts must be considered in the design of the fitting. Refer ESDU Data Sheet 91008 for further details. Ultimate tensile strength of the material = ft. + Breadth of the plate = 2R Thickness of lug = t Diameter of the hole = d Limit load = P Ultimate applied stress f u ( tension ) =
1.5P (2 R − d )t
+
Allowable Tensile is obtained by detail fatigue analysis.
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The above equation assumes that the tensile stress on the cross section is uniform. This is not true as the flow of the stress around the hole causes a stress concentration. To take care of this extra allowance has to be given in reserve factor. Reserve factor = ft/fu (tension). 3.2 Failure by Shear Tear Out:
The above picture illustrates the manner in which failure can occur by the shearing tear out of a plate sector in front of the bolt. Due to load P the bolt presses the plate or lug around bolt - hole edge. Stresses are produced which tends to cause the portion (a) shown in the figure to tear out. Refer ESDU Data Sheets 81006, 71011, 66011 for details. Usually fs for ductile material are 0.6 UTS. Shear out area = As. Ultimate shear stress = fs. Ultimate applied stress f u ( shearout ) =
1.5P 2 xt
It is very common practise to take the shear out area As equal to the edge distance at the centreline of the hole times the thickness t of the plate times two since there are two shear areas.
This is slightly conservative because the actual shear area is larger and area considered is limited by the 40o line as shown in the picture. Reserve factor = fs/fu (shear out)
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3.3 Failure by bearing:
The pull causes the bolt to press against the wall, which in turn presses against the plate wall. If the pressure is high enough the plate material adjustant to the hole will start to crush and flow thus allowing bolt and bush to move which results in elongated hole as shown in the diagram. Ultimate bearing stress = fb. Diameter of the bushing = d Thickness of the plate = t Bearing factor = Kbr. + Ultimate applied stress f u(bearing) =
1.5P dt
Reserve factor = Kbr*fb/fu (bearing) Usually fb is equal to ft. It is good practise to require a reserve factor of 1.5 +
Refer ESDU Data Sheet 81029, 83033 for further details.
3.4 Lug strength analysis under transverse loading:
Cases arise where the lug of a fitting unit is subjected to only a transverse load. Bearing area = Abr. Ultimate Tensile stress = ft. Efficiency failing coefficient = Kt. Allowable load Pu(tension) = K t A br f t Kt depends on 2R/d ratio and upon material used.
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3.5 Lug strength analysis under oblique loading:
Fitting lugs often subjected to oblique loads. It will be resolved in to axial and transverse components. Reserve factor =
(R
1.6 a
1 0.625 + R 1.6 tr )
Where Ra = axial component of the applied load divided by ultimate allowable tensile load Rtr = transverse component of the applied load divided by Pu (tension) of transverse loading. Allowable stresses used in various loading conditions: Table for value of ultimate stresses for various condition: Condition
Ultimate stresses
Tension
ft.
Shear
0.6 ft.
Bearing
2 ft.
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3.6 Lug analysis (AVRO Method):
The Permissible lug stresses listed down are based on the method illustrated in, Av.P.970 chapter 404/2. Allowable stresses used in various loading conditions are listed below. The permissible lug stress for each condition is taken as maximum of values for failing stress. Condition
Ultimate stresses
Tension
0.9 ft.
Shear
0.46 ft.
Bearing
1.85 ft.
For shear if a/d <1.5the allowable shear stress quoted must be factored by K. a/d Vs K: a/d
0.7
0.8
0.9
1.0
1.1
K
0.7
0.725
0.75
0.8
0.85
1.2
1.3
0.875 0.925
1.4
1.5
0.95
1
Actual stresses Condition
Ultimate stresses
Tension
pt 2ct
Shear
pt 2at
Bearing
pt d2 pt dt
Reserve factor = allowable stress/actual stress
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3.7 Problems:
Problem 1: Figure shows a single pin fitting the lug material is AISI steel heat treated to ft = 125000 Psi. The bolt AN steel ft = 125000 Psi. The bushing is steel with Ft = 125000 Psi .The fitting is subjected to an ultimate tensile load of 15650 lb .check the fitting for the design load diameter of bolt is ½’’
A fitting factor is considered = 1.15 Clearance gap = 1/6’’ Applied fitting load = 1.15 × 15650 = 18000lb Check for bolt shear: For diameter = 0.5 Ultimate double shear length of bolt is 29400 Applied load (Ultimate) = 18000 lb Reserve factor = 29400/18000 = 1.62 Failure by bolt bending: There is strength gap between lugs ∴ b = 0.5t1 + 0.25t 2 + g (Or) b =
t1
2
+ t2
4
+g
b = 0.5(7 / 3) + 0.25(3 / 8) + (1 / 64) = 0.218 M =
18000 × 0.218 = 1962 2
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Bending stress =
MD 1962 × 0.5 = = 159824 2I 2 × 0.003069
Allowable bending stress = 180000 Reserve factor = 180000/159824 = 1.13 Failure by lug portion of fitting: Failure by tension tear out: Applied ultimate stress =
18000 = 85333 (1.875 − 0.625) × 0.375
Allowable Ultimate stress = 125000 Reserve factor = 125000/85333 = 1.46 Failure by shear tear out Applied Ultimate stress =
18000 = 85348 2 × 0.375 × 0.2812
Allowable Ultimate stress = 110913 Reserve factor = 110913/85348 = 1.3 Failure by bending: Applied Ultimate stress =
18000 = 76800 0.625 × 0.375
Allowable Ultimate bearing stress = 180000 Reserve factor = 180000/76800 = 2.34 Problem 2: Lug = 7075-T6 = 80ksi Bush = steel-HT = 180ksi Pin = steel min.-HT = 200ksi
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Ultimate applied load = 75 Fitting factor = 1.15 Ultimate applied load = 75 × 1.15 = 86.25 Failure by bolt shear: Allowable ultimate load PU = 2 × A × f s = 2 ×
π × 0.75 2 4
× 200 = 176.7
Ultimate applied load = 86.25 Reserve factor = 176.7/86.25 = 2 Failure by bolt bending: b=
t1 t 2 + +g 2 4
b=
0.375 0.75 + = 0.375 2 4
M = 86.25 ×
0.375 = 16.17 2
Applied bending stress =
16.17 × 0.75 = 390 2 × 0.61553
Allowable bending stress = 200 × 1.15 × 2 = 460 Reserve factor = 460/390 = 1.2
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Failure of lug portion of fitting: Failure by tension tear out: Applied ultimate stress =
8625 = 59.74 (3.175 − 1.25) × 0.75
Allowable ultimate tensile stress = 80 Reserve factor = 80/59.74 = 1.39 Failure by shear tear out: Applied ultimate shear stress =
86.25 = 46 2 × 0.75(1.875 − 0.625)
Allowable ultimate shear stress = 48 Reserve factor = 48/46 = 1.05 Failure by bending Applied ultimate bearing stress =
86.25 = 30.67 3.75 × 0.75
Allowable ultimate bearing stress = 80 Reserve factor = 80/30.67 = 2.6
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4. MEMBER IN BENDING
In this section we will discuss the analysis of structures that are under the action of pure bending. As such, there will be no transverse shear force along the beam section considered. The problems of beam bending considered here are based on the Euler-Bernoulli Beam Theory. In this section we will examine the problems in which the bending moment is applied either symmetrically or unsymmetrically on homogeneous or non-homogeneous beams. In addition, we will discuss the elastic and inelastic bending of beams having symmetric or unsymmetric cross sections. The determination of neutral axis location for elastic and inelastic beams will also be discussed. The variation of bending-induced normal stresses on the beam cross section will be shown in several example problems. Finally, in this section we will discuss the bending of curved beams including the determination of the neutral axis and distribution of normal stresses. 4.1 Introduction 4.1-1 Rectangular Moments of Inertia and Product of Inertia 4.1-2 Parallel Axis Theorem 4.2 Elastic Bending of Homogeneous Beams 4.3 Elastic Bending of Non-homogeneous Beams 4.4 Inelastic Bending of Homogeneous Beams 4.5 Elastic Bending of Curved Beams 4.6 Miscellaneous Problems
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4.1 Introduction To Bending:
The straight beam section under consideration is in the state of pure bending (i.e., transverse shear force is zero along this section). Consequently, as the beam bends, plane sections remain plane but rotate relative to each other as shown in the figure below. As illustrated in the figure the top surface of the beam is shortened due to compression, and the bottom surface is elongated due to tension - both as a result of bending moment M. By examining the figure below, it becomes apparent that at some location between the top and bottom surfaces of the beam there is a surface whose length is the same as the original length of the straight beam. This surface is neither in tension nor in compression, therefore, it is referred to as the "Neutral Surface". The intersection of the neutral surface with the plane of the beam cross section is called the "Neutral Axis". Since the beam cross section only rotates without warping, the slope of the cross sectional plane is constant, indicating that axial deflection due to bending is linear. This finding implies that in originally-straight beams in pure bending, the axial strain must also vary linearly with zero value at the Neutral Axis.
Note: The linearity of axial strain is a consequence of beam being (a) originally straight, and (b)
in pure-bending state. Therefore, no additional restriction such as material property or elasticity is imposed. Now we impose additional restrictions: (c) bending stresses remain below material's elastic stress limit, (d) stress-strain relationship is linear. As a result of restrictions (c) and (d), Hooke's law may be used.
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In summary it can be said that in elastic and homogeneous beams, the Neutral Axis, NA always passes through the centroid of the cross section with its orientation determined according to the shape of the cross section and the orientation of the bending moment. The stress and strain variations for some elastic beams are given below. Examples •
(a) Elastic, homogeneous beam with doubly symmetric cross section.
•
(b) Elastic, homogeneous beam with symmetric cross section.
•
(c) Elastic, non-homogeneous beam with doubly symmetric cross section. I. Elastic, homogeneous beam with doubly symmetric cross section
II. Elastic, homogeneous beam with symmetric cross section
III. Elastic, non-homogeneous beam with doubly symmetric cross section
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4.2 STRESS ANALYSIS OF BEAM:
For robust and stable sections in pure bending the allowable stresses if not give separately may be taken from the 0.2% proof stress in Tension. For sections in pure bending the allowable stresses for ultimate analysis may be increased by applying a form factor. This factor depends upon the shape of the section. It is assumed that the beam section is homogeneous, that is, made of same material
Applied Load = P Cross Section Area = A Neutral Axis = NA Flange Thickness = b For ideal section: Height = h Flange Thickness = t Bending Moment = M Moment of Inertia = I Shear load = V
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4.3 BENDING STRESS: 4.3.1 Limit Stress Analysis:
Working Stress σy = My/I, Allowable proof bending stress = be. Reserve Factor = be/ σy Shear stress at any layer τxy = VQ/IB Average Shear Load τavg = V/th 4.3.2 Ultimate Stress Analysis:
Ultimate working Stress = σy* Factor of safety Allowable ultimate bending stress = fbe* form factor Reserve Factor = fbe form factor/actual ultimate bending stress. Form factor for standard sections: Shape of the section
Form factor
Square
1.5
Rectangle
1.5
Equilateral Triangle Isosceles Triangle Solid circle Solid semicircle Solid ellipse
In x axis 2.343 In y axis 2 In x axis 2.343 In y axis 2 1.698 In x axis 1.856 In y axis 1.698 1.698
For further of form factor refer ESDU data sheets from 01.06.01 to 01.06.03
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4.3.3 Rectangular Moments and Product of Inertia
The second moments of an area, or rectangular moments of inertia, of an area with respect to the x and y axes - shown in the figure above - are defined as
The product of inertia of this general shape is defined as
In this course, the cross-sectional areas consist of simple geometric shapes, therefore, their moments of inertia can be calculated by a simple summation as opposed to the integral equations shown.
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4.3.4 Parallel Axis Theorem
The parallel axis theorem is used for finding the rectangular moment of inertia about an axis different from the centroidal axis of the section. For the figure below the centroidal rectangular moments of inertia are determined using
Using the parallel axis theorem, the moments of inertia about axes XX and YY parallel to the centroidal axes x and y can be determined as
The parallel axis theorem is useful for finding the centroidal moments of inertia of a cross section composed of several simpler geometric parts as shown below.
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The cross section is broken up into several simple shapes, the individual moments of inertia are found, then the moments of inertia about the centroidal axes of the entire section are found using parallel axis theorem. Usually the broken pieces are made up of rectangles. From mechanics of materials, the x and y moments of inertia of a rectangle are
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Example 1
For the cross section shown, determine the (a) Moment of inertia about the horizontal centroidal axis, x (b) Moment of inertia about the vertical centroidal axis, y (c) Product of inertia
EQUATIONS USED
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SOLUTION The cross section is broken up into separate section.
The centroid of the cross section is located at: Note that (2.5) in xbar equation should be (0.25). The final answer for x bar is correct, however.
The moments of inertia for each rectangular section are
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(a) Now, using the parallel axis theorem, we calculate the moment of inertia of the entire cross section about the horizontal x axis:
'y' is the distance from the centroid of the cross section to the centroid of each particular rectangle. (b) Now, using the parallel axis theorem, we calculate the moment of inertia of the entire cross section about the vertical y axis:
'x' is the distance from the centroid of the cross section to the centroid of each particular rectangle.
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(c) The product of inertia is determined as
'x' and 'y' terms in this equation are the same as those defined above. Note that the product of inertia is zero as the cross section has one axis of symmetry, i.e., the y axis. EXAMPLE For the skin-stringer cross section shown, determine the (a) Moment of inertia about the centroidal x axis (b) Moment of inertia about the centroidal y axis (c) Product of inertia
Equations Used
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SOLUTION This is a lumped area cross section. Unless otherwise stated, the moment of inertia of the skin can safely be ignored, this leaves only the moment of inertia due to the lumped areas. Often the actual moments of inertia for the individual lumped areas are not given. This is because the first component in the parallel axis theorem, in comparison to the second, is very small in such configurations.
To demonstrate this, the moment of inertia will be found using the complete equation, then found by ignoring the first component of the parallel axis theorem. With respect to the bottom left stringer, the centroid is located at
The complete moments of inertia are:
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Now by just using the second component in the parallel axis theorem, we get
The error in Ix is 5.9% and that in Iy is 0.5%. In a real situation the difference is around 0.5%. The product of inertia is
As expected, the product of inertia is zero because of symmetry Example
For the skin-stringer cross section shown, determine the (a) Moment of inertia about the centroidal x axis
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(b) Moment of inertia about the centroidal y axis (c) Product of inertia
EQUATIONS USED
SOLUTION With respect to stiffener 'A', the centroid is located at
The rectangular moments and product of inertia are:
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Because the cross section is not symmetrical about the x or the y axis, the product of inertia is not zero. Example
For the skin-stringer cross section shown, determine the (a) Moment of inertia about the centroidal x axis (b) Moment of inertia about the centroidal y axis (c) Product of inertia
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Equations Used
Solution With respect to the yellow stiffener with the area of 2 sq in , the centroid is located at
The rectangular moments and product of inertia are
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Because the cross section is not symmetrical about x or y axis, the product of inertia is not zero. 4.4 Elastic Bending of Homogeneous Beams
The general bending stress equation for elastic, homogeneous beams is given as
(II.1) where Mx and My are the bending moments about the x and y centroidal axes, respectively. Ix and Iy are the second moments of area (also known as moments of inertia) about the x and y axes, respectively, and Ixy is the product of inertia. Using this equation it would be possible to calculate the bending stress at any point on the beam cross section regardless of moment orientation or cross-sectional shape. Note that Mx, My, Ix, Iy, and Ixy are all unique for a given
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section along the length of the beam. In other words, they will not change from one point to
another on the cross section. However, the x and y variables shown in the equation correspond to the coordinates of a point on the cross section at which the stress is to be determined. Sign Convention on Bending Moment Components Mx and My:
As far as the general bending stress equation is concerned, if a moment component puts the first quadrant of the beam cross section in compression, it is treated as positive (see the examples shown below). Notice that this is just a sign convention for the moment components and should not be confused with the sign associated with the bending stress.
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Neutral Axis:
When a homogeneous beam is subjected to elastic bending, the neutral axis (NA) will pass through the centroid of its cross section, but the orientation of the NA depends on the orientation of the moment vector and the cross sectional shape of the beam. When the loading is unsymmetrical (at an angle) as seen in the figure below, the NA will also be at some angle - NOT necessarily the same angle as the bending moment.
Realizing that at any point on the neutral axis, the bending strain and stress are zero, we can use the general bending stress equation to find its orientation. Setting the stress to zero and solving for the slope y/x gives
(II.2) A positive angle is defined as counter clockwise from the horizontal centroidal axis.
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Notice that we can use the equation for orientation of NA to examine special cases. For example, if the cross section has an axis of symmetry, Ixy = 0. In addition if only Mx is applied, then NA will have angle of zero which is consistent with what we would expect from mechanics of materials. From this equation, we see that the orientation of NA is a function of both loading condition as well as cross sectional geometry. Example 1
For the cross section and loading shown, determine •
(a) Neutral axis location and orientation,
•
(b) Bending stress distribution,
•
(c) Location and magnitude of the maximum bending stress. Assume that the stresses due to the applied load do not exceed the elastic limit.
Use the Java screen shown below to modify this example and see the results. Equation Used:
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Solution (a) As seen in the figure above, the cross section is symmetric about the horizontal axis, therefore, the product of inertia is zero in this case. Furthermore, with the bending moment applied about the x axis, the y component of moment is zero. As a result of the previous two conditions, the NA orientation according to eqn. A13.15 will be horizontal - passing through the centroid as expected. This problem is an example of symmetric bending. NOTE: There is no need to find the horizontal position of centroid because there is no need to
calculate the moment of inertia about the y axis as y component of bending moment is zero. (b) Because of the conditions stated in part (a) of solution, eq. A13.13 reduces to
The bending stress distribution will be linear with a zero value at the NA.
(c) In this case, the maximum stress is at the farthest point from the NA. Because of horizontal symmetry about the NA, the stress at the top and bottom of the section will have equal magnitude with the one on top being compressive. To get the maximum value of stress, the reduced equation given previously will be used. The moment of inertia about the x axis is
This makes the maximum bending stress
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Example 2 For the cross section and loading shown, determine •
(a) Neutral axis location and orientation,
•
(b) Location and magnitude of the maximum bending stress. Assume that the stresses due to the applied load do not exceed the elastic limit. Also assume that each lenght shown is measured to the middle of the adjacent member.
Use the Java screen shown below to change the problem data and see the results.
Equations Used:
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Solution (a) As seen in the figure above, this is a symmetric cross section with unsymmetric loading. The response will be an unsymmetric bending. With the product of inertia being zero due to cross-sectional symmetry, we need to calculate the components of the applied bending moment and the rectangular moments of inertia in order to determine the orientation of NA. The components of bending moment are:
By examinning the applied moment, it is clear that both of its components will produce compression on the first quadrant, hence, they are both positive. To calculate the rectangular moments of inertia, it is necessary to know the location of the centroid. Due to horizontal symmetry only the horizontal coordinate of centroid need to be calculated as its vertical coordinate is known due to symmetry.
The moments of inertia about the x and y axes are
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The neutral axis will pass through the centroid with an angle of
(b) The maximum axial stress is at the farthest point from the NA, either at point A, B, C, or D. To get the stress values, use equation A13.13 at all four points. With the product of inertia equal to zero A13.13 reduces to
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The bending stresses at point A, B, C, and D are
The results indicate that the maximum bending stress is at point B
Example 3 For the cross section and loading shown, determine •
(a) Neutral axis location and orientation,
•
(b) Location and magnitude of the maximum bending stress. Assume that the stresses due to the applied load do not exceed the elastic limit. Also assume that each length shown is measured to the middle of the adjacent member.
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Use the Java screen below to change problem data and see the results. Equations Used
Solution This problem requires more analysis as both the loading and cross-sectional shape are unsymmetric. The procedure is similar to the previous example. First need to find the centroid, moments of inertia about the x and y axes, and the product of inertia. The centroid is at
and the moments of inertia and product of inertia are
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The components of the applied bending moment are determined as
The x component is negative because it causes tension in the first quadrant. (a) Since this is a homogeneous section, and it is assumed to be within its elastic limits, the neutral axis will pass through the centroid. Its angle with respect to the x axis is
(b) The maximum bending stress occurs at the farthest point from the NA, either at point A or B.
Therefore, point B is the location of maximum bending stress.
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Example 4 The skin-stringer wing section shown is subjected to a bending moment Mo which acts in the vertical plane putting the top surface in compression. Determine: •
(a) the location of the neutral axis (NA) and its orientation;
•
(b) the largest permissible value of Mo if the maximum stress in the wing is not to exceed 12 ksi;
•
(c) the stresses normal to the plane of the cross section in all stiffeners.
Equations Used:
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Solution Since the cross section is not symmetric, the product of inertia will not be zero. With the reference point at stiffener 14, the centriod of this cross section is obtained first. Then the rectangular moments of inertia and product of inertia are determined.
(a) Since the wing section is elastic, the NA passes through the centroid. Equation A13.15 gives
Note that Mx = Mo and My = 0 in this problem.
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(b) The simplest way to find the largest permissible Mo is to pick out a few candidate stiffeners that appear the farthest from the NA. Then use equation A13.13 to solve for Mo by fixing the stress at 12,000 psi. The lowest moment will be the answer as it will not induce a bending stress at any point above the allowable value of 12,000 psi.
In this case, stiffener 12 will have the maximum stress because it is the farthest from the NA. Therefore, largest permissible moment is
(c) Use Momax for Mx in Eq. A13.13 to solve for stress in each stiffener.
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4.5 Elastic Bending of Non-Homogeneous Beams
Recall that in pure bending the normal strain variation is linear with a value of zero at the neutral axis. Before we discuss the relationship between stress and bending moment, let's first determine the location of NA. The equilibrium condition requires the sum of normal forces to go to zero. Mathematically, this is expressed as
Considering a section made of two different materials with Young's moduli identified by E1 and E2 and with stresses below the elastic limit of each material, we can write
Where yB is the distance from the NA to the bottom surface, yi is the distance from the NA to the material interface and yT is the distance to the top surface. Using the linear equation for strain variation gives
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Factoring out the common terms and E1 gives
A close examination of this equation reveals that the bracketed term must be equal to zero. However, the ratio of E2 to E1 in front of the second term inside the brackets indicates that neutral axis will pass through the centroid of the modified homogeneous section, one with material 2 replaced with an equivalent material 1. This fact is captured in the figure below with the condition that E2 > E1.
Notice that only the width of the section is modified while its height is kept the same. This condition would have been reversed if the bending moment was applied about the vertical axis. The moment equation is written as
For the case of a section made of two different materials, as shown above, the integral is divided into two parts, one for each elastic material. Notice that y is measured from the neutral axis.
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Using the linear normal strain variation we get
Factoring out the constant terms and normalizing with respect to E1 we get
The terms inside the brackets represent the moment of inertia of the modified homogeneous section about the neutral axis, and can be expressed as
Substituting this into the previous equation and solving for the stress gives
This equation can be used to calculate the bending stress only for the portion that is made of material 1. To calculate the bending stress in the portion that is made of material 2, it should be multiplied by the ratio of Young's moduli of the two materials as
The procedure described above is known as the Modified Section Method, and is used in the analysis of elastic non-homogeneous beam sections in bending.
Example Problems •
Example 1 Design of an elastic, non-homogeneous beam under horizontal bending moment
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Example A bending moment of 74,734 N-m is applied to a uniform composite beam shown in the figure below. If the allowable stress for steel is 250 MPa and the allowable stress for aluminum is 100 MPa, what is the minimum width of the 5-mm thick steel plates attached to the aluminum Ibeam. Also determine the neutral axis location and bending stress distribution.
Equations Used:
Solution Since this is a doubly symmetric cross section, the centroid will be located at the center. This is true for both the original and the modified cross sections. With the applied moment acting about the horizontal axis, the NA will be horizontal in this case. From inspection, it is known that the maximum strain will occur in the steel plates. Also, with steel being the stiffest of the two materials it is very likely that it will carry more stress. To calculate the modified moment of inertia, the entire section is modified to all aluminum. The modified moment of inertia in terms of w is
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Let the steel plates be stressed to their limit. The top plate will have the same stress and strain as the bottom plate because of symmetry.
Check the stress at the farthest point in the aluminum part.
Use the stress equation and solve for the width
The bending stress distribution for this cross section is
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Alternate Solution
4.6 Inelastic Bending of Homogeneous Beams
In this section we will first dicuss the inelastic behavior of beams in pure bending, and then elaborate on the method of analysis that we can use in such problems. Inelastic behavior is possible in beams that are made of ductile materials, and as such can be loaded beyond the elastic limit or proportional limit of the material. This implies that the ultimate load carrying capability of a ductile beam is higher than its maximum elastic load. How much higher depends on mechanical properties of the beam material. Naturally, the behavior of a beam in inelastic bending depends upon the shape of the material's stress-strain diagram. If the stress-strain diagram is known, it is possible to determine stress corresponding to a particular value of strain. As in previous discussions we will assume that the material can be idealized as an elastoplastic material with maximum stress being the elastic limit stress, and the maximum strain being considerably higher than the elastic limit strain. It is possible for the elastoplastic material to have different characteristics in tension and in compression. For instance the corresponding
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elastic limit values and even the Young's moduli may be different. This tends to complicate the analysis to a certain degree. Assumptions:
The analysis of an inelastic beam is based on the assumption that plane cross sections of a beam remain plane under pure bending, a condition that is valid for both nonlinear and linear materials. Therefore, normal strain in an inelastic beam varies linearly over the cross section of the beam. Restrictions:
a. Beam has a symmetric cross section. It is not necessary for it to be doubly symmetric. b. Beam is loaded symmetrically, moment is acting about either the x or the y centroidal axis. Neutral Axis Location:The neutral axis of beams in inelastic bending may or may not pass
through the centroid of the cross section.
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The following diagrams show the variations of bending strain and stress across a rectangular beam section ranging from fully elastic to fully plastic condition. Notice that the material is assumed to be elastoplastic with elastic limit in tension equal in magnitude to that in compression. Notice that the N.A. conicides with the horizontal centroidal axis even as the beam becomes fully plastic. If in the previous example the stress-strain variation in compression was different from that in tension, then the position of N.A. would no longer coincide with the horizontal centroidal axis as beam is loaded beyond its elastic limit.
Notice that the resultant axial force is zero as the net compression force balances against the net tension force acting on the cross section.
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In such beam problems, the location of N.A. coincides with the horizontal centroidal axis when the beam is elastic. However, as it is loaded beyond the elastic limit, N.A. shift either up or down relative to the centroid depending on whether the material can carry more tension or compression. The farthest position of N.A. is determined by checking the cross-sectional stress variation for a fully plastic condition. Helpful Observations in Inelastic Bending:
1. If the stress-strain variations in tension and compression are the same, then a. N.A. coincides with the centroidal axis (same as moment axis) if the cross section is symmetric
about
that
axis.
b. N.A. does not conicide with the centroidal axis if the cross section is unsymmetric about that axis. 2. If the stress-strain variations in tension and compression are different, then N.A. does not coincide with the centroidal axis regardless of cross-sectional symmetry about that axis. Determination of a Beam's Moment Capacity:
1. Check the stress-strain variations in compression and tension. Is there a difference between elastic limit stress in tension from that in compression? 2. Is the moment acting about the axis of symmetry or not? 3. Case A. Moment is acting about the axis of symmetry and material properties in compression and tension are the same. •
Maximum elastic moment is determined from the simplified form of Eq. (II.1)
•
Inelastic moment for some given value of maximum strain less than fully-plastic strain is found from the moment equilibrium equation. First determine the strain variation (remember it is
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linear in linearly elastic materials) across the beam. Then relate the strain variation to stress variation by checking the stress-strain diagram. Finally write the integral relating the bending moment to the stress distribution across the beam, and solve for the bending moment. •
Fully-plastic bending moment is obtained by drawing the stress pattern over the beam cross section. Keep in mind that in this case the location of N.A. is the same as centroidal axis or axis of symmetry in this case. Calculate the resultant force in compression, and resultant force in tension. Sum moments about the N.A. and find the total bending moment on the beam. Case B. Moment is acting about the axis of symmetry but material properties in compression and tension are different.
•
Maximum elastic moment is determined from the simplified form of Eq. (II.1).
•
Inelastic moment for some given value of maximum strain less than fully-plastic strain is found from the moment equilibrium equation. However, in this case the N.A. position is unknown. Therefore, an iterative solution based on a guessed position of N.A. is required. Guess a position for N.A. relative to the centroidal axis moment is acting about. From linearity of strain, determine the strain variation, then relate the strains to stresses and use the axial force equilibrium equation, and see whether the sum of forces goes to zero or not. If it goes to zero, then the location of N.A. is correct. Otherwise, guess again, and repeat the procedure. Once the location of N.A. is found, go to the moment-stress integral equation and solve for the value of moment.
•
Fully-plastic bending moment is obtained by drawing the stress pattern over the beam cross section. Here once again the location of N.A. is unknown. However, we know the maxium stress in compression as well as in tension. With the stress being constant in the tension side and constant in the compression side. No iteration is necessary here as the location of N.A. can be determined by summing the axial forces to zero and determining the height of compression and tension portions of the cross section. Once N.A. position is known, then proceed with determining the moment summation about the N.A. to obtain the fully-plastic bending moment. Case C. Moment is acting about a centroidal axis which is not an axis of symmetry, and material properties in compression and tension are different. Example 1 below deals with such a problem. Example For the beam section and loading shown, determine
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•
(a) the maximum elastic bending moment that can be applied
•
(b) the inelastic bending moment resulting in a strain of 0.003 at the bottom edge
•
(c) the maximum (or fully-plastic) bending moment.
Equations Used
Solution We must first find the location of centroid and the moment of inertia about the horizontal centroidal axis (x axis)
The moment of inertia about the x axis is all that is needed because with the product of inertia Ixy equal to zero and no moment about the y axis, the general elastic bending formula reduces to
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(a) Because the elastic-limit stress in the tension side of the stress-strain diagram is less than that in the compression side, the tensile limit will be the maximum stress allowable. Since the bottom side is in tension, let the stress at that location be equal to the limit of 45,000 Pa. This gives a maximum moment of
Now, check the top to make sure the compressive stress doesn't exceed its limit (60,000 Pa).
Therefore, the maximum elastic moment is
The strain and stress diagrams corresponding to the maximum elastic moment are shown below. Notice that the NA in this case coincides with the horizontal centroidal axis (x axis).
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(b) We know the elastic-limit strain in tension and compression to be
With the strain at the bottom now at 0.003, the elastic-inelastic interface, so to speak, is at some location between the NA and the bottom surface. Since stress-strain variations in tension and compression are different in the inelastic region, the NA will not pass through the centroid. Therefore, NA will not coincide with the x axis in this case. However, the NA will be parallel to the x axis. So we know its orientation but not its location. The location of NA is found through an iterative process. Assume NA is @ 68.333 mm from the bottom. Through similar triangles, the elastic-inelastic interface can be found
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If the location of NA is correct, then the summation of forces in the normal direction must be zero.
Based on the stress diagram shown above, the force equation is written as
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Since the sum of axial force components is not zero, the assumed location of NA is incorrect. The negative value indicates that the NA must be moved in the compression side as to reduce the area which is in compression or to increase the area in tension. For the second guess, assume NA to be @ 80 mm from the bottom surface and repeat the procedure.
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The force sum is now positive indicating that the correct location of NA is somewhere between the two locations assumed previously. Using linear interpolation, we come up with the third guess
Using linear interpolation again and finding the summation of forces yields
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The result is satisfactory. Now with the location of the NA known, the moment corresponding to the strain of 0.003 at the bottom side can be calculated.
(c) To find the fully-plastic moment, we must first determine the location of NA resulting in the total axial force to go to zero. In this case, there is no linear stress region. The compression and tension sides are under the state of constant stress equal to the corresponding maximum stress values.
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Using the stress variation shown above and the corresponding force summation, we find the location of NA as follows
The fully-plastic moment is found to be
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The results indicate that in this case the fully-plastic moment is 57% larger than the maximum elastic bending moment. We also saw in this problem that the location of neutral axis changed depending on the magnitude of M. The two factors having the most influence on the results are: (a) Different material elastic limits in tension and compression, and (b) Unsymmetric cross section with respect to the moment axis.
4.7 Elastic Bending of Curved Beams
The discussion, so far of beams in pure bending, has been restricted to beams that are straight before moment is applied. This is no longer true in this section as the beams are initially curved. Curved beams are encountered at various places in aerospace structures. For instance, the frames or rings in airplane fuselage are basically curved beams. To analyze curved beams we must clarify the assumptions and restrictions of the analysis. Assumptions:
a. Cross sections of the curved beam remain plane after moment is applied. b. Material used is linearly elastic. Restrictions:
a.
Beam
must
b.
Beam
segment
have
a
considered
uniform has
cross a
section
constant
along radius
its of
length. curvature.
c. Beam cross section must possess at least one axis of symmetry so the product of inertia is zero. d. Bending moment is applied perpendicular to radius of curvature of the beam. e. Bending stresses must remain below the elastic limit. In general for beams having radii of curvature much longer than their depths, the elastic bending formula gives fairly reasonable results for bending stresses. However, for more accurate calculation of stresses especially when the radius of curvature is of the same order of magnitude as the beam's depth, then we must use Winkler's method of analysis. In curved beams, the assumption of plane sections remain plane after loading is still valid. However, since the length of the beam varies from its top surface to the bottom surface, the strain variation is no longer linear, instead it is hyperbolic. This is evident by the equations below.
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Euqations To Remember These two equations show that stress and strain vary hyperbolically along the cross section.
Unlike in the case of straight elastic beams, the neutral axis does not pass through the centroid of the cross section. Rather, it is located somewhere between the centroid of the cross section and the center of curvature of the beam. Its exact location can be determined by the equation given below.
The differnetial area 'dA' is chosen in such a way that makes the evalution of the integral in the denominator fairly easy. The closed form equation for 'R' for some simple cross-sectional shapes are given in the figure below.
The radius of curvature, measured from the center of curvature to the centroid of the cross section, is determined by
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Using the moment equilibrium equation along with the Hooke's law, the bending stress in terms of moment is obtained as
where or
The sign associated with the moment is positive if the moment tends to increase the curvature. It is negative if it tends to decrease the curvature, or flatten the curved beam. Strain is determined from Hooke's law as
Although the stress and strain variations along the cross section of the curved beam are hyperbolic, Hooke's law is still valid because the material is linearly elastic, i.e., stress-strain relation is linear.
4.8 Miscellaneous Problems:
Problem 1: Material 168T6511 Allowable stress = 370Mpa For simply supported beam
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∑ F = R A + RB = 12100N ∑MA = 0 12500 × 305 − 610RB = 0 RB =
12100 × 305 = 6050 N 610
RA=6050 Bending Moment diagram at B = 0 at c = 6050 × 305 = 1840000N − mm at A = 6050 × 610 − 12100 × 305 = 0 M for limit load at mid section is 1840000N-mm
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b2 <
My My orI > I b2
y = 50 / 2 = 25mm
1840000 × 25 = 124324 = 124000 mm 4 370
Required I x >
Design I x should be greater than require I x Let t = 5, b = 50 Ix =
bd 3 b1 d13 50 4 50 × 40 3 − = − = 254000mm 4 12 12 12 12
Allowable stress = 370Mpa Actual stress =
My 1840000 × 25 = = 181Mpa I 254000
Form Factor
Z x y0 Ix Zx =
td12 5 × 40 2 + bt (d1 + t ) = + 50 × 5 × 45 = 13250 4 4
Form factor =
13250 × 25 = 1 .3 254000
Reserve factor =
370 = 2.04 181
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4.8.1 Bending Stress in Symmetrical Section:
Example 1: Rectangular beam 300mm deep is simply supported for a span of 4m What a UDL of the beam may carry , if the bending stress is not to exceed 120N/mm2 Take I = 8*106 mm4.
Given Data:
d = 300mm y=
l = 4000mm
d = 180 mm 2
I = 8 × 10 6 mm 4
f = 120 N mm 2
M = Bending Moment M for udl and Simply Supported beam at centre beam
wl 2 w(4000) 2 M = = = 2 × 10 6 w 8 8
Also M =
fI y
=
120 × 8 × 10 6 = 6.4 × 10 6 N − mm 150
2000000 × w = 6.4 × 10 6
w = 3.2 N mm 2
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Example 2: A timber beam of rectangular section is to support a load of 8 tonnes of UDL over a span of 3.6m. If the depth is twice the breadth and stressing timber not to exceed 70 Kg/cm2. Find the dimension of cross-section. How the cross-section is modified if there is a concentrated load placed at the centre with same ratio of breadth and depth
Given Data :
w = 2t
f = 70 kg cm 2
l = 3.6m = 360cm
d = Depth = 2 × b
b = Breadth
y=d
2
=b
M = Moment at centre of beam Z = Section modulus =
bd 2 b × (2b) 2 2b 3 = = 6 6 3
For UDL Moment at centre of beam M =
wl 2 wl 2 2 × 3.6 2 = 3.24t − m = = 8 8 8
= 324 × 10 3 kg − cm M = fZ = 70 × 2b 3 3 =
140b 3 kg − cm 3
140b 3 = 90 × 10 3 3 b3 =
b = 12.5
3 × 90 × 10 3 = 1928.6 140
d = 25
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For concentrated load M =
wl = 1.8 = 180 × 10 3 kg − cm 4
140b 3 = 180 × 10 3 3 3 × 180 × 10 3 b3 = = 3857 140 b = 16cm d = 32cm
4.8.2 Bending Stress in Unsymmetrical Section:
Example 1: A beam of I section shown is subjected to a bending moment of 500 kg-m at its neutral axis .Find the maximum stress induced in the beam.
Sl.no.
Area
y
Area × y
1
6 × 2 = 12
12 + 2 2 = 13
156
2
10 × 2 = 20
2 + 10 2 = 7
140
3
10 × 2 = 20
2 2 =1
20
∑
52
21
316
y=
∑( Area × y ) ∑( Area)
153
ssppaatteerr 154
Sl.no.
bd 3 12
A
h
b2
Ah 2
Ah 2 + bd 3 12
1
6 × 23 =4 12
12
13-
49
588
588+4
592
2
2 × 10 3 = 166.67 12
20
7-6=1
1
20
20+166.67
186.67
3
10 × 2 3 = 667 12
20
6-1=5
25
500
500+6.67
506.67
6=7
∑
1285.34
I = ∑( Ah + bd 3 12) × 1285cm 4
Distance of upper extreme fibre
y C = 14 − 6 = 8
Distance of lower extreme fibre
yT = 6
∴y=8
Example 2: A simply supported beam and its cross-section are shown the beam carries a load P=1000kg
self weight is 350 kg/cm (UDL). Calculate normal stress at a-a. Also find out
maximum stress in beam.
Taking moment about A
RB × 3.6 = 1000 × 2.4 + 350 × 3.6 × 3 2 RB =
4668 = 1297 kg 3 .6
R A = (1000 + 350 × 3.6) − 1297 = 963kg
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ssppaatteerr 155
Moment about a-a
M = 963 × 1.2 − 350 × 1.2 × 1.2 2 = 903.6kg − m =90360 kg-cm
y=
Sl.no.
Area
Y
Area*y
1
30 × 20=600
30/2=15
9000
2
− (π 4 × 15 2 ) = −176.7
20
-3534
3
423..3
5466
∑( Area × y ) 5466 = ≈ 13 ∑( Area) 423.3
Sl.no.
bd 3 12
A
h
h2
Ah 2
bd 3 12 + Ah 2
1
20 × 30 3 12 = 45000
600
(15-
4
2400
47400
49
8658.3
-11143
13)=2 2
π × 15 4 64 = 2485
-176.7
(2013)=7
∑
36257
Distance of upper extreme fibre y C = 30 − 13 = 17 Distance of lower extreme fibre yT = 13 ∴ y = 17
M f = , I y
f =
M y I
=
90360 × 17 = 42.6kg / cm 2 36257
To find out maximum stress in beam
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ssppaatteerr 156
M max = 1304.4 f =
1304 × 17 = 61.2kg / cm 2 36257
Example 3: A cross-section of beam is shown in the figure. The beam is made of material with permissible stress in compression and tension equal to 1000 kg/cm2 respectively. Calculate the moment of resistance of the cross-section when subjected to a moment causing compression at top and tension bottom. Also calculate compression at top flange and tension in bottom flange corresponding to this moment. f C = 1000 kg / cm 2
f T = 1400kg / cm 2
156
ssppaatteerr 157
Sl.no.
Area
y
20 × 8=160 32+8/2=36
1
28 × 6=168 4+28/2=18
2
10 × 4=40
3
4/2=2
Area × y
bd 3 / 12
h
h2
Ah 2
bh 3 + Ah 2 12
5760
20 × 8 3 / 12 = 8533
24-
144
23040
23893
36
6048
17024
484
19360
19413
3024 80
y=
368
24-
6 × 28 3 / 12 = 10976
18=6
10 × 4 3 / 12
24-
= 53.3
∑
36=12
8864
2=22 60330
∑( Area × y ) 8864 = = 24cm 368 ∑( Area)
I = ∑(bd 3 / 12 + Ah 2 ) = 60330cm 4
Distance of upper extreme fibre y C = 24 − 40 = 16 Distance of lower extreme fibre yT = 24 y=24 M = fI / y
M1 =
fC 1000 ×I = × 60330 = 3770625kg − cm yC 16
M2 =
fT 1400 ×I = × 60330 = 3519250kg − cm yT 24
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ssppaatteerr 158
Moment of resistance of cross section will be least of M1 and M2 ∴ It is 3519250 kg-cm
for M=3519250 fC =
3519250 M y= × 16 = 933.3 60330 I
Stress at bottom of top flange = 933 × 8 / 16 = 466.5kg / cm 2 Stress at top of bottom flange = 1400 × 20 / 24 = 1167.6kg / cm 2
Compression in top flange: Area = 20 × 8 = 160cm 2 Total compression = Area × Average compression stress = 160 ×
933 + 466.5 = 111960 kg 2
Tension in bottom flange: Area = 4 × 10 = 40cm 2 Total tension =Area × Average tensile stress = 40 ×
1400 + 1167.6 = 51352 kg 2
158
ssppaatteerr 159
4.8.3 Beams of Uniform Strength:
Example 1: A horizontal cantilever beam 3m long is having a rectangular cross section 60mm wide of throughout its length and depth varying uniformly from 60mm at free end to 180mm at fixed end. Find the position of higher stressed section and value of maximum bending stress induced neglect the self weight of cantilever at X. Given Data:
L = 3 × 10 3 mm
W = 4 × 10 3 N
M X = (4 × 10 3 ) × ( x × 10 3 ) = 4 × 10 6 xN − mm Depth = 60 +
180 − 60 x = 60 + 40 x 3
Section Modulus Z = bd 2 6 = 60 × (60 + 40 x) 2 6 = 4000(3 + 2 x) 2 mm 3 M = fZ
f =M
Z
=
4 × 10 6 x 10 3 x N = 2 4000(3 + 2 x) 2 (3 + 2 x) 2 mm
For f to be maximum df =0 dx
d(
10 3 x ) (3 + 2 x) 2 =0 dx
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ssppaatteerr 160
2 (3 + 2 x ) = 0
x = 1.5 f max =
10 3 × 1.5 = 41.67 N 2 mm 2 (3 + (2 × 1.5 ))
4.8.4 Beams of Composite Section:
Example 1: A Composite beam made of steel and timber has a section shown. Determine the moment of resistance of beam.
Given Data: f S = 1000 kg / cm 2
For timber
f T = 50kg / cm 2
For Steel
b = 6cm
b = 1.5cm
d = 20cm
d = 20cm
f = 50kg / cm 2
f = 1000kg / cm 2
Z = bd
2
6
ZT = 2 × (
6 × 20 2 ) = 800cm 3 6
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ssppaatteerr 161
M = f ×Z
M T = 50 × 800 = 40000kg − cm M S = 1000 × 100 = 100000kg − cm
Total Moment = M T + M S = 140000kg − cm Example 2: A Timber beam 100mm wide and 200mm deep is strengthened by a steel plate 100mm wide and 10mm thick screwed at the bottom of the surface as shown. Calculate the moment of resistance of beam, safe stress in timber and steel are 10N/mm2 and 150N/mm2, ES = 20ET.
Given Data: For Timber
For Steel
b = 100mm
b = 100mm
d = 200mm
d = 10mm
f = 10 N
f = 150 N
mm 2
mm 2
E S = 20 ET
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ssppaatteerr 162
Stress in steel will be 20 times the stress in timber resistance offered by steel = 20 × Resistance
offered
by
timber,
So
steel
is
converted
into
equivalent
timber
100 × 20=2000mm. Sl.no.
Area
y
A× y *103
bd 3
12
h
h2
Ah 2
*103
bd 3 + Ah 2 12
*103 1
2000*10
10/2=5
100
=20000 2
100 × 200
10 +
=20000
200/2
2200
=110 ∑
y=
40000
2000 × 10 3 12 = 166666.67
100 × 200 3 12 = 66.67 × 10 6
2300
57.5-
56.25
5625
1291.67
2756.3
55125
121795
50=7.5 11057.5= 52.5 123.09 × 10 6
∑( Area × y ) = 57.5 ∑( Area)
I = 123.09 × 10 6
Distance of upper extreme fibre = 210-57.5 = 152.5
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ssppaatteerr 163
Distance of lower extreme fibre = 57.5 Stress in upper extreme fibre = 10N/mm2 Stress in lower most fibre
=
10 × 57.5 = 3.77 N / mm 2 152.5
Actual stress in steel at this fibre = 3.77 × 20 = 75.4N/mm2
M =
fI 10 × 123.09 × 10 6 = = 8.07 × 10 6 N-mm. y 152.5
4.8.5 Bending of Unsymmetrical Section about Principal Axis:
Example 1: Find the bending stresses at section A-A of beam for unsymmetrical section bending about principle axes.
M X = 2400 × 20 = 48000 M Y = −600 × 20 = −12000 IX =
2 × 63 12
IY =
6 × 23 12
fb = −
M X y M Y x − 48000 y 12000 x − = + = 3000 x − 1333 y IX IY 36 4
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ssppaatteerr 164
x,y
(-1,-3)
(1,-3)
(1,3)
(-1,3)
f b = 3000 x − 1333 y
1000
7000
-1000
-7000
To find Neutral axis , f b = 0
−
M X y MY x − =0 IX IY
3000 x-1333 y = 0 When x = 1, y = 2.25 When x = -1, y = -2.25 4.8.6 Bending of Unsymmetrical Section about Arbitrary Axis:
Problem 1: Find the bending stresses at points at A,B,C of beam cross section shown.
164
ssppaatteerr 165
slno Area
1
4*2
x
11
y
2
Ax
88
Ay
16
bd 3 12
db 3 12
hx
h y bd 3 + 12 Ahx2
2.67
10.6
5
3
202.67
82.67
-120
=8 2
12*
db 3 + 12 Ah y2
Ah x h y
7 6
5
144
120
288
8
0
0
288
0
0
1
8
8
64
2.67
10.6
-5
3
202.67
82.67
-120
693.34
173.34
-240
2= 24 3
4*2 =8
∑
x=
40
7 240
200
∑( Area × x) 240 = =6 40 ∑( Area)
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ssppaatteerr 166
y=
∑( Area × y ) 200 = =5 40 ∑( Area)
I x = ∑(
bd 3 + Ahx2 ) = 693.34 12
I y = ∑(
db 3 + Ah y2 ) = 173.34 12
∑ Ah x h y = 240.00
fb =
M xIxy − M yIx IxIy − I
=
2 xy
x+
M yIxy − M xIy 2
IxIy − Ixy
y
100000(−240) − 10000(693.34) 10000(−240) − 100000(73.34) x+ y 2 (693.34 × 173.34) − (−240) (693.34 × 173.34) − (−240) 2
(−24 × 10 6 ) − (693.34 × 10 4 ) (−24 × 10 6 ) − (17.33 × 10 6 ) fb = x+ y 62583.55 6258.53 f b = −494 x − 315 y
To find Neutral axis fb = 0 − 494 x − 315 y = 0
y=
− 495 x 315
y = −1.568 x
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ssppaatteerr 167
( x, y )
(1,6)
( − 5,6 )
( − 5,4 )
( − 1,−6 )
(5,-4)
(5,-6)
f b = −494 x − 315 y
-2384
580
1210
2384
-1210
-580
167
ssppaatteerr 168
Problem 2: Find shear flow in webs of beams shown in figure .Each of the four flange members has an area of 0.5 sq.in. The webs are assumed to carry no bending stress. Find the shear centre for the given area. Given Data: q = f sb =
V I
∫ ydA
I = n × A× x2 = 4 × 0.5 × 5 2 = 50
qa b =
10000 (5 × 0.5) = 500 50
qb c =
10000 ((5 × 0.5) + (5 × 0.5)) = 1000 50
qc d =
10000 {(5 × 0.5) + (5 × 0.5) − (5 × 0.5)} = 500 50
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ssppaatteerr 169
Taking moment about c
10000e = 500 × 4 × 10 e=2 Similar solid c section of thickness 1 is considered with similar loading case. find the shear flow in members. q = f sb =
V I
∫ ydA
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ssppaatteerr 170
Slno
Area
1
3 ×1 = 3
x
y
Ax
9.5 2.5 28.5
Ay
bd 3 12
db 3 12
hx
7.5
0.25
2.25
4.5 1.65 0.25+60.75 2.25+8.2
hy
bd 3 + 12 Ahx2
=61
10 × 1 = 10
2
3 ×1 = 3
3 ∑
q=
5
0.5
0.5 2.5
16
50 1.5 80
5 1.25 13.75
83.3 0.25
0.83 2.25
0
0.35
db 3 + 12 Ah y2
=10.5
8.33+0
0.83+1.2
=83.3
=2
4.5 1.65 0.25+60.75 2.25+8.2 =61
=10.5
205.3
23
10000 ydA 205.3 ∫
qa b =
10000 (4.5 × x) = 219 x 205.3
qb c =
10000 25 y 2 {[ 4 × 1 × 4.5] + [ − ]} 205.3 2 2
=
10000 [30.5 − y 2 2] 205.3
= 1485.6 − 24.35 y 2 qc d =
10000 [4.5 x] = 219 x 205.3
I = 205.3 x
qa b
0
0
4
876
y
qb c
5
876
0
0
170
ssppaatteerr 171
-5
876
x
qc d
0
0
4
873
Taking moment about c 10000e = 876 × 10 e=
876 × 10 10000
e = 0.876
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ssppaatteerr 172
5. MEMBER IN TORSION
In this chapter we will consider analysis of structural members under pure torsion. The only exception appears in the discussion of members in differential bending where crosssectional warping is prevented. We begin with a review of mechanics of materials and discuss torsion of members with circular cross sections. Both elastic and inelastic stress states are considered for both homogeneous and non-homogeneous members. We then proceed with the elastic torsional analysis of members with solid or hollow non-circular cross sections. This is followed by the elastic torsional analysis of thin-walled members with open cross sections such as an I or a T. Next we examine the torsion of thin-walled structures with closed cross sections. We conclude this section with the torsional analysis of multi-cell, thin-walled stiffened members, typical of wing and fuselage structures. 5.1 Torsion of Circular Bars 5.1-1 Elastic and Homogeneous 5.1-2 Elastic and Non-homogeneous 5.1-3 Inelastic and Homogeneous 5.1-4 Inelastic and Non-homogeneous 5.1-5 Residual Stress Distribution 5.1-6 Power Transmission 5.2 Torsion of Non-Circular Bars • • • • • •
5.3 Elastic Membrane Analogy 5.4 Torsion of Thin-Wall Open Sections 5.5 Torsion of Solid Non-Circular Shapes 5.6 Torsion of Thin-Wall Closed Sections 5.7 Torsion-Shear Flow Relations in Multiple-Cell Closed Sections 5.8 Shear Stress Distribution and Angle of Twist for Two-Cell Thin-Wall Closed Section 5.9 Shear Stress Distribution and θ for Multiple-Cell Thin-Wall Closed Section 5.10 Torsion of Stiffened Thin-Wall Closed Sections 5.11 Effect of End Restraint 5.12. Miscellaneous problems
172
ssppaatteerr 173
5.1 Torsion of Circular Bars: Applications:
Rotating Machinery •
Propeller shaft
•
Drive shaft
Structural Systems •
Landing gear strut
•
Flap drive mechanism
Characteristic of Circular Bars:
When a circular bar is twisted, its cross section remains plane and circular. This characteristic is due to axisymmetric shape of the cross section, and applies to circular bars that are solid or hollow, homogeneous or non-homogeneous, elastic or inelastic. When a straight cylindrical bar is subjected to two equal but opposite Torsional couples, the bar twists and each section is subjected to a shearing stress. Due to these shear stresses there will rotation in the section. This rotation at any section will be proportional from the fixed support. T= (П/16) fs D3 Twist θ = TL/GJ Shear Strain Calculation:
Based on the characteristic stated above, the shear strain variation is linear, and is described by the equation
where r is the radial position measured from the center of the cross section and c is the radius of the cross section. Hollow Bars and Tubes in Torsion:
173
ssppaatteerr 174
For bars and tubes of circular cross section, the Torsional stress is calculated from the formula Torsional − stress =
16 D * Torque Where D and d are respectively the π (D 4 − d 4 )
external and internal diameters of the torsion members. Torsional 0.1% proof stress and Torsional Ultimate stress can be obtained for specific material from data books. Where for specific material do not appear in the tables and no test information is available, an approximate estimate of basic Torsional stresses may be obtained from the values of t1 and ft.
174
ssppaatteerr 175
Rule of Thumb:
The torsion-induced shear strain is always a linear function of r with the maximum value at the edge of the cross section. This is true for all possible conditions, whether the bar is elastic or inelastic, homogeneous or non-homogeneous as shown in these examples.
Shear Stress and Angle of Twist Calculations:
Four possible scenarios are considered for shear stress and angle of twist calculations. In each case the equations used for these calculations are explicitly described. It should be obvious that in all of these calculations we are dealing with members having circular cross sections. As you will see the degree of complexity in the analysis grows as we proceed in the following sequence •
5.1-1 Elastic and Homogeneous
•
5.1-2 Elastic and Non-homogeneous
•
5.1-3 Inelastic and Homogeneous
175
ssppaatteerr 176
•
5.1-4 Inelastic and Non-homogeneous
•
5.1-5 Residual Stress Distribution
•
5.1-6 Power Transmission
Restriction: The applied torque(s) must be in a plane(s) perpendicular to the axis of the bar. 5.1-1 Elastic and Homogeneous:
The torsion-induced shear stress variation in an elastic, homogeneous, and isotropic bar is determined by
where T is the internal torque at the section the shear stress is being calculated, r is the radial position of the point on the cross section the shear stress is solved for, and J is the polar moment of inertia of the entire cross section. For solid bars 0 < r < c and for hollow bars ci < r < co. The above shear stress equation is known as the elastic torsion formula, and we call it ElHocir. This acronym helps us to easily remember under what conditions we can use the elastic
torsion formula, viz, elastic, homogeneous, and circular. The shear stress equation shows that for an elastic bar (i.e., when the maximum shear stress is less than the proportional limit shear stress of the material), the stress varies linearly with radial position. Thus, the maximum shear stress in this case would be at the edge of the cross section (i.e., at the farthest distance from the center). Here are some examples of elastic and inelastic stress variations for homogeneous and non-homogeneous circular bars made of elastoplastic materials.
176
ssppaatteerr 177
The angle of twist at one section relative to another can be found in two ways:
Or
In the first equation we are using geometric relationship between angle of twist and shear strain. This is a very powerful equation as it tells us that we don't need to know the torque or the resulting shear stress in order to calculate the relative twist angle. If we know the maximum shear strain at section B and know the distance L between sections A and B, then we can calculate the relative twist angle using this formula. Notice that this equation can also be written in terms of shear strain at any radial position. What is needed is the value of r and its corresponding shear strain.
177
ssppaatteerr 178
The second equation is useful when we don't know the shear strain, but we do know the internal torque at section B. Then using this equation we can determine the relative angle of twist. Total angle of twist at section B is determined by
5.1.2 Elastic and Non-homogeneous:
The linear strain variation is valid as stated before; however, since the bar is non-homogeneous, i.e., made of two or more materials, we cannot use El-Hocir. This problem is statically indeterminate in the sense that we don't know how much of the torque is being carried by each material. The general torque formula in its integral form must be broken up into two or more parts depending on the material make up of the bar. Also, since the bar is totally linearly elastic (i.e., stresses are below the proportional limit), Hooke's law may be used to relate stress to strain in each respective region in the cross section. Here is what the torque equation looks like for a bar made of two materials:
Where 0 and 1 limits in the first integral denote the radial boundaries of the core material, while 1 and 2 denote those for the shell material. The subscripts 'C' and 'S' correspond to the core and shell materials, respectively. Knowing the maximum shear strain, we can use the above equation to solve for the torque at a given section, or vice versa. Notice that we used the same linear equation for stress-strain relations in both integrals with shear modulus being the only difference. Examining this equation more carefully, we realize that while maximum shear strain always occurs at the farthest distance from the center of the bar, it is not necessary for maximum shear stress to be at the same location. The difference between the shear moduli of the core and the shell could be such that the shear stress in the core at the interface of the shell and core might be higher than that at the edge of the bar near its outside surface.
178
ssppaatteerr 179
Rule of Thumb:
Maximum shear strain occurs at r = c regardless of whether the shell material has higher or lower shear rigidity. Maximum shear stress occurs in the material with the highest shear rigidity regardless of its location. Here are some examples:
Since the core and the shell are assumed to be perfectly attached at the interface, the angle of twist is the same for both materials; therefore,
179
ssppaatteerr 180
5.1.3 Inelastic and Homogeneous:
As mentioned earlier, the shear strain pattern is linear eventhough the member has been stressed beyond the elastic limit and into the inelastic range. For a linearly elastic material, shear stress variation is linear up to the radial distance corresponding to the elastic rim shown in the figures below. Beyond this distance, Hooke's law canNOT be used to relate stress to strain variation. Once again the material is assumed to be elasto-plastic.
Let's see how torsional analysis is performed for a circular bar as the internal torque is increased from zero to that corresponding to the fully-plastic bar. Notice that we are examining only one section (slice) of the bar along its length. Therefore, depending on which section we cosider the internal torque may be different.
180
ssppaatteerr 181
If at the section of interest the maximum shear strain is below the elastic-limit shear strain of the material, then the bar is fully elastic and the corresponding torque can be obtained as shown in (1). If the maximum shear strain reaches the elastic shear strain, then the bar has reached the elastic limit state, and the torque calculated according to (2) is referred to as the "elastic-limit torque". As the torque is increased beyond the elastic-limit torque, the internal shear stress moves into the inelastic range. In that case we must first locate the elastic rim by calculating re. Once that distance is known, then we can calculate the internal torque corresponding to the specified max. shear strain according to (3). Note that in this case, the material is assumed to be elasto-plastic, which implies that beyond the elastic limit, stress remains constant at the elastic limit value. If the torque is increased until the cross-section is completely in the inelastic range, then
the
maximum
inelastic
or
fully-plastic
torque
is
found
according
to
(4).
The angle of twist at a given section in an inelastic circular bar is found by
or
181
ssppaatteerr 182
where
which corresponds to the angle of twist at the onset of yield (when the elastic radius = radius of bar, c ) 5.1.4 Inelastic and Non-homogeneous :
Here, we are dealing with a circular bar that is made of two or multiple materials distributed in the radial direction. The bar is twisted beyond the elastic limit of at least one of the materials. It is not necessary for all materials to be stressed beyond their respective elastic limits. In this case, the shear strain variation is still linear and the torque equation is broken up at the material interface and at the point where the transition from elastic to inelastic region takes place. Let's consider the case where the bar is composed of two different materials. There are several possibilities that need to be investigated: (1) The inner material (core) may become inelastic while the outer shell is still elastic; (2) the outer shell may become fully plastic while the core remains far below its proportional limit; etc. Here is an example of a non-homogeneous circular bar in a state of inelastic stress and strain. It is obvious that the analysis of this type of problem requires more rigor than those dealing with elastic torsion.
182
ssppaatteerr 183
Analysis Procedure:
Analysis of circular non-homogeneous bars in which max. shear strain is specified. Notice that this is a statically indeterminate problem as we don't know what percentage of the resultant torque is being carried by each material. 1. Knowing that the max. shear strain occurs at the farthest distance from the center of circular cross section, we calculate the actual shear strain at the material interface using the linear equation. Now we know the max. shear strain in each material. 2. Next, we calculate the elastic-limit shear strain of each material from the strain-stress diagram or material properties provided in the problem, and compare it with the corresponding actual shear strain. For inelastic torsion problem, the max. shear strain in one of the materials must exceed its corresponding elastic limit value. 3. With this information, we can proceed with plotting the stress variation across the cross section of the bar. This gives us a map that we can use to set up the integral form of the torque equation. The integral would be split into several segments depending on the variations of stresses and materials. 4. First we consider the core. Based on the stress variation in the core, we may need up to two separate integrals. Then, we examine the shell. Again, depending on the stress
183
ssppaatteerr 184
variation, we may need up to two separate integrals - one for the elastic part and another for the inelastic part. 5. Once we identify the limits on each integral, and correctly write the integrand in each case, we can proceed with the integration to find the torque that corresponds to the specified max. shear strain in the problem statement. 6. Notice that the integrals correspond to the portion of the resultant torque carried by the core and the shell. So once we have calculated these values, we can determine the percentage of torque carried by each constituent. 7. The angle of twist is calculated in the same manner specified in the previous section. The main assumption is that the two materials are perfectly bonded at the interface. 5.1.5 Residual Stress Distribution in Homogeneous Circular Bars
When a circular bar is torqued beyond its elastic limit, upon the removal of the applied torque, the shear stress may not completely disappear. This remaining stress is known as residual shear stress and is calculated as follows based on the method of superposition.
In the figure above, the first diagram on the left reflects the elastoplastic behavior of the material during the loading phase (1), and the second diagram in the middle shows the linear behavior of the same material during the unloading phase (in the stress-strain diagram of a linearly elastic material the unloading path is always linear and parallel to the elastic line) (2). By superposing
184
ssppaatteerr 185
the two stress diagrams, we obtain the residual stress distribution (3). Although there is residual stress without the presence of any external torque, the bar remains in the state of equilibrium as the positive area under the curve equals the negative area. Notice that the three diagrams are not drawn to scale. If the bar is torqued beyond the elastic limit, there will be a permanent twist upon removal of the torque. This angle of twist is the difference between the angle of twist due to loading and that due to unloading, and is expressed as
where the recovered angle of twist - upon unloading - is determined as
5.1.6 Power Transmission
Circular bars or shafts are commonly used for transmission of power. Circular cross section is used because, as explained in the previous section, it will not get distorted under torque. From design stand point, it is necessary for the shaft to be strong enough to transmit power safely without exceeding the shaft material's elastic limit. We recall from physics that for angular motion: Work = Torque x Angular Displacement Power = d/dt (Work)
If torque is not a function of time, then the equation for power simply becomes: P=Tw
where w is the angular velocity of the shaft. This is the relationship between the torque in the shaft and the power transmitted through it from one end to the other. It is important that we use consistent units for P, T, and w. Power is commonly specified in horsepower, HP. In that case we can convert HP to ft-lb/sec by multiplying it by 550, and to inlb/sec by multiplying it by 6600. Angular velocity is usually given in revolutions per minute or RPM. It should then be converted to rad/sec. To do this multiply the value in RPM by 2 pi and divide by 60. Therefore, to get the units of in-lb for torque, the conversion factor becomes:
185
ssppaatteerr 186
T = HP x 63025 / RPM. Mechanical Coupling
Mechanical coupling is commonly used to extend the length of a shaft or to connect the shaft to another component. An example of this can be found in flap mechanism as well as in the propeller shaft. The figure below shows that with two disks in contact, the applied torque will tend to rotate one relative to the other. This relative displacement, if prevented, will lead to the creation of shear stress and strain at the surface of contact. These types of disks are typically bolted together. In that case, the torque will put the bolts in the state of plane shear.
In problems of mechanical coupling, it is commonly assumed that the surface of contact is frictionless and that the entire torque is transmitted through the bolts which are distributed in a circular pattern at a particular distance from the center. As in the case of a circular shaft, the shear strain varies linearly from the center of the cross section. Since the diameters of bolts are usually much smaller than that of the coupling plate, it is commonly assumed that the shear strain is constant along the bolt's cross-sectional area. Starting from the original equation for torque we can derive the special form for the mechanical coupling problems as follows:
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Shaft Coupling
In shaft coupling, we are concerned with the number and size of bolts to be used in order to keep the stress in the bolts below the allowable value. As a designer we have a choice of materials, size, number, and distribution pattern for the coupling bolts. This freedom allows us to examine different ways that a coupling can be designed such that it is efficient from the stand point of design as well as manufacturability and cost.
We need to balance the number and size of bolts being used. Is it better to use a large quantity of small bolts or few large bolts. We also need to keep in mind the damage tolerance issue. If we use too few bolts, then what would happen in the case one gets sheared off.
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Will the remaining bolts carry the desired load? Questions like this would have to be answered by the designer before the design can be accepted for production. Below we see the equations used for the coupling analysis.
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5.2 Torsion of Non-Circular Bars
The observations made for torsion of members with circular cross sections do not hold for those with non-circular cross sections. Consider the following facts for members with non-circular cross sections: 1. The shear stress is not constant at a given distance from the axis of rotation. As a result sections perpendicular to the axis of member warp, indicating out-of-plane displacement. 2. The theory of elasticity shows that the shear stress at the corners is zero. 3. Maximum shear strain and stress are not at the farthest distance from the rotational axis of a homogeneous non-circular member. Members with rectangular cross sections:
For a rectangular member under torsion the corners do not distort; the corner square angles remain square after torque is applied. This indicates that shear strain is zero at the corners since there is no distorsion. This fact is illustrated in this figure.
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St. Venant was the first to accurately describe the shear stress distribution on the cross section of a non-circular member using the Theory of Elasticity. Theory of Elasticity shows that: •
the maximum shear strain and stress occur at the centerline of the long sides of the rectangular cross section
•
the shear strain and stress at the corners and center of the rectangular cross section are zero.
•
the strain and stress variations on the cross section are primarily nonlinear.
The theory of Elasticity has been applied to find analytical solutions for the torsion of rectangular elastic members. The resulting equations for shear stress and angle of twist are as follows:
5.3 Elastic Membrane Analogy:
Prandtl showed that the Laplace equation describing the torsion of an elastic member is identical to that used to describe the deflection of an elastic membrane subjected to a uniform pressure.
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The elastic membrane analogy allows the solution of a torsion problem to be determined in a simpler way than that found by the theory of elasticity which requires the availability of the warping function. Elastic Membrane and Twisted Bar Relationships:
Consider a tube which has the same cross-sectional boundary as the bar. For example, if the bar has a solid square cross section of side dimension b, then the tube will have a hollow square cross section of side dimension b as well. Next we stretch an elastic membrane over the tube's cross section and apply internal pressure. The deflected shape of the membrane helps us visualize the stress pattern in the bar under torsion. The analogy can be viewed as follows: 1. Lines of equal deflection on the membrane (contour lines) correspond to shearing stress lines of the twisted bar. 2. The direction of a particular shear stress resultant at a point is at right angle to the maximum slope of the membrane at the same point. 3. The slope of the deflected membrane at any point, with respect to the edge support plane is proportional in magnitude to the shear stress at the corresponding point on the bar's cross section. 4. The applied torsion on the twisted bar is proportional to twice the volume included between the deflected membrane and plane through the supporting edges. Examples
Keep in mind that the slope at a point on the deflected membrane, and not the displacement from the base, is the parameter that is related to the shear stress in the bar. In all of the following examples, the slope is zero at the very top of the membrane, therefore the stress is zero, not the maximum, at the same location on bar's cross section.
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5.4 Torsion of Thin - Wall Open Sections :
The sections shown below are extruded sections
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Example
Use the formulas based on the theory of elasticity to find the maximum shear stress and angle of twist per unit length of a bar having the cross sectional shape shown below.
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EQUATIONS USED
SOLUTION First, need to find the b/t ratio. Because this is a formed section, the lengths 'b' are found this way
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Since b/t is greater than 10, we can let alpha and beta be 1/3. Therefore maximum shear stress is
Angle of Twist per unit length is
5.5 Torsion of Solid Non – Circular Sections:
In a noncircular section the shear stresses are not distributed uniformly. For example when a rectangular bar twists, the shear stresses are not constant at the same distances from the axis of the rotation and thus the ends of segments cut through the bar would not remain parallel to each other when bar twists or in other words warping of the section takes place. Torsional Stress = Tn/J Twist θ = TL/JG J = Polar moment of inertia
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Were J and n are the functions of the cross section. For some standard sections these values are listed below.
Form of cross sections
Formula for shear stress
n=r
2T πr 3
1 4 πr 2
2T πab 2
πa 3 b 3
Solid circular section Solid elliptical section Solid square section
n=
2a 2 b a2 + b2
(equilateral) Symmetrical Aerofoil
inertia
a2 + b2
0.601T a3
2.25a 4
3a 4
20T a3
3a 4 80
Refer a
TC K
n = 1.3519a
Solid triangular section
Polar moment of
Formula for n
n=
4I x 16 I x 1+ AU 2
For further details refer a) Roark’s Formulas for stress and strain table 10.7 page no 401 b) Structural Principles and Data by Dowty table 12.5 page no 122
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Limit stress analysis:
Working stress = Tr/J in n/mm2. Torque = T Polar moment of inertia = J Allowable Proof Torsional Stress = q1 in N.mm2. Reserve Factor = q1/Actual Torsional stress Ultimate stress analysis:
Ultimate working stress = Factor of safety * T*r/J Allowable Ultimate Torsional Stress = fq. Reserve Factor = fq/Actual Ultimate Torsional stress.
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5.6 Torsion of Thin - Wall Closed Sections:
So far, we have considered the torsion of solid and hollow circular sections, thinwalled open sections, and solid noncircular sections. Now we would like to consider the torsion of thin-walled closed sections of various geometry, typical of aircraft wing and fuselage. As before we are interested in calculating the shear stress distribution and overall angle of twist due to applied torque(s). Such analysis can be used to determine, based on a specified material and loading condition, (a) the margin of safety for an existing structure, or (b) the structural sizing such as wall thickness necessary to safely carry the applied loads. Here are some examples of typical shear stress distributions in thin-walled, single- and multi-cell closed sections under pure torsion.
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Derivation
Consider a thin-walled member with a closed cross section subjected to pure torsion.
Examining the equilibrium of a small cutout of the skin reveals that
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Angle of Twist By applying strain energy equation due to shear and Castigliano's Theorem the angle of twist for a thin-walled closed section can be shown to be
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Since T = 2qA, we have
If the wall thickness is constant along each segment of the cross section, the integral can be replaced by a simple summation
Here is an example. Important Observations: In thin-walled structures failure is governed by buckling instability; therefore, the limiting stress is not necessarily the elastic limit stress. Example Problems:
Example 1 Maximum shear stress and angle of twist in a single-cell section Consider the single cell, thin-walled section shown made of aluminum with the given properties. With the torque T=134,400 in-lb, determine (a) the shear flow distribution, (b) the maximum shear stress and its location, (c) the angle of twist if L=100 in.
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EQUATIONS USED
SOLUTION (a) Since this ia a single-cell section, there will only be one shear flow of constant magnitude
along all webs. Note that in calculating the cell area the wall thickness is ignored.
The direction of shear flow is consistent with that of the resultant torque at the section of interest.
(b) Because the skin has two different thicknesses, there will be two different shear stresses.
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Notice that the maximum shear stress occurs in the thinest web. This is always true with pure torsion of single-cell thin-walled sections. (c) Knowing the shear flow distribution, we can calculate the angle of twist as
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5.7 Torsion - Shear Flow Relations in Multiple-Cell Thin- Wall Closed Sections
The torsional moment in terms of the internal shear flow is simply
Derivation
For equilibrium to be maintained at a exterior-interior wall (or web) junction point (point m in the figure) the shear flows entering should be equal to those leaving the junction.
Summing the moments about an arbitrary point O, and assuming clockwise direction to be positive, we obtain
NOTE: It is important to realize that point "O" can be anywhere in space, interior or exterior of the cross-section. As long as we can determine the area enclosed by the ends of each web and point O, we can relate the torque to the shear flow in the web and the enclosed area.
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The moment equation above can be simplified to
Notice that in this equation the shear flow of the interior wall is not explicitly shown because it is a function of the shear flows in the exterior walls. Also notice that the terms in the right hand side of the equation are independent of moment point "O". This is a very important fact, and should be kept in mind when solving single- or multi-cell, closed, thin-walled sections under pure torsion.
5.8 Shear Stress Distribution and Angle of Twist for Two-Cell Thin-Walled Closed Sections
The equation relating the shear flow along the exterior wall of each cell to the resultant torque at the section is given as
This is a statically indeterminate problem. In order to find the shear flows q1 and q2, the compatibility relation between the angle of twist in cells 1 and 2 must be used. The compatibility requirement can be stated as
where
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Note that each closed integral deals with the boundaries of an individual cell, which involve the exterior wall as well as the interior wall separating the two cells. Also notice that the shear flow along the interior wall is the difference between q1 and q2. Now by setting the right hand sides of the two angle of twist equations equal to each other, we arrive at the second equation, which together with the torque-shear flow equation result in the following closed form equations for the two shear flows.
where
and
The shear stress at a point of interest is found according to the equation
To find the angle of twist, we could use either of the two twist formulas given above. It is also possible to express the angle of twist equation similar to that for a circular section
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with the polar moment of inertia expressed as
A careful scrutiny of the equations for shear flow indicates that the first term in the numerator is all that is different between the two. Therefore, if we have a two-cell thin-walled section in which a10A2 = a20A1, then the two shear flows would be equal. Example Problems: •
Example 1 Shear flow distribution, Maximum shear stress, and angle of twist in a twocell section
Consider the same cross section as in example 1 in section I.6, with the same properties and applied torque of 134,400 in-lb. However, an interior web is added as shown in the figure below. Determine: •
(a) the shear flow distribution,
•
(b) the maximum shear stress and its location,
•
(c) the angle of twist if L=100 in.
EQUATIONS USED
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SOLUTION (a) First, write the torque equation.
So far we have obtained one equation in two unknowns. To solve for the two unknowns we need a second equation in terms of the same two unknowns. We use the angle of twist relationship, that is the twist of both cells is the same. Writing the angle of twist equation for both cells and setting them equal to each other gives
Now by substituting this relation into the torque equation we can solve for q1 and q2, and by taking the difference between these two shear flows we can obtain q3 to be
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NOTE: q3 was determined by subtracting q2 from q1, hence, positive value for q3 indicates that this shear flow is in the same direction as q1, which in this problem is found to be counterclockwise. (b) The shear stress along each web is constant, and is equal to the shear flow along the web
divided by the web thickness. There are three different web thicknesses and three different shear flows, hence, there will be three different shear stresses
NOTE: The maximum shear stress occurs in the web which has the maximum q/t value and not necessarily minimum t. As can be seen in this example, the interior web has the minimum
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thickness, but not the maximum stress. This is obviously because, unlike single-cell sections, the shear flow is not constant everywhere. Compared to example 1 in section I.6, the maximum shear stress is in the same place, but it is smaller. In this case the interior web relieved the shear stress carried in wall (2), and increased the stress in wall (1), while reducing the overall angle of twist. The conclusion that can be drawn is that the presence of interior wall increased the torsional stiffness of the thin-walled section. (c) Substituting the values of the shear flows back into the angle of twist equation gives
Suggested Exercise: Rework this example problem by letting t1 = t2 = .06 in., and examine the changes in the results.
5.9 Shear Stress Distribution and Angle of Twist for Multiple-Cell Thin-Wall Closed Sections:
In the figure above the area outside of the cross section will be designated as cell (0). Thus to designate the exterior walls of cell (1), we use the notation 1-0. Similarly for cell (2) we use 2-0 and for cell (3) we use 3-0. The interior walls will be designated by the names of adjacent cells. From section A6.9, the torque of this multi-cell member can be related to the shear flows in exterior walls as follows
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For elastic continuity, the angles of twist in all cells must be equal
The direction of twist chosen to be positive is clockwise.
NOTE: Since the exterior walls in each cell have equal shear flows, we can factor the shear flow out and multiply it by the sum of the quotient S/t of the exterior walls. The above equations for angle of twist together with the torque equation provide a total of 4 linear equations which can be solved simultaneously for the 3 unknown shear flows (i.e., q1, q2, and q3) and 1 unkown angle of twist. 5.10 Torsion of Stiffened Thin-Wall Closed Sections: Types of Stiffeners •
Open
Offer very little torsional rigidity; their contribution may be neglected in the torsional analysis. •
Closed
Offer more torsional rigidity; their contribution may be included in the torsional analysis. Their presence basically leads to a "multiple-cell" box beam. The influence of stiffeners is included in the analysis in the following manner:
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5.11 Effect of End Restraint
The equations derived earlier, for pure torsion of bars with non-circular cross sections, were based on the assumption that at any point along the length the cross section is allowed to have out-of-plane displacement or warping. However, when an end of the bar is fixed, it is restrained from warping - as a result normal stresses are created on the cross section, and the assumption of pure torsion is no longer valid. To better understand this behavior let's consider a thin-walled bar, with an "I" cross section, fixed at one end and free at the other where a torque is applied. In the vicinity of the fixed end the upper and lower flanges of the bar act as short beams, therefore, have considerable bending stiffness, and prevent the cross section from warping under the applied torque. Hence, the torque is carried mainly in the form of transverse shear in these flanges. As a result the upper and lower flanges deflect laterally in opposite directions - referred to as differential bending. In this case, the torque can be represented by a couple of force H acting in each flange
At the other extreme point, near the free end, the applied torque is carried mainly in the form of torsional shear or pure shear as the cross section is free to warp. This is because, near the free end, the flanges are basically long beams with less bending stiffness, therefore, shear rigidity dominates in this case.
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At some distance x between the two ends of the bar, the torque can be represented as the sum of (a) the torque taken up by pure shear and (b) the torque taken up by transverse shear.
For angle of twist due to torsion in (a), from the analysis in section A6.6, we can write
where the length of each rectangular portion of the cross section is assumed to be much greater than its thickness. For angle of twist due to torsion in (b) we use the relationship between the lateral deflection of the flanges and the angle of twist as
Now, from mechanics of materials we can relate the flange deflection to transverse shear force H in the flange as
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Combining the pevious equations we determine angle of twist
Since the two angles of twist are equal, the right hand sides of the two equations can be set equal to determine the ratio between the portion of the torque taken up by transverse shear and portion taken up by pure shear.
Example 1
A wing spar has a cross-sectional shape shown below. It is fixed at the root, i.e., x=0 and free at the tip, i.e., x=200 in. What is the percentage of torque due to bending to that of the total torque, at x=2", x=20", and x=200"? If T=1000 in-lb, what would the angle of twist be at each value of 'x'?
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EQUATIONS USUED
SOLUTION The equation for the ratio of TB to TT was derived earlier in the discussion. With the material properties and geometric parameters known, we calculate the torque ratio for different values of x as follows:
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To find the percentages, it takes a little manipulation, but the results are:
Notice the rapid decrease in the percentage of T carried in the form of differential bending of flanges as x increases. The angle of twist at different spanwise stations is calculated by using the angle of twist equation with different values of x. Note that we can use either the angle of twist based on TT or TB, the result would be the same.
The angle of twist increases going from the fixed end to the free end. This example is used to show that in the presence of an end restraint, the load in the bar is carried differently at the free and fix ends. It also shows how the load is carried in between the extreme points in the problem.
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5.12 Miscellaneous problems: 5.12.1 Problems for Torsion on Circular Section:
Problem 1: A solid shaft is to transmit a torque of 10000kg-m. If shearing stress is not to exceed 450kg/cm2. Find the maximum diameter of shaft Given Data: T=10000kg-m = 1000000kg-cm f q = 450kg / cm 2
D = Diameter of shaft T=
π 16
f s D3
1000000 = D3 =
π 16
× 450 D 3
1000000 × 16 π × 450
D = 22.45cm .
Problem 2: A bar of material BSL168TS11 of diameter 100mm is subjected to an torque of 30 × 10 6 N − mm. Given Data: t 2 = 370 Mpa q 2 = 370 × 0.6 = 222 Mpa
f t = 415Mpa f q = 415 × 0.6 = 249Mpa r = D / 2 = 50mm
J=
π 32
D4
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Limit stress analysis: Allowable torsional stress = 222Mpa. Actual torsional stress
= =
T ×r J
T ×r
π 32
Reserve Factor
=
× 0.4
30 × 10 6 × 50 × 32 = 152.78Mpa. π × 100 4
= 222/152.78=1.45
Ultimate stress analysis: Allowable torsional stress = 249Mpa. Actual torsional stress
= factorofsafety × Tr J = 1.5 × 152.78 = 229.17
Reserve Factor
= 249/229.17 = 1.1
Problem 3: A figure illustrates an aileron control surfaces consisting of circular torque tube outer diameter of 32mm supported on 3 hinge brackets and with control rod fitting attached to the torque tube above the centre support bracket. Find the maximum torsional shearing stress in the tube if the air load on the aileron is shown also calculate the angle of twist of the tube between horn section and end of aileron. A load on strip of aileron 1mm wide = 729.6KN/mm2.
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To find w. 1 76w + [ w × 305] = 729.6 KN / mm 2 2
w = 3KN / mm 2 P1 = 3 × 76 = 228 KN
∴Moment about hinge point = 228 × (76 / 2) + 457 × (1 / 3)305 = 37797.67 Torsional moment = 37848.5 N-mm Max. Torque at centre of aileron = 37848.5 × 661 = 25017858.5 ≅ 25 × 10 6 N − mm . J=
π (D 4 − d 4 ) π (32 4 − 29.5 4 ) 32 D
τ max = θ=
=
32 × 32
= 893.5
Tr 25017858.5 × 160 = = 4479974.5 J 893.5
TL 25Mpa × 661 180 = × = 0.013 deg rees. GJ π 807.6 × 10 6
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5.12.2 Problems for Torsion in Non Circular Section:
Problem 1: A beam of length .300mm having rectangular cross section as shown in diagram is subjected to a limit torque of 1000000 N-mm. Calculate maximum shear stress and angle of twist at that point material BSL 168T b511.
τ max =
T (3a − 1.8b) 8a 2 b 2
τ max =
1000000(45 − 18) 8 × 15 2 × 10 2
= 150Mpa. Allowable stress = 0.6t2 = 0.6 × 370 = 222 kg/m2 Reserve Factor = 222/150 = 1.48 Ultimate stress analysis Ultimate stress = 1.5× 150 = 225Mpa Allowable ultimate stress = 0.6× UTS = 0.6× 415 = 249Mpa Reserve Factor = 249/225 = 1.12
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5.12.3 Problems for Torsion in Open Section Composed of Thin plates:
Problem 1: An beam of angle section and length of 500mm as shown in the figure is subjected to an torque of 150000 N-mm. Calculate τ max and twist of the section material BSL168T6511.
τ max = θ=
T dbt 2
T φbt 3G
α , φ value depends on b/t value b / t = 150/15 = 15 for b/t > 10 ∴τ max =
α , φ = 0.3333or1 / 3
3T 3 × 100000 = = 200 N − mm. bt 2 15 × 10 2
For BSL-168 T6511 E = 70 × 10 3 Mpa ,
G=
Poisson ratio = 0.3
m(70 × 10 3 ) 0.3(70 × 10 3 ) = = 8076 Mpa. 2(m + 1) 2(1 + 0.3)
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θ=
3TL 3 × 100000 × 500 180 = × bt 3G 150 × 10 3 × 8076 π
= 0.0007 deg.
Problem 2: Calculate shear stress and twist for the given section to an torque of 1000000 Nmm. Find τ max and twist.
Stress on leg 1.
τ b1 =
3Tt1 3Tt1 = 3 3 b1t1 + b2 t 23 + b3 t 33 ∑ bt
=
3 × 100000 × 5 = 53.3 25(100 + 75 + 50)
τ b2 =
3 × 100000 × 5 = 53.3 25(100 + 75 + 50)
Unit twist
θ=
3T 3 × 1000000 180 = × = 0.0756 deg . Gt (b1 + b2 + b3 ) 8076 × 125(225) π 3
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5.12.4Problems for Torsion on Thin Walled Closed Section (Symmetrical Section):
Problem 1: Calculate the shear stress in the member of the section shown in the diagram (Berth Batho Method).
Area of section 1. =
πr 2 2
=
π × 25 2
= 39.3
2. = b 2 = 100 3. = b 2 = 100 Line integral a= ∫ a10 =
ds t
π × 10 2 × 0.025
= 629
a12 =
10 = 200 0.05
a20 =
20 = 667 0.03
a23 =
10 = 333 0.03
a30 =
20 10 + = 917 0.03 0.04
Twist
θt =
q ds ∫ 2 AG t
for cell 1.
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2Gθ 1 =
1 [629q1 + 200q1 − 200q 2 ] 39.3
eq.1
for cell 2. 2Gθ 2 =
1 [200q 2 − 200q1 + 6679q 2 + 333q 2 −333q3 ] 100
eq.2
for cell 3. 2Gθ 3 =
1 [333q3 − 333q 2 + 917q3 ] 100
eq.3
Considering θ is equal through out i.e. θ 1 = θ 2 = θ 3 Equating equation 1 and 2,2 and 3,3 and 1. 1 1 [629q 1 +200q1 − 200q 2 ] = [200q 2 + 200q1 + 667q 2 + 333q 2 − 333q3 ] 39.3 100 1 1 [200q 2 − 200q1 + 669q 2 + 333q 2 − 333q3 ] = [333q3 − 333q 2 + 917q3 ] 100 100 1 1 [333q3 − 333q 2 + −917q3 ] = [629q1 + 200q1 − 200q 2 ] 100 39.3 16q1 + 5q1 − 5q 2 = 2q 2 − 2q1 + 6.67 q 2 + 3.37q 2 − 3.33q3 − 200q1 + 1200q 2 − 33q3 = 1250q3 − 333q 2 12.5q3 − 3.3q 2 = 16q1 + 8q1 − 5q 2 comparing above 3 equation we get q1 = 143.4lb / in q 2 = 234.1lb / in
q3 = 208.8lb / in T = 2 A1 q1 + 2 A2 q 2 + 2 A3 q3 = 2 × 39.3 × 143.4 + 2 × 234 × 100 + 2 × 208.8 × 100 = 100000 lb-in.
≈ 11300000 N-mm.
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Stress in members: m10 = q1 / t10 = 143.5 / 0.025 = 5720lb / in 2 = 41.6 N / mm 2 m12 = (q1 − q 2 ) / t12 = (234 − 143) / 0.05 = 1812lb / in 2 = 13.2 N / mm 2 m20 = q 2 / t 20 = 234 / 0.03 = 7800lb / in 2 = 56.6 N / mm 2 m23 = q3 − q 2 / t 23 = (234 − 208.8) / 0.03 = 840lb / in 2 = 6.1N / mm 2 m30 = q3 / t 30 = 208.8 / 0.03 = 6900lb / in 2 = 50.6 N / mm 2 Problem 2: Determine the internal shear flow of the system for the two cell tube shown in figure. When subjected to an torque of 20000lb-in. (Method of successive connection)
C.o.factor cell 1 and cell 2
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Co.(1− 2) =
Co( 2−1) = q1 =
q2 =
10 0.05 ( L / t ) web(1 − 2) = = 0.40 10 30 ∑( L t )cell 2 + 0.05 0.1
( L t ) web(2 − 1) 10 0.05 = = 0.25 ∑( L t )cell1 40 0.05
2 A1 2 × 100 = = 0.25 ∑( L t )cell1 40 0.05
2 A2 2 × 100 = = 0.4 10 30 ∑( L t )cell 2 + 0.05 0.1
q1I = (CoF ) 2−1 × q 2 = 0.25 × 0.4 = 0.1 q 2I = (CoF )1− 2 × q1 = 0.4 × 0.25 = 0.1 q1II = (CoF ) 2−1 × q1I = 0.25 × 0.1 = 0.025 q 2II = (CoF )1− 2 × q 2I = 0.4 × 0.1 = 0.04 q1III = (CoF ) 2−1 × q 2II = 0.25 × 0.04 = 0.01 q 2III = (CoF )1− 2 × q1II = 0.025 × 0.4 = 0.01 q1IV = (CoF ) 2−1 × q 2III = 0.25 × 0.01 = 0.0025 q 2IV = (CoF )1−2 × q1III = 0.40 × 0.01 = 0.004 Final q 2 = q1 + q1I + q1II + q1III + q1IV = 0.25+0.1+0.025+0.01+0.0025=0.3875 Final q 2 = q 2 + q 2I + q 2II + q 2III + q 2IV = 0.4+0.1+0.04+0.01+0.004=0.554 2 A1 q1 = 2 × 100 × 0.3875 = 77.5
2 A2 q 2 = 2 × 100 × 0.554 = 110.8 T = 77.5+110.8 = 188.3 q1 for T = 20000 =
0.3875 × 20000 = 41.2 188.3
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q 2 for T = 20000 =
0.554 × 20000 = 58.5 188.3
Shearstress1 =
Shearflow 41.2 = = 824 Thickness 0.05
Shearstress 2 =
Shearflow 58.5 = = 585 . Thickness 0.1
Problem 3: Find the shear flows in all webs of box beam shown in figure .The section is symmetrical about horizontal axis through centroid Moment of inertia about neutral axis I = 4 × 0.5 × 5 2 + 2 × 1 × 5 2 = 100 Difference in bending stress between two cross section unit distance a part is Vy 10000 × 5 = = 500 I 100 To obtain shear flow moment about f is taken Area3 =
Area1 =
1 × 10 × 10 = 50 2
π × 10 2 8
+
1 × 20 × 10 = 139.25 2
Areea2 = 239.25-189.25=50
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Taking moment about 1000 × 8 = 2 × 139.25 × q 0 + 2 × 50(90 − 250) + 2 × 50(q0 − 500) 8000 = 278.5q 0 + 100q0 − 25000 + 100q0 − 25000 135000 = 178.5q 0 q0 =
135000 = 324 478.5
5.12.5 Problems for Torsion in Thin walled Sections (Unsymmetrical conditions):
Problem 1: Find the shear flow in the webs of the beam shown in the diagram.
I x = 2 × 3 × 52 + 2 × 1× 52 = 200 I y = 2 × 3 × 10 2 + 2 × 1 × 10 2
= 800 I xy = 1 × 2 × 10 × 5 − 2 × 3 × 10 × 5 = -200
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Vz = 10000 Vx = 4000 Change in bending moment Δf b =
=
V z I xz − V x I x V I − Vz I z x + x xz z 2 I x I z − I xy2 I x I z − I xz − 800000 − 8000000 10000(−200) − 4000(200) x+ z 2 120000 200 × 800 − 200
Δf b = −23.3x − 73.33 z
Loading in flanges =(-23.33x-73.33z)Af Flange
Flange Area
Coordinates x
-23.33x
-73.33z
Δf b
Loading on
z
flanges a
3
-10
5
233.3
-366.7
-133.3
-400
b
1
10
5
-233.3
-366.7
-600
-600
c
3
10
-5
-233.3
366.7
133.3
400
c
1
-10
-5
233.3
366.7
600
600
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Taking moments about centroid 2 × 50 × q 0 + 2 × 50 × (q 0 − 400) + 2 × 50 × (q 0 − 1000) + 2 × 50 × (q 0 − 600) 400q 0 − 40000 − 100000 − 60000 = 0 400q 0 = 200000 q0 =
200000 = 500 400
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6. TRANSVERSE SHEAR LOADING OF BEAMS WITH SOLID OR OPEN CROSS SECTIONS:
The topics discussed in this section deal with transverse shear loading of beams that have (a) solid; or (b) open cross sections. In the case of (b) the beam could also be either thin-walled or stiffened as in skin-stringer sections. Cross-sections with one, two, or no axis of symmetry are analyzed for shear flow distribution, shear center location, and maximum shear stress. Topics covered: Flexural shear stress, shear flow, and shear center Restrictions: The beam sections considered in this chapter are all homogeneous and elastic.
6.1 Introduction 6.2 Shear Center 6.3 Flexural Shear Stress and Shear Flow 6.4 Shear Flow Analysis for Symmetric Beams 6.5 Shear Flow Analysis for Unsymmetric Beams 6.6 Analysis of Beams with Constant Shear-Flow Webs (i.e., skin-stringer type sections)
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6.1 Introduction:
In previous discussion on pure bending of beams the transverse shear forces and corresponding shear stresses were absent. That absence simplified the flexure problem as the applied loads (i.e., bending moments) only produced flexural normal stresses. In this chapter our focus is on a more general loading condition - one that involves both bending moment and transverse shear force. The existence of the latter indicates that we no longer have pure bending problem. Recall from mechanics of materials that in beam bending transverse shear forces and bending moments are related to each other. This relationship was displayed in drawing the shear and moment diagrams for various beams and loading conditions. We first are going to revisit the problems that we encountered in mechanics of materials course. In order to do that we must recall problem restrictions and the method of analysis that was used in the solution of those problems.
Restrictions:
1. Shear stress at every point in the beam must be less than the elastic limit of the material in shear. 2. Normal stress at every point in the beam must be less than the elastic limit of the material in tension and in compression. 3. Beam's cross section must contain at least one axis of symmetry. 4. The applied transverse (or lateral) force(s) at every point on the beam must pass through the elastic axis of the beam. Recall that elastic axis is a line connecting cross-sectional shear centers of the beam. Since shear center always falls on the cross-sectional axis of symmetry, to assure the previous statement is satisfied, at every point the transverse force is applied along the cross-sectional axis of symmetry. 5. The length of the beam must be much longer than its cross sectional dimensions. 6. The beam's cross section must be uniform along its length.
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6.2 Shear Center:
Consider the figure below showing a cantilever beam with a transverse force at the tip. Under the action of this load, the beam may twist as it bends. It is the line of action of the lateral force that is responsible for this bend-twist coupling. If the line of action of the force passes through the Shear Center of the beam section, then the beam will only bend without any twist. Otherwise, twist will accompany bending. The shear center is in fact the centroid of the internal shear force system. Depending on the beam's cross-sectional shape along its length, the location of shear center may vary from section to section. A line connecting all the shear centers is called the elastic axis of the beam. When a beam is under the action of a more general lateral load system, then to prevent the beam from twisting, the load must be centered along the elastic axis of the beam.
The two following points facilitate the determination of the shear center location. 1. The shear center always falls on a cross-sectional axis of symmetry. 2. If the cross section contains two axes of symmetry, then the shear center is located at their intersection. Notice that this is the only case where shear center and centroid coincide.
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If the cross section contains no axis of symmetry or only one axis of symmetry, the determination of the exact location of shear center requires a more detailed analysis which we will discuss in following sections. 6.3 Flexural Shear Stress and Shear Flow: Derivation:
Consider a cantilever beam with a symmetric cross section subjected to a lateral (or transverse) force P at the tip. The shear and normal stresses induced by this force are required to be below the corresponding elastic limits of the material.
Next, we consider a transverse section located at some distance x from the free end of the beam as shown above. The bending moment at this section is related to the normal stress according to the equation
In order to go any further, we need to know how
x varies along the section. Recall from
previous chapter that if the section is in the state of pure bending, then the normal strain vaires linearly. Furthermore, if the section is in elastic condition and the material is linearly elastic, then according to Hooke's law, normal stress would vary linearly as well. With that in mind, we need to determine whether the normal strain varies linearly in the presence of transverse shear force. Without getting too deep into the theory of elasticity, it suffices to say that if the beam is long compared to its cross section, then the normal strain does vary linearly. On the other hand, if the beam is short, then normal strain variation is in fact nonlinear when transverse shear force is present. So we either have to limit our analysis to long beams or assume that the normal strain variation is linear regardless of the length of the beam. With the requirement of elastic condition and the
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use of linearly elastic materials, the general elastic bending formula described in the previous chapter can be used. In the presence of one moment component and zero product of inertia, the normal stress equation reduces to
The distance y is measured from the centroidal z axis (which in this case coincides with the neutral axis), Iz is the moment of inertia of the section about the z axis. Relating the bending moment to the applied shear force P gives
The maximum normal stress, at section x, will occur at the farthest distance from the neutral axis. In this case the stress distribution indicates that distance to be to the top surface of the section as shown below.
To examine the shear force and associated shear stress created as a result of transverse force P, let's consider a beam segment highlighted in the figure below. While the left edge of the beam segment is free from any force in the x direction, the right edge is not due to the presence of
x.
The question is how can equilibrium be maintained in this case? This points to the existance of a horizontal shear force H along the bottom surface (longitudinal surface) of the highlighted segment. The important point to notice first is that a transverse shear force results in the creation of a longitudinal shear force.
Now the equation for equilibrium of forces in the x direction is written as
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where A* represents the area of the beam section from y1 to c. Making appropriate substitutions and simplifying the resulting equation gives
The integral term represents a geometric quantity referred to as the moment of area,and is denoted by Qz. The subscript z indicates that the moment of area is obtained about the z axis. The equation for H can be written as
This equation indicates that the longitudinal shear force is an explicit function of x and an implicit function of y through the moment of area. At the tip section x is zero, hence H is zero for any value y. However, for any positive value of x, the variation in H depends on the location of y1. It can be shown that for y1 = 0, H reaches its maximum value whereas at y1 corresponding to the top and bottom surfaces, H goes to zero. Therefore, the longitudinal shear force H varies linearly with respect to x and quadratically with respect to y. Now, divide both sides of the equation by x
Note that P times x represents the moment Mz at position x, and that in the limit as x approaches zero, Px/x simply represents the change in the bending moment which is simply equal to the transverse shear force at that location. Therefore, we can rewrite this equation as
The ratio of H over x represents the quantity known as the shear flow denoted by q. In this case, shear flow is acting along the longitudinal surface located at distance y1. For the problem described here, the shear flow can be written as
At a given position x, q varies according to the variation of Q. The equation for Q reveals that at the NA corresponding to y1 = 0, Q reaches its maximum value, and at y1 corresponding to the top and bottom surfaces, it will be zero. To obtain the flexural shear stress along the longitudinal surface corresponding to y = y1, we simply divide q by the width of the longitudinal surface. If this dimension is denoted by t, then
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The subscripts on
indicates that the longitudinal surface has a normal vector in the y direction
and that the shear stress component is in the direction of x axis. Since shear stress in the longitudinal plane must be equal to that in the transverse plane, then the two subscripts can change places without any change in the right hand side of the equation. The equation above is known as the average flexural shear stress formula. Although the value of shear flow is always maximum at the neutral axis location, we cannot conclude that shear stress will also be maximum there. The reason is that t may vary from one point to another, and that it is possible for a point other than the neutral axis to have the maximum shear stress. In order to calculate the maximum shear stress, the following relationship should always be followed.
Restrictions:
Before we can apply the average flexural shear stress formula, we must consider the restrictions that apply to this equation: 1.
The
section
has
to
be
homogeneous
(made
of
a
single
material).
2. The section has to have an axis of symmetry, therefore, product of inertia is zero. 3. The shear force V passes through the shear center of the section, and is parallel to one of the two
principal
centroidal
axes.
4. The shear stress at every point is below the corresponding elastic limit of the material. Pop Question:
What if the longitudinal surface we considered in above derivation was not parallel to z axis? What would change in that case? Answer:
To answer these questions, consider the figure shown below.
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In the case of horizontal longitudinal surface, we will be calculating the yx or xy component of shear stress. To calculate Qz we use the area above the longitudinal surface. t in this case is equal to the distance from the left edge to the right edge of the top section. In the case of the vertical longitudinal surface, we will be calculating a different component of shear stress, i.e., is the same as
xz
zx
which
on the transverse surface. In this case Qz can be found by considering the area
to the right side of the longitudinal surface, and t would be equal to the flange thickness. This difference is demonstrated in the following discussion: If V is along z direction, then the moment of inertia is that about the perpendicular axis or y axis in this case. For Vz loading, the flexural shear stress formula can be written as
Therefore, at a given point on the cross section we can calculate two components of shear stress depending on the orientation of the longitudinal surface. Generally speaking, one component is always much larger than the other thus, that would be the one of most interest in the analysis. Table below gives equations for maximum shear stress for specific cross sections. The distance e is defined as the distance from the neutral axis to the point of maximum shear stress. The shear force V is normal to the NA and it acts along the vertical axis of symmetry, hence through the shear center. The maximum shear stress is given in terms of the ratio V/A.
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Shear Stress Variation:
If we obtain the equation that describes the shear stress as a function of position on the cross section, it would be easy to find how shear stress varies from point to point as shown in the example figure below.
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Knowing that the ratio V/I is constant for a given section, we need to only concentrate on the variation described by Q/t. For example, if we follow the derivation steps, we find the average flexural shear stress variation for a beam with rectangular cross section, as shown above, will be parabolic according to the equation
Shear Flow Variation:
The shear flow distribution at a given cross section is determined by writing the equation VQ/I, which is basically the shear stress multiplied by 't'. The shear flow distribution calculation can be seen in the example problems given at the bottom of this page. Here are some examples of shear flow diagrams for sections with at least one axis of symmetry. Each of these cases show the shear flow when the shearing load passes though the shear center. In each of these shear flow diagrams, the maximum shear flow occurs at the neutral axis, which passes through the centroid of the section. This will also be the location of the maximum shear stress as the wall thickness is constant. Also notice how the shear load is perpendicular to the NA.
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Example 1 A cantilever beam with T cross section is loaded at the tip by a vertical force. Determine at section n-n: •
(a) Maximun compressive stress.
•
(b) Maximum tensile stress.
•
(c) Maximum shear stress.
Note: Stresses induced by the tip load do not exceed the elastic limits of the material used.
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EQUATIONS USED
Solution Using line b-b as reference, with positive direction downward, the y centroidal coordinate is determined as
Since the beam is elastic, the netrual axis will pass through its cross-sectional centroid. The moment that force P produces is about the horizontal axis, therefore, neutral axis coincides with the horizontal centroidal axis in this problem as shown in the figure below. The applied load puts the portion above the NA in tension and the one below in compression.
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The moment of inertia about the horizontal centroidal axis is the only one that is needed to be calculated here, and is determined as
(a) The maximum compressive stress will occur at the farthest point from the NA, on the compression side. At section n-n the bending moment is
The corresponding compressive stress will be
(b) The maximum tensile stress will occur at the farthest point from the NA on the tension side.
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(c) As was explained earlier, the shear stress should be viewed as the ratio (V/I) times the ratio (Q/t). At a given section along the length of the beam (V/I) is constant. This means that the maximum shear stress will occur at the point on the cross section which results in the largest (Q/t) value. Without any exception, the first moment of area, Q is always maximum at the NA. Since in this cross section the width t is also minimum at the NA, we can conclude that the maximum shear stress will occur there. Side Note:
Based on the above discussion, it is quite possible for the maximum shear stress to occur at a point other than the NA. You must always be aware of this fact when analyzing a beam under transverse shear loads. To determine Q, cut the cross section horizontally at the NA and choose either the top or the bottom portion, it doesn't matter which because the same answer will result using either portion. Therefore, it makes sense to pick the portion with the simplest shape to calculate Q, in this case the bottom portion.
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The maximum shear stress on the transverse plane n-n is found to be
The shear stress distribution over the entire cross section was not asked for in the problem statement, but it is shown below for completeness.
Notice that the shear stress distribution shows only one component of shear stress (i.e., the vertical component). While this component is the dominant one along the vertical flange, it is the smallest one along the horizontal flanges. To determine the dominant component of shear stress along the top flanges (i.e., horizontal component), we must make a vertical cut along either the left or the right flange (due to symmetry) and calculate Q/t along the flange. The variation of shear stress will be linear with the largest flange shear stress occuring near the junction of the three flanges. We will see this difference in later example problems. However, in this problem the overall maximum shear stress at section n-n is the one found earlier. It is also important to point out a couple of things about the shear stress distribution shown. First, notice that the variation of shear stress is quadratic. This is governed solely by the equation for
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Q. Second, notice the discontinuity at the junction. This is because as we go from top to bottom surface, the thickness suddenly changes at the junction. The sharp reduction in thickness is responsible for the sharp increase in the shear stress.
Example 2 A beam with the cross section shown is acted upon in its plane of symmetry by a 3 kN force. For points A and B located at section n-n, shown in the figure below, determine the average shear stress. The moment of inertia and centroid location are given in the figure. Note: Stresses induced by the load do not exceed the elastic limits of the corresponding material.
Equation Used
Solution First the vertical shear 'V' at n-n is determined from the free body diagram shown below
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The moment of area Q at point A, for a horizontal cut is given as
Thickness 't' is defined by the width of the 'cut' which in this case is 100 mm. Hence, the average shear stress at point A is
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Notice this is the vertical component of shear stress at point A. The moment of area Q at point B, for a horizontal cut is given as
Point B is at the intersection of the web and the flange. The width of the cut is 20 mm, making the shear stress
The shear stress at a point is mostly dependent on the thickness of the member at that point. Usually, in thinner members more shear stress will be present as demonstrated by this example. Shear Stress Components τYx And τZx
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The T section shown below is used to explain the difference between the two components of shear stress, i.e.,
yx
and
zx.
Let's assume that the transverse shear force is along the y axis
(passing through the shear center of the section) and that Vy = 1000 lb.
Solution: With the shear force along the y axis, the average flexural shear stress formula for the two components of interest is written as
Although the general form of the equation is the same for both components, there is a difference in the way Qz and t are found in each case. With Vy given, we proceed with the calculation of centroidal position.
The centroid is located at 3.25 in. from the bottom edge of the section. The moment of inertia about the z axis is found as
To calculate the
yx
component of shear stress, the longitudinal surface has to have a normal in
the y direction. The intersection of this longitudinal surface and the transverse surface is shown by a solid blue line in the figure below.
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Depending on the position of the longitudinal surface along the y axis, the corresponding value for Qz is found as
The equation for Qz is quadratic. Qzmax = 5.28 in3, and it occurs at the NA (y2 = 0). The variation of Qz is shown below.
With the variation of Qz known,
yx is
found as
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Keep in mind that t is the width of the longitudinal surface. Along -3.25 < y < 0.75, t = 1 in. and along 0.75 < y < 1.75, t = 4 in. As is shown in the figure above, there is a discontinuity in the shear stress at y = 0.75 in. due to the discontinuity in t. The maximum value of
yx
is found to be 290.64 psi and it occurs at the NA. The values of
surface are equal in magnitude to the corresponding It is crucial to pay attention to
yx
xy
yx
over the longitudinal
values on the transverse surface.
variation at the junction. The value of 68.81 psi found in the
analysis refers to the shear stress in the horizontal flanges just above the junction point. As there can be no shear stress on a free surface, the shear stress points B and C but not at point A. As indicated by its plot,
yx
and its counterpart yx
xy
are zero at
is more significant in the vertical
flange as opposed to horizontal flanges. To calculate the
zx
component of shear stress, the longitudinal surface has to have a normal in
the z direction. The intersection of this longitudinal surface and the transverse surface is shown by a solid blue line in the figure below.
Depending on the position of the longitudinal surface along the z axis, the corresponding value for Qz is found as
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The equation for Qz is linear in this case. At the section intersecting the corner of the horizontal and vertical flanges i.e., zz = +/-0.5, Qz = 1.875 in3. The variation of Qz is shown below.
The blue dashed line gives the Qz variation at the junction of the horizontal and vertical flanges. This value is used to find the shear flow at that location. With the variation of Qz known,
zx is
found as
Keep in mind that t is the width of the longitudinal surface. Along -2 < z < -0.5, t = 1 in. As is shown in the figure above,
zx
reaches its maximum value of 103.21 psi in the horizontal flanges
at the junction with the vertical flange. In the vertical flange this component of shear stress is zero as shown below.
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Notice that there is a counterpart shear stress ( zx
xz)
on the transverse surface. Also in this case
is more significant in the horizontal flanges as opposed to the vertical flange.
key observation:
At any point on the boundary the resultant shear stress is parallel to the boundary. This fact helps us to sketch the figures shown below as a general depiction of shear flow and shear stress variations in this example.
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Derivation of the average flexural shear stress equation for a rectangular section under a vertical transverse shear force:
We start with the formula for the average flexural shear stress
where in this case 'V' is the vertical shear force at the section, 'I' is the moment of inertia about the horizontal centroidal axis, 't' is the width of the section, and 'Q' is the first moment of area about the horizontal centroidal axis. For this derivation we need to isolate a portion of the section. We make a horizontal cut at a distance 'y' from the horizontal centroidal axis (in this case also the neutral axis) as shown in the figure below.
'V' force is usually a known quantity so we do the derivation in terms of V. The moment of inertia about the horizontal centroidal axis of the rectangular cross section is
Next we write the equation for the moment of area of the shaded portion.
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Substituting these equations into the average flexural shear stress equation we get
The cross sectional area of the beam is 'A=2bc', this gives the final form of the equation
The plot of this shear stress variation is shown below.
For this narrow beam, the maximum shear stress occurs at the neutral axis. Also, at the very top and bottom the shear stress is zero.
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6.4 Shear Flow Analysis for Symmetric Beams: Shear Flow Calculation:
To calculate the shear flow over a section of interest we must have the value of transverse shear force V that is acting along a principal axis. This force is either given or should be obtained from the shear diagram. Then we need to have the moment of inertia about an axis that is perpendicular to the direction of the transverse shear force. For example, if V is along y, we need to have Iz, or if V is along z, we need to calculate Iy. With Vz/Iy or Vy/Iz known, we calculate the first moment of area, Q. If V is along the y direction, we need to calculate Q about the z axis. To do this a segment of the cross section is isolated from the rest, and its moment about the z axis is calculated. The way we isolate a segment is by cutting it perpendicular to its thickness. We will see how this is done in the example problems at the end of this section. An example of a shear flow diagram is shown below.
Notice that in this example, the transverse shear load is in the vertical direction. Thus, the moment of inertia about the horizontal centroidal axis is used for the calculation of shear flow. The shear flows along the top flange and the web are calculated as
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The direction of shear flow has to be consistent with that of the resultant shear force, in this case Vy. We can also calculate the shear force acting in each member. To do this, we simply integrate the shear flow along each member.
Key Observation:
If we examine the internal shear force distribution we observe the following facts. Since the resultant shear force is in vertical direction, the two flange forces (in horizontal direction) will add up to zero as they should. However, if we compare the force in the vertical web, F2 to Vy we find that they are not the same. Although this may indicate that force equilibrium is violated, the fact is otherwise. We know from previous discussion that shear stress at a given point is represented in terms of its two perpendicular components. We also said that usually one component is much larger than the other. In this case the vertical component of shear stress in top and bottom flanges is much smaller than its horizontal component but not necessarily zero. If we were to consider the vertical component of shear stress in the two flanges and calculate the corresponding shear force, we will find that they added with F2 will be equal to Vy. The reason we don't go through the trouble of calculating these forces is because they are much smaller than F2. In fact we can see that if b > > t, then F2 is > 90% of Vy. It is important to know this fact when doing bending analysis under transverse loads. Shear Center Calculation:
The shear center is found using moment equilibrium. We show the resultant transverse shear force acting at the shear center which is at some distance e from the point of reference, usually the centroid of the cross section. We then write the moment produced by the resultant shear force V, set it equal to the sum of moments produced by individual internal force components, F1, F2 and F3, and solve for the unknown distance, ez in this case. Paying attention to the direction of each moment, we can write
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Beam Sections Not Loaded Through the Shear Center:
If the applied shear force does not pass through the shear center, it will force the beam to twist as it bends. This eccentricity produces a torque, that will cause an additional shear flow and shear stress. The analysis used for torsion of beams with open cross sections (I.4) can be used here to find the constant shear flow and corresponding shear stress at a desired point on the cross section. In analysis of such sections, the shear force is replaced by an equivalent force-couple at the shear center. The final shear stress diagram will be the superposition of 1. the shear stress with the shear force passing through the shear center,and 2. the shear stress induced by the associated torque about the shear center. An example of this kind of loading is shown below. Note that in this case, the maximum shear stress occurs at a point on the neutral axis which is on the left edge of the vertical flange.
Shear Force in Fasteners:
In many applications, beam sections consist of several pieces of material that are attached together in a number ways: bolts, rivets, nails, glue, weld, etc. In such so called built-up sections we are interested in knowing the amount of shear stress and the resulting shear force at the cross section of fasteners or over the glued surface.
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The figure shown below gives an example of two rectangular members that are attached by means of mechanical fasteners. In this case, we want to know the amount of shear stress as well as shear force carried by each fastener. The fasteners are spaced evenly at a distance of s. Each fastener has a cross-sectional area denoted by Af.
Note that the surface of contact or joint is treated as a frictionless surface. Therefore, the shear flow is carried entirely by the fasteners. If at a given section, there are more than one fastener, the shear flow will maintain the same value, but the shear force and shear stress will change depending on the number and size of fasteners used. For example, if at a given section there are two identical fasteners as shown below, then the force in each is found as shown below.
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Example Problems Example 1 For the beam and loading shown, determine: •
(a) the location and magnitude of the maximum transverse shear force 'Vmax',
•
(b) the shear flow 'q' distribution due to 'Vmax',
•
(c) the 'x' coordinate of the shear center measured from the centroid,
•
(d) the maximun shear stress and its location on the cross section.
Stresses induced by the loads do not exceed the elastic limits of the material. Assume the transverse force passes through the shear center of the beam at every cross section.
Equations Used
Solution
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(a) To find the maximum transvere shear, the shear at each section along the beam between sections A and D must be found. This means that we need to draw the shear diagram. To do this we have to take the following steps: From the free body diagram, the reactions at A and D are
The transverse shear variation along the section with distributed force can be found by using the integral equation
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The resulting shear diagram shown below indicates that the maximum shear force is Vmax = 55.8 kN and it occurs along portion CD.
(b) For the shear flow analysis we must consider a beam section with maximum shear force. In this case any section along portion CD will satisfy this requirement. To obtain the shear flow, first draw the picture of the cross section. Since the moment of inertia about the horizontal centroidal axis (perpendicular to the direction of V) is already given, we can go directly to the moment of area (Q) calculation. Begin at the free end of the top right flange where we know the shear flow is zero, and consider a portion of length s as shown in the figure below.
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Write the equation for the shear flow in terms of the moment of area of the portion of length s.
Notice that the moment of inertia is that of the entire I cross section about the horizontal centroidal axis, which is perpendicular to the direction of V. Due to symmetry about the vertical centroidal axis, the shear flow in the top left flange will be the same as the flow in the top right (same magnitude, opposite direction). Also due to symmetry about the horizontal centroidal axis, the shear flow in bottom flanges will be the same as the ones on top (same magnitude, opposite direction).
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In the web, the shear flow doesn't start from zero, like it did in the flanges. This is because the starting point of the web is a junction and not a free end. The starting shear flow for the web is shown in the figure below.
Using the fluid flow analogy the flow coming into the web-flange junction from flanges must leave the junction through the web. To determine the shear flow variation along the web, isolate a section of length w along the web as shown in the figure below.
Write the shear flow equation by writing the starting value which is 128.96 kN/m, and adding to it the contribution of the web. V/I ratio stays the same (as it should), and all that remains to be calculated is the moment of area of length w along the web.
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The complete shear flow diagram is shown below. The shear force in the web is found by calculating the area under the q diagram. This gives a value of 54.626 kN which represents approximately 98% of the total shear force which is 55.8 kN. The rest of it (not shown), as explained before, will be carried by the top and bottom flanges.
(c) Since this is a doubly symmetric cross section, the shear center is located at the intersection of the axes of symmetry which is the centroid. (d) The maximum shear stress will occur at the point on the cross section where the ratio (q/t) is maximum. In this problem that point is located on the neutral axis.
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In summary, the maximum shear stress in the flange is 6.829 MPa and occurs at the intersection of the flange and the web (i.e., s = 61 mm). The minimum shear stress in the web is 21.493 MPa and the maximum shear stress is 30.511 MPa. Example 2 For the beam and loading shown, determine: •
(a) the location and magnitude of the maximum transverse shear force 'Vmax',
•
(b) the shear flow 'q' distribution due to 'Vmax',
•
(c) the 'x' coordinate of the shear center measured from the centroid,
•
(d) the maximun shear stress and its location on the cross section.
Stresses induced by the load do not exceed the elastic limits of the material. The transverse force passes through the shear center of the beam at every cross section.
Equation Used
271
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Solution (a) To find the maximum transvere shear, the shear at each section along the beam between sections A and D must be found. This means that we need to draw the shear diagram. To do this we have to take the following steps: From the free body diagram, the reactions at A and D are
The transverse shear can be found by using the integral equation from mechanics of materials
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The resulting shear diagram shown below indicates that the maximum shear force is Vmax = 55.8 kN and it occurs along portion CD.
Notice that the answer to part (a) is the same as the one found in Example 1. This is because the beam supports and loading condition dictate the transverse shear force variation along the length of the beam. The shape of the beam cross section has no effect on this variation.
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(b) For the shear flow calculation, with the shear force and the moment of inertia about the horizontal centroidal axis known, we again begin from a free end where we know the shear flow is zero. In this case, we start from the free end of the top right flange. Note that we could have chosen any of the three free ends of the cross section. The final variation would come out to be exactly the same regardless of which flange tip we started from.
For the segment of length s, write the moment of area equation and put it in the shear flow equation as given below.
Due to symmetry about the vertical centroidal axis, the shear flow in the top left flange will be the same as the flow in the top right (same magnitude, opposite direction).
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For the web, begin at the web-flange junction. What pours into the web is what enters the junction from each flange.
Consider the segment of length w along the web. Write the equation for its moment of area, and put it in the shear flow equation as in Example 1.
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The final shear flow diagram is shown below. Note the linear variations in horizontal flanges and quadratic variation in the vertical flange.
If we integrate the shear flow equation for the vertical flange we obtain a shear force of 55.02 kN. This is approximately 99% of the total shear force of 55.8 kN. The remaining 1% is carried by the horizontal flanges. (c) Do to symmetry about the vertical axis, the shear center falls along that axis, but not at the centroid as was the case with doubly symmetric I cross section of beam in Example 1. With the shear center falling on the vertical axis of symmetry its x-coordinate measured from the centroid is obviously zero.
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(d) The maximum shear stress will occur at the point on the cross section where (q/t) is maximum. In this case maximum shear stress occurs at the neutral axis and its value is
To summarize, the maximum shear stress in the horizontal flanges occur at their respective junction with the vertical flange. This shear stress is 9.02 MPa. The minimum shear stress in the vertical flange is 28.39 MPa and the maximum shear stress is 35.99 MPa. EXAMPLE 3 For the beam and loading shown, determine: •
(a) the location and magnitude of the maximum transverse shear force 'Vmax',
•
(b) the shear flow 'q' distribution due the 'Vmax',
•
(c) the 'x' coordinate of the shear center measured from the centroid,
•
(d) the maximun shear stress and its location on the cross section.
Stresses induced by the load do not exceed the elastic limits of the material. NOTE:In this problem the applied transverse shear force passes through the centroid of the
cross section, and not its shear center.
Equations Used
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Solution (a) To find the maximum transvere shear, the shear at each section along the beam between sections A and D must be found. This means that we need to draw the shear diagram. To do this we have to take the following steps: From the free body diagram, the reactions at A and D are
The transverse shear can be found by using mechanics of materials
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The resulting shear diagram shown below indicates that the maximum shear force is Vmax = 55.8 kN and it occurs along portion CD. Notice once again that the diagram is exactly the same
as those found in the previous two example problems as the loads and boundary conditions have not changes.
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(b) For the shear flow analysis we need to have the moment of inertia about the centroidal axis perpendicular to the shear force, in this case the horizontal centroidal axis. The dimensions of the cross section are measured to the middle of each adjacent member for simplicity. Hence, the cross section can be treated as
The moment of inertia is found as
Notice that in this problem the applied lateral load does not pass through the shear center. This means that as a result of this loading, the beam will twist while bending. The principle of superposition allows us to break this problem into two parts. In the first part we will determine
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the shear stress due to V through the shear center, and in the second part we will determine the shear stress due to accompanying torque as was explained earlier. For the shear flow due to V passing through the shear center, we start at the free end of the top right flange by isolating a portion of length s along the flange, same as in previous examples.
Write the equation for the moment of area of the isolated section, and put it in the shear flow equation as given below.
For the web start from the web-flange junction and write the shear flow equation considering the moment of area of w portion along the web.
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The bottom flange will have the same shear flow as the top flange due to horizontal centroidal symmetry of the cross section.
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NOTE: always check your final value of q to make sure that at the free end you get zero. The shear flow diagram shown is only due to V through the shear center and does not represent the final or true shear flow variation. However, this is the variation we need in order to find the location of shear center.
(c) Because of symmetry about the horizontal centroidal axis, the shear center must be located on that axis. Therefore, its 'y' coordinate is known. The 'x' coordinate of shear center can be determined by summing moments about the centroid. First, we need to find the force in each flange and the web. The force along each member is just the area under the shear flow curve.
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The force in the flange is the area of the triangle (the height and length would be 488.5 and .12). But as long as the equations are available, it's easier to just integrate.
Always calculate the summation of vertical and horizontal forces, and check them against the resultant force. With only a vertical force acting on the section, the sum of horizontal forces should add up to zero, as they do. Notice that the vertical force along the web is almost 99.7% of V with the remaining 0.3% carried by the flanges as explained before. Summing moments about the centroid gives the horizontal coordinate of the shear center. Since the answer came out positive, it means that the shear center is to the left of the web as assumed.
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(d) The shear flow diagram shown above is only the contribution of the force through the shear center. There is also a shear flow induced by the torque about the shear center. Knowing the location of shear center, we can now calculate the torque as shown in the figure below.
The value of torque is determined as
Recall from the discussion of thin-walled sections in pure torsion and the elastic membrane analogy that the shear stress contours along each member can be shown as in the figure below.
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To use the equation for torque-induced maximum shear stress, we first need to calculate the a/t ratio. Since all members have the same thickness, the maximum shear stress due to torque occurs in the middle of the longest member, in this case the web.
Looking at the shear stress contours, we observe that near the inside edge of the web the shear stress is downward, therefore, it would add to the one caused by the shear force through the shear center. With both components being maximum at the neutral axis, the calculation of the overall maximum shear stress would be performed as
The maximum shear stress occurs at the NA, on the inside edge of the web. 6.5 Shear Flow Analysis for Unsymmetric Beams:
In chapter II, a general equation for the bending stress was introduced. It was further shown that this equation could be used on any cross section, symmetric or unsymmetric, under symmetric or Unsymmetric bending moment. There is a similar general equation for shear flow. It too can be used for any cross section, symmetric or unsymmetric, under symmetric or unsymmetric transverse shear force. This equation is derived from Eq. (II.1), and is given as
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When the section is symmetric there will be no product of inertia, and if only a vertical shear force is applied, this equation reduces to what we have used previously for determining shear flow distribution at a given section.
The analysis procedure for unsymmetrical cross sections is as follows: •
Determine, as needed, the centroid, moments of inertia, and the shearing load in the problem.
•
Consider the two parts of the general shear flow equation separately.
•
Determine the shear flow resulting from each part.
•
Combine the two parts together to get the final shear flow distribution.
When using this equation, it is important to consider the 'signs' on the transverse shear force components, Vx and Vy. The sign convention used is as follows:
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Example For the beam and loading shown, determine: •
(a) the location and magnitude of the maximum transverse shear force,
•
(b) the shear flow 'q' distribution due to 'Vmax',
•
(c) the 'x' coordinate of the shear center measured from the centroid of the cross section.
Stresses induced by the load do not exceed the elastic limits of the material. The transverse shear force is applied through the shear center at every section of the beam. Also, the length of each member is measured to the middle of the adjacent member.
Equations Used
Solution
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With the cross section being unsymmetric about the centroidal axes x and y, we need to calculate the product of inertia. However, first we need to know where the centroid is located. Then we can calculate the moments of inertia Ix, and Iy, and the product of inertia, Ixy. The calculations lead to the following answers:
(a) To find the maximum transvere shear force, the shear at each point between A and D must be found. From the free body diagram, the reactions at A and D are
The transverse shear can be found from mechanics of materials
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From the diagram, the maximum shear force is found to be
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(b) To find the shear flow use equation A14.14. With only a vertical force acting on the section, whatever the sign, the magnitude of the shear flow will not change with the choice of sign. But to be consistent with the discussion on the sign convention, the shear force is negative along portion CD of the beam, as shown in the shear diagram.
Notice that in the above equation Vx is zero, as there is no horizontal force acting on the section. Work on one part of the general shear flow equation at a time.
Part 1
Obtain the shear flow variation by considering the summation of xA which constitutes the moment of aera Q in this case.
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The equation for the vertical flange is linear, and it would be easy to plot. The horizontal flange would be a little harder to plot accurately as it is parabolic, we also need to obtain more information. For example, what are the roots of the shear flow equation (corresponding to a zero shear flow, which indicates a change in shear flow direction), and where is the slope of the shear curve zero (corresponding to the location of maximum shear flow along the horizontal flange).
With all the necessary information determined, the shear flow distribution corresponding to part 1 of the general shear flow equation can be plotted as shown below.
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Notice that the shear flow never took a zero value, except at the free ends, hence no change in sign (i.e., direction) was observed. Part 2 For this part we need to calculate the moment of area due to summation of yA.
With the picture shown above, the moment of area can be calculated and put into the part 2 of the general shear flow equation as shown below.
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With the flow variation along the vertical flange being parabolic this time, we need to examine the equation along the vertical flange to see if shear flow goes to zero anywhere along its length.
The shear flow from part 2 is shown below.
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Now, combine the two parts together by adding (algebraically) the shear flow components along each flange as.
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Vertical Flange Check to see if there is a change in direction of shear flow along this flange.
Horizontal Flange Check to see if there is a change in direction of shear flow along this flange.
As the calculations indicate, there is infact a point along the horizontal flange where the shear flow switches direction. This is captured in the final shear flow distribution shown below.
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To make sure that the values found are correct. We must check the equilibrium of forces in x and y directions, individually.
The equilibrium is satisfied. Therefore, we can presume that our analysis is correct. (c) The 'x' component of the shear center is determined by summing the moments about an arbitrary point, here we choose point 'G' at the intersection of vertical and horizontal flanges.
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With the resultant force along each flange passing through the corner of the angle section, they produce no moment about point G, therefore
If we look at this angle section carefully, we observe that with the two flanges being identical, and separated by 90 degrees, an axis of symmetry is created along the 45 degree line between the two flanges passing through the corner. Hence, as before the shear center is located on the axis of symmetry. Notice also since this axis of symmetry was not along x or the y axes, the product of inertia Ixy did not vanish. Something to ponder upon! 6.6 Analysis of Beams with Constant Shear Flow Webs:
The discussions of flexural shear stress and shear flow, so far, have been focused on sections which are fully effective in bending. This means that the entire cross section was used in calculation of centroid and moments of inertia. The shear flow varied either linearly or non-linearly along each member of the cross section. As such the calculation of shear flow was somewhat involved. In this section our focus shifts to built-up beams (with open section) that are composed of thin web(s) supported by stiffeners (or stringers). In the analysis of such beams the following assumptions are made.
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Assumptions:
1. Calculations of centroid, symmetry, moments of area and moments of inertia are based totally
on
the
areas
and
distribution
of
beam
stiffeners.
2. A web does not change the shear flow between two adjacent stiffeners and as such would be in
the
state
of
constant
shear
flow.
3. The stiffeners carry the entire bending-induced normal stresses, while the web(s) carry the entire shear flow and corresponding shear stresses. Analysis:
Let's begin with a simplest thin-walled stiffened beam. This means a beam with two stiffeners and a web. Such a beam can only support a transverse force that is parallel to a straight line drawn through the centroids of two stiffeners. Examples of such a beam are shown below. In these three beams, the value of shear flow would be equal although the webs have different shapes.
The reason the shear flows are equal is that the distance between two adjacent stiffeners is shown to be 'd' in all cases, and the applied force is shown to be equal to 'R' in all cases. The shear flow along the web can be determined by the following relationship
Important Features of Two-Stiffener, Single-Web Beams: •
Shear flow between two adjacent stiffeners is constant.
•
The magnitude of the resultant shear force is only a function of the straight line between the two adjacent stiffeners, and is absolutely independent of the web shape.
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•
The direction of the resultant shear force is parallel to the straight line connecting the adjacent stiffeners.
•
The location of the resultant shear force is a function of the enclosed area (between the web, the stringers at each end and the arbitrary point 'O'), and the straight distance between the adjacent stiffeners. This is the only quantity that depends on the shape of the web connecting the stiffeners.
•
The line of action of the resultant force passes through the shear center of the section.
For multi-stiffener, multi-web beams the shear flow never changes direction in the web between the adjacent stiffeners. But it can change direction at a stiffener separating two adjacent webs. Example 1 For each of the constant-shear-flow web sections shown, determine: •
(a) the magnitude of the resultant shear force,
•
(b) the location of the resultant force (same as shear center), with respect to point 'O' shown with each section.
•
(c) the direction of the resultant force.
Let q=100 lb/in in each section.
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Equations Used
Solution At a glance all of the three sections look very different, but once we do the analysis, we will be able to see the similarities. Maybe this is a good place to make an educated guess. With respect to magnitude, location, and direction of the resultant force how similar are these sections? Section 1: The straight distance between the stringers at the ends of the web is
The magnitude of the force is therefore given as
The direction of the force is parallel to 'h'. The location of the force from point 'O' is
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The area A is shown by the hashed region. Notice that the line of action of the resultant force passes through the shear center of the section.
You may ask the question, how would I know by using the scalar equation for 'e' which side of line 'h' the force should be shown? The answer is simple. Remember that the resultant force should produce the same moment about point 'O' as the components. Looking at the shear flow direction shown for section 1, the shear flow produces a counter clockwise moment about point 'O'. Therefore, the resultant shear force must do the same. The arrow on the resultant force vector is consistent with the direction of the shear flow along the web, or the shear force in each segment of the web (i.e., the vertical parts and the horizontal part). Section 2: The magnitude of the force is given as
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The direction of the force is parallel to 'h'. The location of the force measured from point 'O' is
The area 'A' is shown by the hashed region.
Notice that the same reasoning as before can be used to show the correct direction and location of the resultant shear force. Section 3: The magnitude of the force is given as
The direction of the resultant shear force is parallel to 'h'. The location of the resultant shear force requires the use of equation for 'e'. Note that for the previous two sections when we drew straight lines from each stringer to reference point 'O', the resulting region produced a single enclosed area. With the shape we have here, when we connect the ends of the web to point 'O' by two straight lines, the result is two instead of one area. Hmmm! Interesting! What should we do now?
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Well we don't give up this easily. We just have to interpret equation we use for 'e' a bit more carefully. In situations such as this, when more than one cell is created, the direction of moment of shear flow in the two areas may be opposite. As is the case here. So we look at the expanded form of equation for 'e'. We calculate 'e' associated with each cell from the equation e=2A/h, then we add them together by paying attention to the direction of moment the shear flow in each cell produces about point 'O'.
Here, the bigger cell produces a counter clockwise moment, whereas the smaller one produces a clockwise moment. The location of the resulant force is, hence, determined as
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Now, when we compare the three sections, we realize in all three, the magnitude of the resultant shear force was exactly the same. The difference in these sections appeared only in magnitude of 'e'. Also note that in all sections, the shear flow was directed in northwest direction, and as can be seen so was the direction of the resultant shear force. Finally, the resultant force in each section passes through the shear center of the section. Example 2 For the multi-web, multi-stringer open-section beam shown, determine •
(a) the shear flow distribution,
•
(b) the location of the shear center.
Equation Used
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Solution The location of the centroid with respect to stringer 4, and moments of inertia are
(a) With no horizontal shear force, and with the above information, equation A14.14 reduces to
Starting at stringer 1 and working around to stinger 6, the shear flow due to the vertical shear is
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The shear flow diagram and the location and direction of each force component for this section are shown below.
It is important to pay attention to the location of each force, especially the ones for the semicircular webs.
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(b) To determine the 'x' coordinate of the shear center the web forces need to be calculated. Then the moments are summed about stringer 4 with clockwise positive.
To find the 'y' coordinate of the shear center, a fictitious horizontal shear force 'Vx' is applied to the beam, then the same analysis is repeated.
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Now calculate the force in each web.
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Use the moment equilibrium to determine the y-coordinate of the shear center.
The shear center location is shown in the figure below.
7. TRANSVERSE SHEAR LOADING OF BEAMS WITH CLOSED CROSS SECTIONS:
The topics discussed in this section deal with transverse shear loading of unstiffened or thin-walled stiffened beams with CLOSED cross sections. Such beams are generally referred to as BOX Beams. The methods of analysis discussed here are relevant to airplane wing, tail, and fuselage structures. However, more reference is given to aircraft wing structures composed of a single-cell or a multiple-cell closed box beam.
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Topics covered:
Flexural shear stress, shear flow, shear center of unstiffened and thin-walled stiffened sections. Sections 1 through 4 deal with uniform beams whereas sections 5 and 6 deal primarily with no uniform and tapered beams in which the cross-sectional moments of inertia are not constant. Restrictions: The beams are considered to be homogeneous and elastic.
7.1 Single-Cell Unstiffened Box Beam: Symmetrical About One Axis 7.2 Statically Determinate Box Beams With Constant-Shear-Flow Webs 7.3 Single-Cell Multiple-Flange Box Beams. Symmetric and Unsymmetric Cross Sections 7.4 Multiple-Cell Multiple-Flange Box Beams. Symmetric and Unsymmetric Cross Sections 7.5 The Determination of the Flexural Shear Flow Distribution by Considering the Changes in Flange Loads (The Delta P Method) 7.6 Shear Flow in Tapered Sheet Panels 7.1 Single Cell Beams:
Shear flow distribution in beams having open cross sections. We used the general shear flow equation to determine the shear flow variation starting from a free edge where we know q = 0. In beams having closed cross sections, however, there is no free edge. Consequently, the shear flow analysis for closed-section beams is slightly more complicated than that for the open-section beams. For this analysis we will rely, to some extent, on the techniques discussed previous chapters.
Two types of closed-section shear flow problems are discussed here. In case 1, the resultant shear force passes through the shear center of the closed section. Therefore, we know that in such a loading, the beam will bend without twisting. This is referred to here as the bending problem.
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In case 2, the resultant shear force does not pass through the shear center. As a result, the beam will twist as it bends. This is referred to here as the bend-twist problem. There are two different methods to solve these types of problems. Method 1 is known as the direct method, and method 2 is known as the indirect or shear center method. These two
methods of analysis are explained below. Case 1: Bending Problem:
In these problems we know that the transverse shear force passes through the shear center of the section. How do we know that? There are two possibilities. This fact may be explicitly stated in the problem statement. Else, it may be realized by examining the geometric shape of the beam cross section and where the applied load is shown to be acting. Therefore, it is obvious that beam should bend without twisting. To start the analysis, assume the shear flow is zero at an arbitrary point. This implies that the section has been "cut" longitudinally at that point, thus creating a free edge. Then, we use the general shear flow equation to find the "preliminary" shear flow distribution along each wall. We designate this shear flow by q'. The reason we call this the preliminary shear flow is that it is based on the assumption that shear flow is zero at the selected point. Also, with this being a bending problem, the shear flow should result in the angle of twist of zero. If we were to use q' to solve for the twist angle, it would not result in zero twist. This tells us that the shear flow q' is not the true shear flow distribution. To satisfy the zero-twist requirement, a constant shear flow of unknown magnitude q0 is added to the shear flow distribution q' found previously. Recall the angle of twist equation for closed sections discussed in Chapter I. We use that equation with q along each side being q' of that side + or - q0 depending on whether they are both in the same or opposite direction. By setting the angle of twist to zero we find the constant shear flow q0. This value is then added to q' values found previously to determine the final shear flow distribution, q. These steps are graphically shown in the figure below for a doubly symmetric cross section under a vertical transverse shear force through the shear center (which in this case coincides with the cross-sectional centroid). Note the symmetry in the shear flow pattern. This is the consequence of force acting along an axis of symmetry.
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Case 2: Bend-Twist Problem: 1. Direct Method
If we are just interested in determining the shear flow distribution due to the resultant shear force with no interest in the location of shear center, then this is the easiest method to use. In using this method, it does not matter whether or not the transverse shear force is acting through the shear center. As in the previous case, we begin the analysis by selecting a point and assuming that q' = 0. Then we use the general shear flow equation to determine the preliminary shear flow distribution q' all around the section. This shear flow distribution will satisfy the force equilibrium but not the moment equilibrium. To satisfy the moment equilibrium a constant shear flow q0 must be added to the preliminary shear flow system. The magnitude of this constant shear flow is found
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from moment equilibrium. That is the sum of moments due to the shear flow system of q'+q0 about an arbitrary point must balance the moment produced by the resultant shear force about the same point. With one equation in one unknown, the magnitude of q0 will be determined. Note that the shear center location will remain unknown or unimportant throughout the solution. For our rectangular section example with the shear force shifted near the right edge, this approach yields the final shear flow pattern as shown below
2. Shear Center Method
We use this method if in the process of calculating the shear flow distribution due to the applied load we also want to determine the location of the shear center. To explain this procedure let us consider the same example used for the direct method. We can represent the resultant force acting at some arbitrary location in terms of an equivalent force-moment system acting at the shear center of the section. We don't know the magnitude of the moment as we do not know the location of the shear center, but we can write it as the force times its moment arm,
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measured from the point of application of the force to the shear center as shown in the figure below. Now we solve the problem in two parts. First we determine the location of the shear center by determining the shear flow distribution due to force through the shear center. The procedure is exactly the same as the one described above in the bending problem. At the conclusion of this part, we know the location of the shear center. If the location of shear center is exactly along the line of action of the applied force as specified in the problem (i.e., d = 0), then the solution is finished. Otherwise, we calculate the moment shown in the figure as T = Vd. This moment has a shear flow associated with it which can be found using the relation T = 2qA. From this relation we solve for the constant shear flow shown in the figure as qbar. This shear flow has to be superimposed over the one found previously, in the shear center calculation, in order to get the final shear flow distribution corresponding to the actual loading condition.
Example
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For the thin-walled single-cell rectangular beam and loading shown, determine •
(a) the shear center location (ex and ey),
•
(b) the resisting shear flow distribution at the root section due to the applied load of 1000 lb,
•
(c) the location and magnitude of the maximum shear stress.
Equations Used
Solution (a) Due to vertical symmetry of the cross section, the shear center will lie on the centroidal y axis, hence ex = 0 in this case. However, with no symmetry about the horizontal centroidal axis, the value of ey is unknown. Here is how we calculate the vertical coordinate of the shear center: We consider the root section and apply a fictitious horizontal shear force Vx through the shear center. Note that the applied vertical load must be removed for this analysis. In this part we demonstrate the solution procedure described in case 1. With no vertical force and no product of
inertia, the general shear flow equation reduces to
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Because of vertical cross-sectional symmetry and symmetric loading, the neutral axis will pass through the centroid as shown below. In order to use the general shear flow equation, we assume q = 0 at an arbitrary point. This assumption models the cross section as an "open" section. The resulting shear flow distribution is designated by q' to indicate that it only represents a preliminary and not the final shear flow distribution. We decided to assume q' = 0 at the upper left corner. We start from that point and go around the section and determine the preliminary shear flow distribution.
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The resulting preliminary shear flow distribution due to Vx is shown below.
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Since Vx passes through the shear center, the section will not twist. Using the angle of twist equation from chapter I and setting it to zero we obtain the constant shear flow q0 required to keep the section from twisting.
The negative sign on the shear flow q0 implies that its direction is opposite to that assumed in the figure above. The constant shear flow q0 is added to the preliminary shear flow system found previously to determine the final shear flow distribution (not shown here) due to Vx. The direction of positive shear flow is arbitrary, in this case we chose it to be in the counter clock-wise direction. The fact that the shear flow found in the analysis using the general shear flow equation and the shear flow found using the angle of twist equation were in opposite directions is important. They oppose each other.
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Now, by summing the moments about point 'A' at the corner, we determine the vertical location of the shear center. Remember that the force associated with the shear flow is just the area under the shear flow curve. It can either be found by integrating the shear flow equation or, if the curve is simple, like a triangle, it can be solved for directly.
The shear center is located at (0", 0.6") from the centroid. For this part the direct method will be used. The resultant shear force will be equal to the vertical load at the tip. With no horizontal shear force and no product of inertia, the general shear flow equation reduces to
Again, assume q = 0 at a point (i.e., cut the section at corner 'A') and work around the cross section, finding the preliminary shear flow distribution.
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Each one of these equations should be checked for its maximum value location and whether there are any sign changes, indicating a change in the shear flow direction. To find a change in sign, set the equation to zero and solve for 'w'. If the result is within the limits of 'w' along that particular web, then there is a sign change. To find the location of the maximum value, take the derivative of the equation with respect to the 'w' and set the equation equal to zero. This part is just for quadratic or higher degree equations. The maximum for the linear equations will be at one end or the other. The analysis of one equation is shown here.
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With this part done, the shear flow for the open section can be drawn as shown below.
Sum the moments about corner 'A' to get the shear flow q0 needed to balance the moments. This time the forces along 'CD' and 'DA' are found by integration.
Add this constant shear flow to those found previously to get the final q distribution.
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These equations should be analyzed again to determine any shear flow direction changes and locations of maximum shear flow. Two in particular are segments 'BC' and 'DA'. The location for the change in direction should be the same. Notice that in the above shear flow equations w2 and w4 are measured in opposite directions. The final shear flow distribution is
Notice that although the cross section is symmetric about the y axis, the shear flow distribution is not. This is because the vertical shear force is not on the axis of symmetry. It is always a good idea to check equilibrium equations to make sure they are satisfied. In this case the summation of horizontal forces add up to zero as they should, and the summation of vertical forces add up to very close to 1000 lb. The slight difference is due to the fact that the vertical component of the shear in the horizontal webs have been ignored. This idea was first discussed in the discussion of transverse shear loading of open sections in chapter III. Also if we were to take moments about any point, we will find the sum of moments due to shear forces balance the one due to external force. By examining the shear flow distribution and thickness variation, it is clear that the maximum shear stress is in segment 'AB' at w=3.377".
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7.2 Statically Determinate Box Beams with Constant-Shear-Flow Webs:
The box beam shown in the figure below has three stiffeners and three constantshear-flow webs, as a result, there are 6 unknown loads (i.e., 3 shear flows qAB, qBC, qAC and 3 axial forces FA, FB, and FC).
Since for a non-concurrent, non-coplaner (general) force system we have 6 equilibrium equations, we should be able to use these equations to solve for the 6 unknown loads.
Hence, in statically determinate box beams, we do not need to use the procedure involving the general shear flow equation as described in previous section. If we were to modify the box beam by including an additional stiffener, then the problem becomes statically indeterminate as there would be 8 unknown loads to solve for as opposed to 6 in the previous case.
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Under these conditions, we rely on the procedure described in the previous section based on the general shear flow equation. The solution procedure becomes much simpler, however, as the shear flow along each web is constant. The example below describes the method of solution for statically determinate stiffened box beams. The discussion of statically indeterminate box beams will be given in the next section. Example For the single cell, 2-flange cantilever beam shown, determine: •
(a) The shear flow distribution at the root section due to the vertical load of 500 lb acting at the tip.
•
(b) The axial load carried by flanges A and B if the beam has a length of 100 in.
•
(c) The horizontal position of the shear center, ex.
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Equations Used
and equations of equilibrium Solution In this problem there are 4 unknown force quantities, 2 shear flows and 2 axial loads. Therefore, we can use the equations of equilibrium. But first, the force in each web must be determined. Recall from section A14.9 that the magnitude and location of the resultant force of the shear flow in a web are found using
For this problem, it is nessecary to leave 'R' in terms of the shear flow 'q'.
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Now sum the moments about point 'o'. Note that since R1 passes through this point, q1 will not appear in the moment equilibrium equation. Hence, we can directly solve for q2.
Now sum the vertical forces to solve for q1.
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The shear flow diagram is shown in the figure below.
The calculation of axial forces in the flanges or stringers is fairly straight forward. We first draw the free-body diagram of the box beam by separating the webs and flanges as shown in the figure below.
Knowing the direction of shear flow in each web, the direction of shear flow in the flanges can be determined. The axial force in each flange will be the sum of shear flows acting on it multiplied times the length. Notice that there is no axial force at the tip end of each flange.
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Now we have determined all the unknowns asked for in the problem statement. Take a moment to go back and review the solution procedure to make sure you understood all the steps involved. You should have noticed that we only used three of the six equilibrium equations. It would have been possible to calculate the forces PA and PB without having to draw the free-body diagrams. We could have just summed forces along the length of the box beam to obtain one equation in terms of PA and PB. This equation would have indicated that PA = PB immediately. Then, we would have used a moment equation about a horizontal line perpendicular to the length of box beam and passing through either point A or B. This way we could find the magnitude of the unknown force. Try this approach to make sure you will get the same answer as shown in the solution. To determine the horizontal position of the shear center, we assume that the vertical force of 500 lb is acting at a distance of ex to the right of stiffeners. The following procedure is used to determine the value of ex. Recall that by definition, if a transverse force passes through the shear center of a section, it will cause the beam to bend without twisting. Therefore, if the 500 lb force is passing through the shear center, as considered in this section, then the angle of twist, theta, should be zero. We begin by writing the summation of moments about point A 500(ex)
-
R1(3)
-
R2(2
pi)
=
0
=> 500 ex - 24 q1 - 50.27 q2 = 0 (1) Next, we sum forces in the vertical direction. R1
-
R2
-
500
=
0
=> 8 q1 - 8 q2 - 500 = 0 (2) The last equation is that for the angle of twist (5+5)q1/.05
+
(4pi)q2/.032
=
0
=> 200 q1 + 392.7 q2 = 0 (3) Solving equations (1) through (3) simultaneously gives
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ex = -.133 in, q1 = 41.41 lb/in, & q2 = -21.09 lb/in. Now, we have found the horizontal position of the shear center. The negative sign indicates that the shear center is to the left of stiffener A and not to the right as originally assumed. We have also determined the values of the shear flows in the two webs when the transverse shear resultant passes through the shear center. 7.3 Single Cell- Multiple Flange Beams: Symmetric and Unsymmetric Cross Sections:
This section deals with statically indeterminate box beams with constant-shearflow webs. For example, a single cell box beam with four stiffeners and four webs. The shear flow analysis in such box beams is basically the same as that discussed in section IV.1 for closed sections. We cut the cross section at a particular web in order to start the solution at a point with a known shear flow (i.e., q' = 0), and work around the cross section using the general shear flow equation. Then, we replace the cut web and show a constant shear flow qo around the closed cell. This unknown shear flow is the actual value of shear flow in the web that was originally cut. Both the shear center and direct method can be used to solve for the unknown shear flow. The shear flow between stringers can be calculated as in previous section. Remember that for these problems, the shear flow between two adjacent stringers is constant in both magnitude and direction. The change in direction can only occur at a stringer location. The example below shows the analysis procedure. Example For the single-cell skin-stringer closed section shown, determine (a) the shear flow distribution for a vertical transverse shear force of Vy passing through the shear center, (b) the horizontal position of the shear center ex, (c) the shear flow distribution for a horizontal transverse shear force of Vx passing through the shear center, (d) the vertical position of the shear center ey. Note: show the shear center location position relative to the centroid.
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Equations Used
Solution The centroid of the cross section is located at
Notice that since the stiffeners are assumed to carry the bulk of bending in skin-stringer sections, only the areas of the stiffeners are used in calculating the centroid as well as the moments of inertia. (a) Apply a load Vy in the downward direction. We begin by removing the skin between stringers #7 and #1 so we can start from a point with a known shear flow (i.e., zero) and be able to use Eq. A14.14. With Vx = 0, equation A14.14 reduces to
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The moments of inertia are
Start at stringer #1, and work around the section.
Note that the value of q' found for each web would have been the true value of shear flow if the original section was an OPEN section (i.e., with no web between stiffeners 1 & 7). This would
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be similar to the problems discussed previously in Chapter A14. However, in this problem we are dealing with a closed section so the q values found from Eq. A14.14 are NOT the actual shear flows, and that is why we use the variable q' to make that distinction. Now, replace the skin between stringers #7 and #1. Since the resultant shear force passes though the shear center, the twist angle must be zero. We use the angle of twist equation set to zero. This equation allows us to find the value of shear flow qo that once is algebraically added to q' values would give the true shear flow in each web.
This is the required shear flow to keep the closed section from twisting. Note that it is in terms of Vy. By algebraically adding qo to q' in each web, we will come up with the actual shear flow distribution for the case when Vy is passing through the shear center
To find the horizontal location of the shear center, we sum the moments about the position of one of the stringers. Here we choose stringer #2.
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ex is measured from the centroid of the cross section. To determine the shear flow distribution for V = Vx and Vy = 0, we pretty much follow the same procedure as used in part (a). Again, remove the skin between stringer #7 and #1. With Vy = 0, equation A14.14 reduces to
As before, work all the way around the section to find q' in each web.
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Use the angle of twist equation set to zero to solve for qo.
By algebraically adding qo to q' in each web, we will come up with the actual shear flow distribution for the case when Vx is passing through the shear center
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Note the direction of shear flow around the section. To check the validity of the shear flows, we can sum forces along the x direction to make sure it comes out to be equal to Vx which is the resulant force of all shear flows. Finally, we sum moments about the position of stringer #2 to solve for dy and ey.
Therefore, the shear center is located at 5.85" to the left and .08" below the centroid as shown in the figure below. Just for the curiosity sake, where would the shear center be if there really was no skin between stringers #7 and #1, making this an open skin-stringer section? It can be easily found by using the q' values in the above analysis in the equation for the summation of moments. Use the q' in part (a) to find the ex and use the q' in part (c) to find ey in this case. The shear center would be located at
This location is consistent with the location of the shear center for a channel section.
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7.4 Multiple-Cell Multiple-Flange Beams: Symmetric and Unsymmetric Cross Sections:
For finding the shear flow distribution and shear center location in two-cell or multiple-cell (i.e., more than 2 cells), multiple-flange beams the same methods demonstrated in section IV.1 can still be used. Keep in mind that such beams are always statically indeterminate, and we must use either the direct method or the shear center method to solve for the shear flow distribution.
However, following the procedure described in section IV.1 does not provide us with sufficient number of equations to solve for all the unknown shear flows. An important fact to remember in these problems is that all the cells will twist the same amount.
This will help us establish sufficient number of equations to solve for the unknown shear flows. The equation for the angle of twist is the same as the one used before for closed-section beams with constant-shear-flow webs.
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Example For the uniform two-cell, skin-stringer cantilever box beam and loading shown; •
(a) determine the shear flows in the webs and the axial forces in the stringers at the root section,
•
(b) determine the shear and normal stresses at the root section,
•
(c) determine the shear center location,
•
(d) determine the deflection at the tip which includes vertical and horizontal displacements and twist.
Equations Used
Solution At first glance, we can tell that this is a statically indeterminate problem. There are 7 unknown forces and shear flows, but we only have 6 static equilibrium equations to work with.
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However, if we include the angle of twist equation, there will be sufficient equations to solve for all the unkowns in the problem. Start by summing the vertical and horizontal forces and the moments about stringer A using the view shown in the figure below.
The equilibrium equations are
Now write the angle of twist equation for each cell. Since the angle of twist in the two cells must be equal, we can set the right hand sides of these equations equal to each other and obtain an additional equation in terms of the unknown shear flows.
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We now have 4 equations and 4 unknowns. Solving them gives the shear flows as
The axial forces are found by breaking up the wing section and drawing the shear flows on the panels and stringers.
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The shear stresses in the webs are found simply by dividing the shear flow in the web by the corresponding web thickness. The normal stress in each stringer is found by dividing the axial force by the area of that stringer.
To find the x coordinate of the shear center, we must apply a vertical force of arbitrary magnitude. In this case, we apply a vertical force of 2500 lb. The location of the shear center is assumed to be to the left of the vertical web between stringers A and B as shown below. We write the force and moment equilibrium equations along with the equation for angle of twist in each cell. Since the force is passing through the shear center, there cannot be any twist. Therefore, each twist equation is set equal to zero.
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We have 5 equations and 5 unknowns. Solving for ex yields
To find the y coordinate of the shear center, we apply a horizontal force, which in this case is chosen to be 500 lb in +x direction. Following the same procedure as used previously, we find ey as follows.
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(d) With the cross section having no centroidal axis of symmetry, we cannot directly solve for the vertical tip deflection based on the formula
This equation gives the correct answer only if x and y are principal axes. In this problem x and y are not the principal axes as Ixy is not zero. Hence, what we need to do is to first locate the principal axes, then calculate the corresponding moments of inertia. Then by transforming the forces in the directions of principal axes, we solve for the tip deflection. The cross-sectional centroid is located at (7.2",5") from flange B. The corresponding moments of inertia are calculated to be Ix = 25 in4, Iy = 207.36 in4, and Ixy = -36 in4. Using the equations for the principal plane and corresponding moments of inertia, we obtain Ixp = 214.21 in4 and Iyp = 18.15 in4 with yp axis being at 10.77 degrees counter-clockwise from the x axis. Transforming the applied forces in xp and yp directions gives Pxp =-2,362.49 lb, and Pyp = 958.47 lb. Now
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using the deflection equations for the tip-loaded cantilever beam, we obtain 4.34" and 0.15" in xp and yp directions, respectively. These values correspond to E = 10 Msi. These results translate into a tip deflection of (x,y) = (-0.67,4.3"). Therefore, the tip would deflect upward and to the left (as viewed from the root) as a result of the loads and the cross-sectional geometry. If we calculate the tip deflection by disregarding the facts stated above, we would end up with a tip deflection of (x,y) = (.08",3.33"). This would result in error of 22.6% in vertical deflection. The twist at the tip is found by using the angle of twist equation for either of the two cells; the answer will come out the same, regardless of which cell is used.
NOTE:
In this example problem we used the equations of equilibrium along with the angle of twist equation to solve for the unknown shear flow and axial forces. Here is the procedure for solving for the shear flows using the direct method: As before we need to remove some webs to create an open section. The choice of which webs to remove is arbitrary so long as once these webs are removed the section will be fully open. For example if we remove the curved and flat webs connecting stringers A and B, we will have an open section consisting of only two webs and three stringers. Now, we use the general shear flow equation to solve for the preliminary shear flow q' along the remaining webs. Note that we have to do this once with only Vy as the resultant shear force, and another time with Vx as the only resultant force. Once the q' values are found, we will replace the webs that were removed earlier. Because of having two cells, we show two constant shear flows qo1 and qo2 in cells one and two, respectively. We write the angle of twist equation for each cell with q along each web being the algebraic sum of q' and qo. By setting the angle of twist in both cells equal, we will have one equation in two unknows qo1 and qo2. Next, we write the equation for the summation of moments about an arbitrary point. This will give another equation in terms of qo1 and qo2.
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Solving them simultaneously gives the two constant values. The actual values of the shear flows are found by algebraic addition of qo's and q' values in all webs. We have to pay attention to the sign of each component in finding the actual shear flows. Having found the shear flows in the webs, we can then draw the free-body diagram of the beam to solve for the axial forces in the stringers as was done in the solution above. The described procedure is what we would use in the case of multiple-cell multiple-flange box beams to be discussed next. 7.5 The Determination of the Flexural Shear Flow Distribution by Considering the Changes in Flange Forces (The Delta P Method):
The method of solution based on the general shear flow equation is valid for beams with uniform cross section (constant moments of inertia). However, in aircraft structures it is common to encounter nonuniform beams. The wing box is a good example. In such problems the general shear flow equation could lead to erroneous values for shear flows. This is because this equation assumes Ix, Iy, and Ixy are all constant along the length of the beam whereas in a nonuniform beam that is not the case. Therefore, we must use a different method of analysis when dealing with nonuniform beams. The Delta P method described in this section is what we would use for nonuniform beams. Note that nonuniformity could be in different forms. For example, we could have a skin-stringer box beam with stringer areas varying along the length of the beam causing the moment of inertia to be a function of position along the length. We could also have webs of varying dimensions, i.e., tapered . The shear flow analysis using the Delta P method is more tedious than using the general shear flow equation and is not recommended for uniform beams. The example problem below shows the procedure for using the Delta P method. Let's consider a square beam of 100 in. in length subject to a 100 lb load at its tip passing through the centroid. There are four stringers at the corners, each with an area of 1 sq. in. We'd like to find the shear flow pattern by using the Delta P method.
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The final shear flow pattern is shown above. The following solution steps show how this pattern was obtained. With Delta P method we must identify two adjacent sections along the length of the beam. The section at which the shear flows are sought is called section A-A, and the adjacent section, some distance away along the length, is called section B-B. We proceed by calculating the normal force in each stringer at section A-A. To do this we need to use the general bending stress equation. First, we need to identify whether there is any symmetry with respect to horizontal or vertical centroidal axes. This is done by examining the stringer areas and their distribution. Also determine how many components of bending moment are present at section A-A. In this case the cross section is doubly symmetric, so obviously the product of inertia is zero. Also with the force at the tip acting in the vertical direction, there is only one moment at the root, that is Mx. Once the stress in each stringer is determined, the force in each stringer is determined by the product of stress times the cross-sectional area. These steps are repeated at section B-B. Notice that this section is closer to the tip, therefore, the bending moment at this section is less than that at the root, section A-A. Since there is no change in cross-sectional geometry, the centroidal location and the moment of inertia do not change. The forces in stringer at section B-B are obtained and shown below.
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The difference between the axial force at each end of the stringer is determined next. This is where the name Delta P comes from.
As in previous method of solution based on Eqn. A14.14, we must begin at a point with known shear flow. So we cut the box beam along one web which would make shear flow zero there. Then we start with stringer A and calculate the shear flow along the 5" length (i.e., the distance
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between sections A-A and B-B) by dividing the Delta P of stringer A by 5". This shear flow has to be equal to that in the adjacent web at the top. Since the web is untapered, the shear flow along all four sides would be equal. We march around the entire section and find the shear flow in each web. Notice that equilibrium is maintained in each stringer and in each web as it must. Please note carefully the directions of the forces and shear flows. You can click on the picture for a more detailed drawing of the stringers. Each stringer and web is in equilibrium, this can be proven by summing the forces.
Next, we close the section by replacing the cut web. To calculate the shear flow along the web that was set to zero, we use the moment equilibrium equation. This equation gives the constant shear flow qo. This value is then added (with attention to direction) to other shear flows to obtain the final shear flow pattern at the root, section A-A.
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What we just did is the Delta P method in a nut shell. Notice how the shear flows are exactly the same as found before. Because this beam has no taper along its length, we could have taken station B-B to be anywhere along the length and still get the same shear flows at the root. This, however, is not the case in tapered beams, as we will discover next. Delta P METHOD: Application Procedure:
1. Identify the section (ie station A-A) along the beam where the shear flow is to be determined (usually the root for a cantilever beam). 2. Choose another station (ie station B-B) in the vicinity of the first one. For tapered beams, the closer the better. 3. Determine the bending moments at station A-A and station B-B.
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4. Determine the normal stress in each stringer, using equation A13.13, at each station. 5. Determine the normal component of the force in each stringer at each station. 6. Calculate the difference between the axial forces at the two stations for each stringer. This is the dP for each stringer. 7. If the beam has a closed cross section, modify it to an open section. 8. Evaluate the q' variation at station A-A based on Delta P/Delta L 9. Replace the removed web(s), and calculate the necessary constants (i.e., q01, q02, q02,...q0n, where n = no. of cells). 10. The vector addition of q' and q0 in each cell gives the final shear flow distribution at station A-A. Example For the uniform single-cell, skin-stringer cantilever box beam and loading of example 1 in sections A15.9 & A15.10 determine the shear flows in the webs at the root section using the Delta P method.
Equations Used
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Solution First define stations A-A and B-B. Since we are looking for the shear flow at the root, let that be station A-A. This is a cantilever beam with constant cross section, therefore station B-B can be anywhere between station A-A and the tip. Since the cross-sectional properties are constant along the length of the beam, the magnitude of the shear flows will be independent of the location of station B-B. In this example, station B-B is placed at 50 inches from the tip.
Using equation A13.13, the normal stress in each stringer is found (remember that the stringer coordinates are in reference to the centriod).
The normal force is just the stress times the area.
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Now repeat the procedure for B-B. If this cross section and/or the stingers varied with the length, then the moments of inertia and centroid would need to be calculated again. But in this case they don't. We just need to recalculate the bending moments as the moment arms are different for station B-B.
The delta P's (i.e., the difference in the axial forces at the ends of each stiffener between stations A-A and B-B) are:
Remove the necessary webs to make this an open section. Webs AB (the quarter circle) and web AB (the interior web) are removed in this example. Now calculate the shear flow for the cross
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section. Here is the way of finding the shear flows. Starting at 'A', show the dP found earlier in the proper direction.
In order to maintain equilibrium, there must be another force in the opposite direction. This force can either be represented as a force or a shear flow along the right edge of stringer A where it is attached to a web.
The shear flow in the web that is attached to stringer A is of equal magnitude but of opposite direction due to equilibrium requirements. The same is seen in stringer C. Note that since the web is rectangular, the shear flow along all four sides would be equal.
Stringer C has a shear flow of 250 lb/in on the left side. In addition, it has a dP of 694.44 lb. Hence, in order to maintain equilibrium there is a need for another shear flow along the right edge of stringer C. This shear flow is calculated in the figure below.
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This new shear flow is passed onto the bottom web, then onto stringer 'B' which already has a dP acting on it.
Since B is the last stringer having a free edge on the left, the force acting along that edge should come out zero. If it doesn't, then something is wrong in the calculations.
It is important to note that the shear flows calculated so far are not the final answers. This is evident by the fact that these shear flows satisfy only the force equilibrium and not the moment equilbrium. Now, we replace the webs that were removed earlier and write the angle of twist equation for each cell in terms of the two unknown constant shear flows (one for each cell) q1 and q2. Since the angle of twist in cell one is the same as that in cell two, we get one equation at the end in terms of q1 and q2 only.
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Next, we sum the moments about 'A' to get another equation in terms of the two constant shear flows. Notice that the moment due to external forces is zero due to the location of the moment center.
Solving the two equations simultaneously gives q1 and q2. The final shear flows at the root section are given below:
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7.6 Shear Flow in Tapered Sheet Panels:
From aerodynamic, structural, and performance efficiency standpoints it is very common to find tapered wings in both civilian and military aircraft. Tapered wings have an advantage over non-tapered wings. From an aerodynamic standpoint, they have lower drag which allows for faster speeds and a better lift distribution over the surface of the wing. Tapered wings are structurally more efficient with root section, having to support more load, being the section with the largest moment of inertia. Tapered wings are also used for improved maneuverability in military aircraft. In tapered web-stringer sections, the stringers relieve some of the shear load carried by the webs. This can be shown by the following example. Below is a planview of a flat, rectangular stiffened web. By simple summation of moments and forces, the axial forces in the stringers and shear flow in the web can be determined.
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Now, for a tapered stiffened web of the same length we have
We observe that while stringer B has no slope in x-y plane, stringer A does. As a result, the axial force in stringer B is the same as its x component, while the axial force in stringer A is the summation of its x and y components shown above. As shown below, the shear force in the web at the left end is no longer equal to P as in the unhampered section, but rather is less - with the difference picked up by stringer A.
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In tapered box beams, we must pay close attention to the slope of each stringer to determine its axial force. Also in tapered webs the shear flow is no longer the same along two adjacent edges, as is the case with untapered webs. With respect to the use of Delta P method for calculating the shear flow distribution at a desired section, we realize that the accuracy of the method is decreased if the two adjacent sections along the beam (i.e., sections A-A and B-B shown in the figure below) are too far apart. This is especially true when we have a beam with nonuniform taper (one with the stringers and webs changing dimensions at a different rate from root to tip).
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8. COMBINED TRANSVERSE SHEAR, BENDING, AND TORSION LOADING
The problems of combined loading are fairly common in aircraft structures. For example, if we consider the aerodynamic loads acting on a wing. We realize that in many cases the line of aerodynamic centers does not coincide with the elastic axis (i.e., the line passing through the shear center of every spanwise section of the wing). As a result the wing bending will be combined with twisting action as well. So in this case the loads at the aerodynamic center line can be replaced with an equivalent force couple system acting at the elastic axis. The force component will produce transverse shear loading and bending moment along the length of the wing while the couple produces a torque. The example problem below is taken from the previous section and it basically highlights the procedure for analyzing such problems in general. One must keep in mind that depending upon the complexity of the beam cross section the bending and shear stresses may require the use of general bending stress and general shear flow equations. In addition to the loadings stated earlier it may be possible to have axial tension as well. The rotating wing problem (i.e., helicopter blade) is a good example of such a problem. 9. INTERNAL PRESSURE:
In this section we will consider the analysis of pressurized structures. Examples of these structures are found in fuselage of high-flying aircraft that are required to be pressurized for passenger comfort. We will consider both monocoque and semi-monocoque pressure vessels. The method of analysis is confined to plane-stress problems of cylindrical and spherical pressure vessels with the shell treated as an elastic membrane incapable of supporting any bending loads. 9.1 Membrane Equations of Equilibrium 9.2 Special Problems in Pressurized Cabin Stress Analysis 9.1 Membrane Equations Of Equilibrium For Thin-Walled Pressure Vessels
Pressure vessels are shell-type structures. Thin walled implies skin thickness is much less than the radius of curvature. The most common shapes used for pressure vessels are cylinders and spheres. Spherical pressure vessels are the most efficient. The skin is usually very thin, it cannot support any bending, only membrane stress. In these problems, the plane stress condition is assumed.
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Unstiffened (Monocoque) Cylindrical Pressure Vessels
Hoop Stress
Axial Stress
Maximun Shear Stress
Unstiffened Spherical Pressure Vessels:
Due to symmetry, the stresses exerted on all four faces of the element must be equal.
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Example 1 For the monocoque cylinderical pressure vessel shown, determine the •
(a) hoop stress
•
(b) axial stress
•
(c) maximun shear stress
If the pressure was increased to 20 psi, what should be the thickness of the skin to maintain the same level of hoop stress as found in part (a)?
The pressure shown is the gauge pressure. Equations Used
Solution (a) The hoop stress is given as
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(b) The axial stress is found by
(c) The maximum shear stress is determined as
If the pressure was increased to 20 psi, the new skin thickness to maintain the same level of hoop stress would be found as follows
Therefore, it takes 36% more skin thickness to carry 36% more pressure for the same level of stress in skin.
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Example 2 A spherical pressure vessel is made of aluminum that has the following properties: the ultimate normal stress is 68,000 psi and the ultimate shear stress is 41,000 psi. Determine the minimum skin thickness if the gauge pressure inside is 14.7 psi and the diameter is 6 ft. Equations Used
Solution The skin thickness based on ultimate axial stress is
The skin thickness based on ultimate shear stress is
The largest of the two will be the minumum skin thickness that would be used based on the material ultimate strength values.
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Notice that no factor of safety is used in this example. Normally there is a factor of safety associated with every design. For example, if we were to use a factor of safety of 2 for normal stress and 3 for shear, we would use 68000/2 = 34000 psi for max allowable normal stress and 41000/3 = 13666.67 psi for max allowable shear stress. Under these conditions the minimum allowable skin thickness would be twice as large as the one found earlier. i.e., tmin = .007782"
Example 3 An idealized transport aircraft fuselage, shown in the figure below, is to be fabricated from aluminum alloy with the ultimate stresses shown. Knowing that a maximum allowable gauge pressure of 14.7 psi is desired and factors of safety of 1.5 in tension and 2 in shear are required, determine the minumum allowable skin thickness. The diameter is 72 inches. The end caps are spherical, and have the same thickness as the cylindrical section.
Equations Used
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Solution First, we determine the allowable stresses as
For each section of the fuselage we find the skin thickness based on each equation. The largest value will be the minimum allowable skin thickness. Cylindrical Portion:
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Spherical Portion:
The minimum allowable skin thickness is
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9.2 Special Problems In Pressurized Cabin Stress Analysis: Axially-Stiffened (SemiMonocoque) Pressure Vessels:
In an aircraft, the vertical and horizontal tail produce loads that put the fuselage in bending. An unstiffened pressure vessel cannot carry these loads efficiently. Hence, stringers are commonly added longitudinaly to support the skin and create a semi-monocoque pressure vessel. At the point of attachment, the stringer and skin, or shell, have the same strains, but different stresses. This difference is due to the fact that skin is in a 2-D state of stress while the stringer is in a 1-D state of stress.
The stress equations are:
where
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Example Consider a semi-monocoque cylindrical pressure vessel with a diameter of 78 inches. The skin and stringers are made of the same material with an ultimate normal stress of 60,000 psi, an ultimate shear stress of 8,000 psi, and a Poisson's ratio of 0.3. We are using 10 identical stringers with the total area equal to 40% of the skin cross sectional area. For internal gauge pressure of 15 psi with yield factors of safty of 1.5 in tension and 2 in shear, determine: •
the pressure vessel's minimum skin thickness
•
the cross sectional area of each stringer
Equations Used
Solution We find the skin thickness based on the stress allowables in the skin.
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We will repeat the procedure based on the stress allowables in the stringer.
The minimum thickness is then found to be the one that does not violate any of the stress criteria
The area of each stringer is found using
Real Aircraft Structures
Here are some slides of several fuselage sections of a Boeing 727 showing the frames and stringers with different cross sectional shapes. Slide #1: Notice in this slide the use of different shaped stringers in the fuselage. The top one is a hat, below it is an angle, below that is a Z, and at the bottom of the slide is another hat stringer. Also notice how closely (approximately 1") the rivets are placed next to each other.
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Slide #2: This is a close up view of a frame-stringer joint. Notice the use of a clip attaching the stringer to the frame. Also notice the use of skin doubler between the fuselage skin and the frame.
Slide #3: This slide shows what is commonly referred to as a "mouse hole" - the opening in the frame allowing the stringer to pass through it.
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Slide #4: This slide shows the use of hat stringers in the fuselage section. Notice that the stringers are mounted with the top web as opposed to the flanges touching the skin. This form of attachment requires only a single row of rivets, whereas otherwise two rows of rivets would have to be used. This is a good example of how manufacturing requirements have been incorporated in the structural design.
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10. ANALYSIS OF JOINTS:
A complete aircraft structure is manufactured from many parts. These parts are made from sheets, extruded sections, forgings, castings, tubes or machined shapes which must be joined together to form subassemblies. The subassemblies must then joined together to form larger assemblies and then finally assembled in to complete aircraft. Many parts of the complete aircraft should be arranged such that they can be disassembled for shipping, inspection, repair and replacement and usually joined by fasteners. In order to facilitate the assembly and disassembly of the aircraft, it is desirable to design connections containing as few fasteners as possible. 10.1 Analysis of Riveted Joints:
Rivets are low cost, permanent fasteners well suited to automatic assembly operations. Initial cost of rivets is substantially lower than that of threaded fasteners. But tensile and fatigue strengths of rivets are lower than for comparable bolts or screws. Riveted parts cannot be disassembled for maintenance or replacement without destroying the rivet. Although the actual state of stress in a riveted joint is complex, it is customary to ignore such considerations as stress concentration at the edge of rivet holes, unequal division of load among fasteners and non uniform distribution of shear stress across the section of the rivet and of bearing stress between rivet and plate, simplifying assumptions are made, which are summarized as follows 1)
The applied load is assumed to be transmitted entirely by the rivets, friction between the connecting plates being ignored.
2)
When the centre of the cross sectional area of each of the rivets is on the line of action of the load, or when the centroid of the total rivet area is on this line, the rivet joins are assumed to carry equal parts of the load if of the same size: and to be loaded proportionally to their section area other wise.
3)
The shear stress is assumed to be distributed uniformly across the rivet section.
4)
The bearing stress between the rivets and plates is assumed to be distributed uniformly over an area equal to the rivet diameter times the plate thickness.
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5)
The stress in the tension member is assumed to be uniformly distributed over the net area.
6)
The stress in the compression member is assumed to be uniformly distributed over the gross area. The possibility of secondary failure due to the secondary causes, such of the shearing or
tearing out of a plate rivet and edge of plate or between adjacent rivets , the bending or insufficient upsetting of long rivets, or tensile failure along zigzag line when rivets are staggered, are guarded against in standard specifications by provisions summarized as follows 1)
the distance from a rivet to a sheared edge shall not be less than 1 (3/4) diameters, or to a planned or rolled edge, 1(1/2) diameters.
2)
The minimum rivet spacing should be 3 diameters.
3)
The maximum rivet pitch in the direction of the stress shall be 7 diameters, and at the ends of a compression member it shall be 4 diameters for the distance equal to1 (1/2) times the width of the member.
4)
In case of diagonal or zigzag chain of holes extending across a part, the net width of the part shall be obtained by deducting from the gross width the sum of diameters of all holes in the chain and adding, for each gauge space in the chain.
5)
The shear and bearing stresses will be calculated in the basis of the nominal rivet diameter, the tensile stresses on the hole diameter.
6)
If the rivets in a joint are so arranged that the line of action of the load does not pass through the centroid of the rivet areas then the effective eccentricity must be taken in to account.
Failure of riveted joints: Failure of riveted joints may be due to following 3 reasons 1. Shear failure of rivets 2. Tensile failure of plate along rivet line 3. Failure by double shear of plates
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10.1.1 Shear Failure of Rivets:
Applied Ultimate load = P Number of rivets = n Allowable ultimate shear strength of nth rivet = Psn. in N In general Ultimate load on single rivet (nth rivet) Psrn =
Psn P in N Ps1 + Ps 2 + Ps 3 + ...Psr
Reserve factor = Psn/Psrn. For ultimate shear strength of rivets refer Design Allowables for LCA Airframe tables 5.3 - 5.10 page no 109 - 116 Hal report no Hal/MP/R&D/03/82 for Russian specifications. Hal report no Hal/MAT/6/75 for SP, LN, AN, L2 specifications 10.1.2. Tensile Failure of Plate along Rivet Line: 10.1.2.1 For Single Row And Rivets Having Uniform Diameter And Equal Pitch:
Allowable ultimate tensile stress = ft.N/mm2. Ultimate tensile stress of the plate σ =
Pp in N/mm2. ( p − d )t
Stress is calculated for pitch distance. Ultimate applied Load = P in N/mm
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Pitch of the river = p in mm. Diameter of rivet = d in mm. Thickness of the sheet = t in mm. Reserve factor = ft/σ. 10.1.2.2 More Than One Row and Rivets With Equal Spacing:
For top plate the possibility of failure of plate is at row 1 and for bottom plate it is row 3 Ultimate applied load = P N/mm. Ultimate applied load per row = P/nr. Number of rivet rows = nr. Pitch of the rivet = p in mm. Diameter of the rivet = d in mm Thick ness of the sheet = t in mm. Ultimate tensile stress in sheet σ =
(P / n r ) p ( p − d )t
in n/mm2.
Allowable ultimate tensile stress = ft in n/mm2. Reserve factor = ft/σ
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10.1.2.3 Rivet Lines With Varying Number Of Rivets (all rivets are same type):
Allowable ultimate tensile stress in sheet = ft in n/mm2. Ultimate tensile stress in row n σ =
Ps rn Pp in N/mm2. Ps rt ( p − d )t
Number of rivet rows = n Applied ultimate load = P N/mm. Diameter of the rivets = d mm Pitch of the rivets = p mm Thickness of the sheet = t mm Number of rivets in row n = rn. Total number of rivets = rt. Reserve factor = ft/σ.
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10.1.2.4 Rivets Of Varying Diameter, Sheet Thickness And Pitch Distance:
In this case the ultimate tensile stress of all rivets are calculated separately and reserve factor is calculated for each rivet separately minimum reserve factor is taken as reserve factor of the joint Allowable ultimate tensile stress = ft in N/mm2. Ultimate tensile stress in the plate near nth rivet σ =
Psn Pp n Ps1 + Ps 2 + Ps 3 + .. + Psr ( p n − d n )t n
Allowable ultimate load of nth rivet = Psn. in N Pitch of the nth rivet = pn in mm Diameter of the nth rivet = dn in mm Thickness of the sheet at nth rivet = tn in mm Reserve factor = ft/σ For Allowable refer 1. ADA/LCA/050000/137 2. G065 Guide for selection of fasteners 10.1.3 Failure by Double Shear of Plate:
Diameter of the rivet = d Distance between rivet centre and edge of the plate = S in N
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Ultimate working stress τ =
( fos ) P in N/mm2. ⎛ ⎞ ⎛d⎞ 2t ⎜⎜ S − ⎜ ⎟ ⎟⎟ ⎝ 2 ⎠⎠ ⎝
Allowable ultimate shear stress = fs in N/mm2. Reserve factor = fs/τ Generally the sheet will be cleared for tear out of the rivets if the edge distance measured from the centre is at least 2d. (Refer Aircraft structures by Perry page number 301) Refer page 15 of Hal report no Hal/MAT/6/75 for edge margin data. 10.1.4 Rivets subjected to tensile loads:
Generally rivets are not subjected to tensile load. But in some cases for example the skin on upper surface of the wing, due to upward suction, air forces places the rivet that hold the skin to the stringers and ribs in tension, angle section riveted on both flanges subjected to an eccentric loading, rivets connecting floor boards, however these tension loads in most of the cases is relatively small. If tension load is incidental or major a bolt shall be used. Applied load = P Number rivets = n Actual load per rivet = (fos) P/n in N Allowable ultimate load Pt in N Reserve factor = Pt/(P/n) For ultimate tensile strength refer Design Allowable for LCA ADA/LCA/050000/137 Airframe tables 5.11 - 5.12 page no 117 – 118
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10.1.5 Rivets subjected to Eccentric loads:
Centre of Gravity of the rivet system = Cg Eccentricity of the load from the Cg of the Rivet System = e in mm Applied Ultimate load in the System = P in N Number of Rivets in the system = n Area of the Individual Rivet = A in mm2. Moment due to Eccentricity of the load M = Pe . in N-mm. Distance of rivet from Cg in X-axis is x in mm. Distance of rivet from Cg in Y-axis is y in mm Arm distance of nth Rivet from Cg of the Rivet System Rn = x 2 + y 2 in mm. Shear force due to the moment of the load in nth Rivet Vn1 =
MRn in N. ∑ Rn2
Direct Shearing force due to the load P Vn 2 = P n in N Combined Shear Force in nth Rivet Vn = Vn21 + Vn22 + 2Vn1Vn 2 cosθ in N. Allowable Ultimate Shear Load = Ps in N Applied Ultimate Shear Load = Vn in N Reserve factor = Ps/Vn.
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10.1.6 Inter-Rivet Buckling:
In rivet sheet panels the effective area is considered to act monolithically with the stiffener. However if rivets that fasten the sheet to the stiffener are spaced too far apart, the sheet will buckle between the rivets before the critical stress is reached, which means that sheet is less effective in helping the stiffener to carry a subjective compressive load. It is called inter-rivet buckling. To calculate inter-rivet buckling stress, it is assumed that adjacent rivets act as a column with fixed ends. If Et is known then Cπ 2 Et t 2 Inter-rivet buckling stress f IR = 12 p 2 Et = Tangent modulus of Elasticity Pitch of the rivet = p in mm Thickness of sheet = t in mm Fixity coefficient of rivets = C Applied stress f a =
My in N/mm2. I
Bending moment in the section = M Distance between neutral axis and edge = y Moment of inertia of the section = I Type of attachment Flathead rivets Spot welds Mushroom or Snap head Countersunk or dimple head
Fixity coefficients 4 3.5 3 1.5
Reserve factor = f IR f a For further reference refer 1. ESDU DATA sheet 02.01.08 2. NACA tech note 3785, 1957
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10.1.7 Calculation Of Fir Using Nomograms Given In ESDU DATA Sheets:
Strains at two points = εR, εR’. Corresponding compressive proof stresses = fR, fR’. m = material property and is given by the formula
log (ε R ε R' ) (Or) log( f R f R' )
Use ESDU DATA sheet 76016 figure – 5 to find out m Calculate
( mεf E ) and use figure – 6 to find out f R
n
f R in sheet 76016
R
1
p ⎛ f ⎞2 Calculate fn and ⎜ n ⎟ t ⎝ CE ⎠ Use ESDU DATA sheet 02.01.08 to find f IR f n Calculate fIR. 10.1.8 Problems:
Problem 1: Figure shows a lap joint involving two rows of rivets as shown. Sheet material is 2024-T3 clad rivet are 8/32 diameter and 2117-T3 material and of protruding head type. Sheet thickness is 0.04. Applied load Ultimate = 1000lb/in. Ft for 2024-73 clad =60000 Psi.
Shear in Rivets: For two rivets 2 × 596 = 1192
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Applied load =1000 Reserve factor = 1192/1000 = 1.192 Tensile failure of sheet: Allowable Ultimate stress f t = 60000 Psi Ultimate tensile stress σ =
1000 = 29629 (1 − 5 )(0.04) 32
Reserve factor = 60000/29629 = 2.03
Failure of double shear of plate: Ultimate working stress =
1000 = 53333.33 2 × 0.04[5 16 − 5 64]
Allowable Ultimate shear stress = 72000 Reserve factor = 72000/53333.3 = 1.35 Problem 2: A Single rivet lap joint is made in 15mm thick plates with 2mm diameter rivets. Determine the strength of the joints if pitch of rivets is 6cm material is BSL 168. Applied load is 15000N edge is 6cm away from rivet line. Shear failure of Rivets: Ultimate load per rivet P =
1.5 × 15000 = 11250 2
For given material Allowable Ultimate shear strength =28270N Reserve factor = 28270/11250 = 2.51 Tension failure of plate along rivet line: Ultimate tensile stress in sheet =
1.5 × 15000 = 375 (6 − 2) × 15
Allowable tensile Ultimate stress for sheet = 415 Reserve factor = 415/375 = 1.11 Failure of double shear of plate:
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Ultimate working stress z =
1.5 × 15000 = 150 2 × 15 × 5
Allowable Ultimate shear = 415 × 0.6 = 249 Reserve factor = 249/150 = 1.66 Problem 3: Find Reserve factor for shear failure of rivets and tension failure of plate as shown if pitch = 0.375 in. The sheet is 24S-TAlclad Ft = 60000Psi subjected to an ultimate load 1200 lb/in. For pitch = 0.5
Shear failure for rivets: Ultimate load per rivet = 1200/4 = 300 Allowable Ultimate load = 374 Reserve factor = 1.25 Tensile failure in plate along rivet line:
∑σ =
1200 = 50000 2 × (0.5 − 0.125) × 0.032
Allowable Ultimate tensile stress = 60000 Reserve factor = 60000/50000 = 1.2
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For pitch = 0.375 Shear failure for rivets load per rivet =
1200 × 0.375 = 225 2
Allowable load = 374 Reserve factor = 374/225 = 1.66 Tensile failure in plate along rivet line: Ft =
1.2 × 0.375 = 56000 (0.375 − 0.125) × 0.032
Allowable Ultimate tensile load = 60000 Reserve factor = 60000/56000 = 1.070 Problem 4: In the lap joint shown in the figure discuss the possible modes of failure of joint allowable tension 142 N/mm2, Applied load 10t (ultimate). Distance of rivet line from edge is 80mm.
Possible failure: Tearing of plate along rivet line Shear failure of rivet Failure by double shear of rivet
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Bearing failure of rivet Shear in Rivets: Applied load = 10t = 100000N Allowable load = 4 × 142 × π × 20 2 / 4 = 178442.46 Reserve factor = 178442.46/100000 = 1.78 Tension failure of sheet: P = 100000N Ultimate tensile stress =
100000 (120 − 24) × 14
Allowable stress = 142 Reserve factor = 142/71.42 = 1.98 Failure by double shear: Ultimate shear stress =
100000 = 59.29 2 × 14 × (80 − 20)
Allowable shear stress = 0.6 × 142 = 85.2 Reserve factor = 85.2/59.49 = 1.43 Problem 5: Diagram shows a rivets joint with multiple rivets with a concentrated load of 30000N.material BSL 168 for rivet.
σ Br for rivet = 440Mpa Thickness of sheet t = 3mm Load bearing capacity for given rivet Cr = d × t × σ Br For rivet (1) Cr1 = 4 × 3 × 440 = 5280N
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For rivet (2) Cr2 = 4.8 × 3 × 440 = 6336N For rivet (3) Cr3 = 5.6 × 3 × 440 = 7392N For rivet (4) Cr4 = 4.8 × 3 × 440 = 6336N For rivet (5) Cr5 = 4 × 3 × 440 = 5280N Ultimate load in every rivet
Psr1 =
5280 × 30000 5280 = × 30000 = 5172N 5280 + 6336 + 7392 + 6336 + 5280 30624
Psr 2 =
6336 × 30000 = 6206 N 30624
Psr 3 =
7392 × 30000 = 7241N 30624
Psr 4 = 6206 N Psr 5 = 5172 N Allowable load in every rivet
Ps1 = 6825 N Ps 2 = 8839 N Ps 3 = 10297 N Ps 4 = 8839 N Ps 5 = 6825 N
Reserve factor for rivet (1) =
6825 = 1.32 5172
Reserve factor for rivet (2) =
8839 = 1.42 6206
Reserve factor for rivet (3) =
10297 = 1.42 7241
Reserve factor for rivet (4) =
8839 = 1.42 6206
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Reserve factor for rivet (5) =
6825 = 1.32 5172
Reserve factor for the joint = 1.32 For same problem now rivets with equal diameter 4 is considered, Ultimate load in every rivet =
30000 = 6000 N 5
Allowable load in every rivet = 6825N Reserve factor in every rivet =
6825 = 1.1375 6000
Problem 6: Bruhn Method: With reference to the Problem 5 Ft thickness of sheet 3mm Psn allowable tensile load for rivets of diameter 4mm = 6825N 4.8mm = 8839N 5.6mm = 10297N
∑ Psn = 6825 + 8839 + 10297 + 8839 + 6825 = 41625 N Ultimate load in every rivet (1) =
6825 × 30000 = 4920 N 41625
Ultimate load in every rivet (2) =
8839 × 30000 = 6370 N 41625
Ultimate load in every rivet (3) =
10297 × 30000 = 7421N 41625
Ultimate load in every rivet (4) =
8839 × 30000 = 6370 N 41625
Ultimate load in every rivet (5) =
6825 × 30000 = 4920 N 41625
Reserve factor for rivet (1) =
6825 = 1.38 4920
Reserve factor for rivet (2) =
8839 = 1.48 6370
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Reserve factor for rivet (3) =
10297 = 1.38 7421
Reserve factor for rivet (4) =
8831 = 1.38 6370
Reserve factor for rivet (5) =
6825 = 1.38 4920
Problem 7: For the rivet joint shown
ft = 415N/mm2 P = 700 N/mm2 t = 3mm d = 4mm p = 12mm Calculate reserve factor ? Allowable ultimate tensile stress in plate = 415 N/mm2 Ultimate tensile stress in plate σ =
700 × 12 (12 − 4) × 3
Reserve factor = 415/350 = 1.2 Problem 8: For the rivet joint shown. Calculate R.F. p= 12mm
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P = 1000 N/mm ft = 415 N/mm2 t = 3mm d = 4mm nr = 3 Allowable ultimate tensile stress in plate = 415 N/mm2 For top plate most likely the failure to occur is row1 for hollow plate the failure will occur mostly at row3 Ultimate tensile stress in plate = =
( P / nr ) p N / mm ( p − d )t (1000 / 3) × 12 = 166.67 N / mm 2 (12 − 4) × 3
Reserve factor = 415/166.67 = 2.49 Problem 9: For rivet joint shown .Calculate R.F? ft = 415 N/mm2 d = 4mm P = 1200 N/mm2 t = 3mm
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p = 12mm Allowable ultimate tensile stress in plate ft = 415 N/mm2 Load bearing capacity per line = d × t × σ ho × n for row (1) = 4 × 3 × 415 × 5 = 24900 for row (2 = 4 × 3 × 415 × 4 = 19920 for row (3) = 4 × 3 × 415 × 5 = 24900 Actual stress of row 1 =
24900 1200 × 12 × = 214.28 N / mm 2 24900 + 19920 + 24900 (12 − 4) × 3
Actual stress of row 2 =
19920 1200 × 12 × = 171.5 N / mm 2 24900 × 2 + 19920 8×3
Actual stress of row 3 =214.28 N/mm2
R.F. (1) =
415 = 1.93 214.3
R.F. (2) =
415 = 2.42 171.5
R.F. (3) =
415 = 1.93 214.3
Problem 10: Calculate the Reserve factor. t = 3mm,d = 4mm,P = 1500N,S = 10mm,fs = 249 N/mm2 Allowable ultimate shear stress = 249 N/mm2
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Ultimate shear stress =
P 1500 = = 31.25 N / mm 2 2t ( s − d 2) 2 × 3 × (10 − 2)
Reserve factor = 249/31.25 = 6.8 Problem 11: In a rivet connection shown in the diagram. Calculate shear force in each rivet. Rivet diameter = 20mm Strength = 2197 Material = BSL168 ft = 415 Centre of gravity of rivets = ∑
Ax A
Let area of rivet be ‘a’
∑ A = 4 × a × (20) + 4 × a × (34) = 216a
∑ A = 12a x=
216a = 18 12a
∴ Eccentricity e = 44+14+2 = 60 M = pe = 4 × 10 = 240 t R12 = R42 = (18) + (12) = 468 2
2
R1 = R4 = 468 = 21.63
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ssppaatteerr 393
R22 = R32 = (18 2 + 4 2 ) = 340
R2 = R3 = 340 = 18.44 R62 = R72 = (2 2 + 4 2 ) = 20
R6 = R7 = 20 = 4.47 R52 = R82 = (2 2 + 12 2 ) = 148
R5 = R8 = 148 = 12.16
(
)
Ra2 = R122 = 16 2 + 12 2 = 400
Ra = R12 = 20
(
)
R102 = R112 = 16 2 + 4 2 = 272
R10 = R11 = 272 = 16.49 ∑ R 2 = 2[468 + 340 + 148 + 20 + 400 + 272] = 3296 Shear force due to moment of 240 t-cm3 Shear force in 1 and 4
MR1 = 1.575t ∑ R2
in 2 and 3
=
240 × 18.44 = 1.34t 3296
in 5 and 8
=
240 × 12.16 = 0.885t 3296
in 9 and 12
=
240 × 20 = 1.455t 3296
in 10 and 11
=
240 × 16.49 = 1.2t 3296
Direct shearing force due to load P in each rivet = P/no. of rivet = 0.33 tonne Tan angles: tan θ 1 and tan θ 4 = 12 / 18 = 0.667,θ = 33o 45' tan θ 2 and tan θ 3 = 4 / 18 = 0.222,θ = 12 o 32' tan θ 5 and tan θ 8 = 12 / 2 = 6,θ = 80 o 30' tan θ 6 and tan θ 7 = 4 / 2 = 2,θ = 63 o 23'
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tan θ 9 and tan θ 12 = 12 / 16 = 0.75,θ = 36 o 52' tan θ 10 and tan θ 11 = 4 / 16 = 0.25,θ = 14 o 2'
Combined shear force =
F12 + F21 + 2 F1 F2 Cosθ
For rivet 1 and 4
=
(1.575) 2 + (0.33) 2 + 2 × 1.575 × 0.33 × Cos146 o19'
=
2.48 + 0.109 − 2 × 1.575 × 0.33 × 0.321
= 1.670 = 1.29tonne For rivet 2 and 3
= 1.34 2 + 0.332 + 2 × 1.34 × 0.33 × Cos167 o 28' = 1.045 = 1.02t
For rivet 5 and 8
For rivet 6 and 7
For rivet a and 12
=
0.885 2 + 0.332 + 2 × 0.885 × 0.33 × Cos80 o 33'
=
0.8836 = 0.94t
=
0.326 2 + 0.332 + 2 × 0.326 × 0.33 × Cos63o 26'
=
0.311 = 0.558t
= 1.455 2 + 0.332 + 2 × 1.43 × 0.33 × Cos36 o 52' = 1.73t
For rivet 10 and 11
= 1.2 2 + 0.332 + 2 × 1.2 × 0.33 × Cos14 o 2 ' = 1.524t
Ultimate shear load in rivet is 1.73 tonne For diameter 20 ultimate allowable load = 2713N2 Total shear load = 2713 × 12 = 32556 = 3.26 tonne Reserve factor = 3.26/1.73 = 1.88 Problem 12: Inter rivet buckling stress at sheet of 1.2mm thickness and aluminium alloy of BSL165 T6511 buckles with snap head riveting at 15mm pitch, bending moment of section is 24000 N-mm. Moment of Inertia of section = 800mm4 For material BSL 165 T654 E = 63400 N/mm2
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ssppaatteerr 395
0.05% compression proof stress = 304 N/mm2 0.1% compression proof stress = 319 N/mm2
tR 0.001 = =2 ' t R 0.0005 f R 319 = = 1.049 f R' 304 m=
log(2) log(4)
= 14
mt R F 14 × 0.001 × 63400 = = 2.78 319 fR for m = 14,
mt R E = 2.78 fR From Nomograms 5 f n f R = 0.92 f n = 0.92 × 329 = 293.48 N / mm 2
For snap head rivet C=3 p ⎛ fn ⎞ ⎜ ⎟ t ⎝ CE ⎠
1
2
15 ⎛ 293.48 ⎞ = ⎜ ⎟ 1.2 ⎝ 3 × 63400 ⎠
1
2
= 0.49
for m = 14 P ⎛ fn ⎞ ⎟ ⎜ t ⎝ CE ⎠
1
2
= 0.49
f 1R = 1.01, f1R = 1.01 × 293 = 295.93N / mm 2 fn Applied ultimate stress =
24000 × 8 = 240 N / mm 2 800
Reserve factor = 295.93/240 = 1.23
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Problem 13: A Simple riveted butt joint made up of 12.7 thick main plates and cover plate of 9.5mm thick rivet diameter is 20.6mm pitch of rivet is 76.2mm. The Ultimate stress of material is 385N/mm2 in tension load acting is 50000N
ft = 385 N/mm2 Shear failure of rivets: fs = 0.6ft = 0.6*385 = 231
fos × P
Z=
n×
π 4
×d
1.5 × 50000
=
4×
2
Reserve factor =
π 4
× 20.6
= 56.25
2
231 = 4.1 56.25
Tensile failure of plate along rivet line: f t = 385 N / mm 2
σ =
1.5 × 50000 fos × P = neff × t (P − d ) 2 × 9.5 × (76.2 − 20.6 )
Reserve factor =
385 = 5.4 71
Bearing of Rivet: f be = 2 × f t = 770
σb =
1.5 × 50000 = 383.2 9.5 × 20.6
Reserve factor =
770 =2 383.2
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ssppaatteerr 397
10.2 Analysis of Bolted Joints:
The aircraft bolt is used primarily to transfer relatively large loads from one structural member to other. Bolt threads should not be placed in bearing or shear and bolts should be used in double or multiple shears if possible in order to increase strength, efficiency in bolt shear and to decrease bending tendency in bolt. Methods of failure of single bolt fitting: As the load on the fitting is transferred from one side of fitting to the other, internal stresses are produced which tend to cause the fitting to fail in several ways. 1. Failure by bolt shear 2. Failure by bolt bending 3. Failure by lug portion of fitting 10.2.1 Failure by bolt shear:
As the pull P is placed on the fitting, it tends to shear the blot at the sections (1-1), (2-2). Let P be the ultimate load applied on the bolt. Applied ultimate load = P in N Applied allowable load = Pa in N Reserve factor = Pa/P For allowable refer 1. ASSSA/AEROFAS/206/190\/002 2. ADA/LCA/050000/137 page numbers 119-122
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10.2.2 Failure by Bolt Bending:
It is important that sufficient bending strength is provided in bolt to prevent permanent bending deformation of fitting and bolt under the limit loads, so that bolts can be readily removed in maintenance operation. Furthermore bolt bending weakness can cause peaking up of a non uniform bearing pressure on fitting lugs, thus influencing the lug tension and shear strength. The unknown factor in bolt bending is the true value of the bending moment on the bolt because the moment arm to the resultant bending force is unknown.
Moment arm b = 0.5t1 + 0.25t 2 + δ in mm δ = clearance gap between lugs in mm δ = 0 if no gap between lugs ⎛b⎞ Maximum bending moment M = P⎜ ⎟ in N-mm ⎝ 2⎠
Bending stress =
MD in N/mm2. 2I
Reserve factor =
f bt (MD 2 I )
Diameter of the bolt = D in mm Moment of inertia = I in mm4. Refer MIL-HDBK-5 for further details A fitting factor of 1.15 will be used which is standard practise for military aircraft. It is multiplied with Ultimate applied load. For failure of lug portion refer chapter 3 of this manual.
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10.3 Analysis of Welded Joints:
Welding plays an important role in joining or connecting structural units in aircraft. The major structural unit where welding plays an important role are engine mounts, landing gears and the attachment of plates and machined fitting to such structure. Generally welds are avoided in sections subjected to tensions, since they produce weakening effect. In some connection, it is not avoidable. In such case stresses is to be kept low. Standard splices or joint overlapping will be used. There may be failure due to shear or due to tension depending upon loading condition. 10.3.1 Shear Failure in Weld:
Applied ultimate load = P in N Allowable ultimate load Pa = f s Lt eff in N Reserve factor = Pa/P Allowable ultimate shear stress = fs. Length of weld = L in mm teff = minimum of throat thickness or plate thickness in mm Allowable stress multiplied by weld efficiency factor of 0.8
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10.3.2 Tensile failure of weld:
Applied ultimate load = P in N Allowable ultimate load Pa = f t Lt eff in mm Reserve factor Pa/P Allowable ultimate tensile stress = ft. Allowable stress multiplied by weld efficiency factor of 0.8 in welding For further references Refer 1. MIL-W-6858 and MIL-W-6873.
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11. COMBINED STRESS THEORIES FOR YIELD AND ULTIMATE FAILURE:
Aircraft structures are subjected to many types of external loadings. The loads often cause axial bending and shearing stresses acting simultaneously. If the structures are to be designed satisfactorily, combined stress relationship must be known. Assumption of uniform stress is made while doing analysis. A failure theory is needed to define onset of failure. 11.1 Determination of Yield Strength of Structural Member Under A Combined Loading:
The flight vehicle structure must carry the limit load without yielding which in general means the yield strength of the material cannot be exceeded when the structure is subjected to the limit loads. In some flight vehicle structure part involving compact unit of pressure vessel, biaxial or triaxial stress conditions are often produced and it is necessary to determine that any yielding will occur under such combined stress action when carrying the limit loads. For where no elastic instability occurs the following well-known theories of failure have been developed. 1. Maximum Shearing Stress Theory. 2. Maximum Strain Theory. 3. Maximum Energy of Distortion Theory. 4. Octahedral Shear Stress Theory. 11.1.1 Maximum Shearing Stress Theory:
Otherwise called as Tresca condition, According to this theory, failure occurs in a material when Principal stress difference equals uniaxial yield shear stress. f p1 = q 2 (Or )
f p 2 = q 2 (Or )
f p1 − f p 2 = q 2
fp1 = Major Principal Stress fp2 = Minor Principal Stress f p1 = f p2 =
σ x −σ y 2
σ x −σ y 2
1 + 2 1 − 2
(σ
−σ y )
2
x
+ τ 2 in N/mm2.
2
(σ
−σ y )
2
x
2
+ τ 2 in N/mm2.
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Where σx = stress in X direction in N/mm2 σy = stress in Y direction in N/mm2 τ = shear stress in N/mm2 11.1.2 Maximum Energy of Distortion Theory:
It is also called Von-Mises Theory. According to this theory failure occurs in a material Von-Mises stress equals uniaxial yield stress of the material. fvon = Von-Mises stress in N/mm2.
[[
] [
] [
For Principal Stress in 3-D f von =
1 2 2 f p1 − f p 2 + f p 2 − f p 3 + f p 3 − f p1 2
For Principal Stress in 2-D f von =
f p1 + f p 2 − f p1 f p 2 in N/mm2. 2
] ] in N/mm2. 2
2
For axial stresses and shear stresses
f von =
[
(
1 (σ x − σ y )2 + (σ y − σ z )2 + (σ z − σ x )2 + 6 τ xy2 + τ yz2 + τ zx2 2
)] in N/mm2.
Where σx, σy, σz = stresses in corresponding X, Y, Z directions in N/mm2. τxy, τyz, τzx = shear stresses in corresponding XY, YZ, ZX planes in N/mm2. Reserve Factor = t2/fvon. 11.1.3 Maximum Strain Energy Theory:
Otherwise called as Haigh’s Theory. According to this theory, , the failure occurs at a point in a member when the strain energy per unit volume in a biaxial system reaches the limiting strain energy per unit volume as determined by the simple tension test. This Theory is used for the Ductile Materials. Strain Energy due to applied load U = Allowable Strain Energy U a =
2 f p1 f p 2 ⎤ 1 ⎡ 2 2 ⎢ f p1 + f p 2 − ⎥ 2E ⎣ m ⎦
t 22 2E
Reserve factor = Ua/U.
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11.1.4 Octahedral Shear Stress Theory:
Also called as Equivalent stress Theory. According to this theory, in an elastic body failure occurs at a Point when Octahedral Shear Stress is equal to yield tensile strength of the material. Octahedral Shear stress for tensile stress f oct =
1 ( f p1 + f p 2 + f p3 ) 3
Reserve factor = t2/foct.
Octahedral shear stress for shear stress f octs =
1 3
(f
− f p 2 ) + ( f p 2 − f p 3 ) + ( f p 3 − f p1 ) 2
p1
2
2
Reserve factor = q2/focts.
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11.2 Determination of Ultimate Strength of a Structural Member Under Combined Load:
Since the structural designer of flight vehicles must ensure that the ultimate load can be carried by the structure without failure. It is necessary that reliable methods be used to determine the ultimate strength of a structure. Generally Maximum Principal Stress Theory is used to determine the onset of failure at ultimate loading conditions. 11.2.1 Maximum Principal Stress Theory:
According to this theory, failure occurs at a point in a member when the Maximum Principal stress in a biaxial stress system reaches Ultimate Allowable strength of the material.
f p1 =
f p2 =
σ x +σ y 2
σ x +σ y 2
2
1 + 2
⎛σ x −σ y ⎜⎜ 2 ⎝
⎞ ⎟⎟ + τ 2 in n/mm2. ⎠
1 − 2
⎛σ x −σ y ⎜⎜ 2 ⎝
⎞ ⎟⎟ + τ 2 in n/mm2. ⎠
2
Where σx, σy, are applied ultimate stresses in X and Y directions respectively in N/mm2. Τ is applied ultimate shear stress in N/mm2. Reserve factor =
ft ft (Or ) (Or ) f p1 f p2
fs ⎛σ x −σ y ⎜⎜ 2 ⎝
2
⎞ ⎟⎟ + τ 2 ⎠
ft = Allowable ultimate tensile stress in N/mm2. fs = Allowable Ultimate Shear Stress in N/mm2. 11.3 Problem:
Problem 1: A load on a bolt consist of axial pull of 10 KN together with a transverse stress force of 5 KN , diameter of bolt is 15mm permissible tensile stress at elastic limit is 150 N/mm2 = t2 Poisson ratio 1/m = 0.3 f t = 180 N / mm 2 Given:
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Pt = 10 KN Ps = 5 KN t 2 = 100 N / mm 2 1/m = 0.3
σ =
Pt 10000 = = 56.58 N / mm 2 A π 2 15 4
τ=
PS 5000 = = 28.29 N / mm 2 A π 2 15 4
Principal stress f P1 = f P2 =
σ 2
σ 2
+
1 σ 2 + 4τ 2 2
−
1 σ 2 + 4τ 2 2
f P1 =
56.58 1 + 2 2
(56.58)2 + 4(28.29)2
= 62.93N / mm 2
f P2 =
56.58 1 − 2 2
(56.58)2 + 4(28.29)2
= −6.35 N / mm 2
Maximum shearing stress theory: Applied stress according to theory = ( f P1 − f P 2 ) = 62.93 + 6.35 = 69.28 N / mm 2 Allowable stress = 0.6 × t 2 0.6 × 150 = 90 N / mm 2 Reserve factor =
90 = 1.3 69.28
Strain Energy Theory: Strain energy due to applied load U =
2 f − f P2 ⎤ 1 ⎡ 2 f P1 + f P22 − P1 ⎢ ⎥ m 2E ⎣ ⎦
Strain energy due to allowable load Ua =
1 2 t2 2E
Both numerator and denominator ,1/2E is present so we can drop 1/2E U = (62.93) + (− 6.35) + 2(62.93)(6.35)(0.3) = 4177.59 2
2
U 0 = 100 2 = 10000
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Reserve factor =
10000 = 5.4 4177.59
Maximum energy of distortion theory: f von =
=
( f P1 )2 + ( f P 2 )2 − f P1 × f P 2 (62.93)2 + (6.35)2 + (62.93)(6.35) =
4410 = 66.33
Allowable stress = 100 Reserve factor =
100 = 1.5 66.33
Octahedral shear stress theory: Stress due to applied load = =
1 ( f P1 + f P 2 + f P 3 ) 3 1 (62.93 − 6.35) = 18.86 N / mm 2 3
Allowable stress = 100N/mm2 Reserve factor =
100 >3 18.86
Principal stress theory: f P1 = PtU = 1.5 × pt = 1.5 × 10000 = 15000 PSU = 1.5 p S = 1.5 × 5000 = 7500
σ =
PtU 15000 = = 84.88 N / mm 2 A ⎛π 2 ⎞ ⎜ 15 ⎟ ⎝4 ⎠
τ=
PSU 7500 = = 42.44 N / mm 2 A ⎛π 2 ⎞ ⎜ 15 ⎟ ⎝4 ⎠
f P1 =
84.88 1 + 2 2
Reserve factor = f P2 =
84.88 1 − 2 2
(84.88)2 + 4(42.44)2
= 102.44 N / mm 2
180 = 1.75 102.44
(84.88)2 + 4(42.44)2
= −17.52
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Principal shear stress = f PS = Reserve factor =
1 2
(84.88)2 + 4(42.44)2
= 60 N / mm 2
180 × 0.6 = 1.8 60
Problem 2: A mild steel shaft of 5cm diameter subjected to a bending moment of 1900000 Nmm. and torque of 2000000 N-mm. Yield point of steel in tension is 400 N/mm2 and diameter of shaft is 50mm Take 1/m = 0.3 D = 50mm M = 1900000 N-mm T = 2000000 N-mm 1/m = 0.3 t2 = 400N/mm2 Stress due to bending moment σ = Shear stress due to torque τ =
M 1900000 × 32 = = 154.82 N / mm 2 Z π × 50 3
16T 16 × 2000000 = = 81.5 N / mm 2 3 3 πd π × 50
Maximum principal stress f P1 =
154.82 1 + 2 2
(154.82)2 + 4(81.5)2
= 77.41 + 112.4 = 189.8 N / mm 2
f P2 =
154.82 1 − 2 2
(154.82)2 + 4(81.5)2
= 77.41 − 112.4 = −35 N / mm 2
Maximum Shearing Stress theory: Applied Shearing stress f P1 − f P 2 = 189.8 + 35 = 224.8 N / mm 2 Allowable Shearing stress = 400 × 0.6 = 240 N / mm 2 Reserve factor =
240 = 1.06 224.8
Strain Energy theory: Strain energy due to applied load = (189.82) + (35) + 2 × 189.82 × 35 × 0.3 = 41234.84 2
2
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Strain energy due to allowable load = 160000 Reserve factor =
160000 = 3.8 41234.81
Maximum energy of distribution theory: Stress due to applied load f VON =
(189.8)2 + (35)2 + 189.8 × 35 = 209.5 N / mm 2
Allowable stress = 400 Reserve factor =
400 = 1.9 209.5
Octahedral shear stress theory: Stress due to applied load =
1 (189.8 − 35) = 51.6 3
Applied stress = 240 Reserve factor = 240/51.6 >3
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12. CUTOUTS IN PLANE PANELS:
The aircraft structure is continually faced with requirements for opening up webs and panels to provide access for other members such as control rods, hydraulic lines, electrical wire bundles etc pass through. Other cutouts such as windows, doors, service panels, hatches, bomb bays, inspection access holes etc are provided in fuselage. Transport and fighter aircraft fuselages contain numerous cutout areas of different sizes and shapes located in various regions of fuselage body. However cutouts such as those in large passenger openings, cargo openings, service openings emergency exits windows etc often occur in regions where high loads must be resisted and there fore additional structure is needed to carry the loads around the openings. Such openings require strengthening of abutting internal structure. There are several ways of providing cutouts. Some of them are listed here. 1. Providing suitable framing members around the cutouts. 2. Providing Doubler or bent where framing cannot be done. 3. Providing standard round-flanged holes, which have published allowable. 4. Providing lightning holes having flanges 5. Providing non-circular and non-rectangular cutouts.
12.1 Framing Members Around The Cutouts:
A cutout is shown. The first step is to assume a shear flow equal and opposite to that of Present shear flow with no cut. Then second is determining corresponding balancing loads in the framed area. Adding this load system to original one will give the final loads all shear flows are shown as they act on edge members. (On flanges and stiffeners)
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A. Considering no cutout:
B. Applying equal and opposite shear flow at the cutout area:
C. Final shear flow distribution:
There is self balancing internal shear flow in area of framing cutout, so no external reactions outside of the framing area are required, (i.e.) To eliminate the redundancies, it is usually assumed that the same shear flow exists in panel above and below the cutout area it is also assumed that shear flows are same in panels to the left and right of the cutout. Shear flow in the panels above and below the centre panel must statically balance the force due to the q.
qH =
q (H 2 ) in N/mm. (H 1 + H 2 )
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Shear flow in the panels to the left and right of the centre panel must statically balance the force due to the q.
qL =
q ( L2 ) in N/mm (L1 + L2 ) Shear flow in the corner panels must also balance the force due to the shear flow
in the panel between them
q HL = q H
( L2 )
(L1 + L3 )
in N/mm
Final shear flow in cell sheet is given in diagram C. this is the way changes always occur in the area formed about the cutout Next is calculate axial loads developed in all of the framing members due to the cutout. This will add or subtract depending on their direction to any loads before the cutouts have been made. At A-A
At E-E
For sheet: Allowable Ultimate Shear Stress = fs in N/mm2. Actual Ultimate Shear Stress = τ =
q in N/mm2. t
Reserve Factor = fs/τ t = thickness of sheet in mm
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q = shear flow in sheet in N/mm. In frames or stingers: Allowable ultimate tensile stress = ft in N/mm2. Actual ultimate tensile stress σ = P/A in N/mm2. Reserve factor = ft/σ A = area of cross section of frame or stinger in mm2. P = Load at the point considered in N
12.2 Framing Cutouts with Doublers and Bents:
Frequently a cutout in the web of a beam must be so deep that it nearly removes the entire web. In this case a heavy doubler or bent is provided around the cutout to carry the shear. It is illustrated in the diagram.
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The doubler is made to fit around the cutout. Appropriate internal radii are in the cut web and doubler at corners to keep the stresses due to curved beam bending reasonable. The loading imposed on doubler is the shear flow q. The doubler should be analysed as frame.
At A-A Due to shear flow q section A-A experiences Bending, Tensile and Shear stresses. Bending moment at A due to shear flow M =
wV qhw in N-mm. = 2 4
Force in X direction due to shear flow Fx = V = Force in Y direction due to shear flow Fy =
qh in N. 2
qw in N. 2
Bending: My in N/mm2. I
Actual Ultimate bending stress be =
Allowable Ultimate Bending stress = fbe=ft in n/mm2. Reserve factor = fbe/be.
Tension: Ultimate Tensile load Fx =
qh in N. 2
Actual Ultimate Tensile Stress σ =
Fx in N/mm2. A
Allowable Ultimate Tensile Stress = ft in N/mm2. Reserve factor = ft/σ
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Shear: Ultimate Shear Load = Fy in N. Actual Ultimate Shear Stress τ =
Fy A
in N/mm2.
Allowable Ultimate Shear Stress = fs in N/mm2. Reserve factor = fs/τ Where q = shear flow in the member in N/mm. w= width of the doubler in mm. h = height of the doubler in mm. y = distance of the extreme fibre from neutral axis in mm. A = cross sectional area in mm2. I= moment of inertia for the cross section in mm4.
12.3 Ring or Donut Doublers for Round Holes:
Frequently a cutout size requirement is such that a standard round-flanged hole will provide the needed open space and strength. This is done by providing Donut Doubler.
w = ring width in mm. R = inner radius of ring in mm. R1 = R +
w in mm. 2
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D = 2R diameter in mm. q = web shear flow in n/mm. tr ring thickness in mm tw = web thickness in mm. Donut doubler reinforcements are capable of maintaining the original gross area strength of the web. The data presented here consists of round holes with diameter up to 50% of the beam depth and ring reinforcements W/D ratios of 0.35 to 0.50. Bending and Tensile stresses will act at A-A. At A-A: Bending stress be =
3qRR1 in N/mm2. tw w2
Tensile stress σ t =
qd in N/mm2. wt r
Stress Interaction at failure Applied Ultimate Tensile Stress σ = σ be + σ t Allowable Ultimate Tensile Stress = ft in N/mm2. Reserve factor = ft/σ Analysis for Rivets:
⎛ ⎞ tr ⎟⎟ in N/mm. Shear flow due to the Ring Doubler = 2q⎜⎜ ⎝ t r + 0.8t w ⎠ ⎛ ⎞ tr ⎟⎟ p max in N. Load on the rivet Pr = 2q⎜⎜ t t + 0 . 8 w ⎠ ⎝ r Pmax is the maximum pitch in rivet set up in mm. Actual Ultimate load on rivet due to shear flow = Pr in N. Allowable Ultimate Load for Rivet = Pa in N. Reserve Factor = Pa/Pr.
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12.4 Webs with Lightning Holes Having Flanges:
This is a simple easily formed web. Figure shows various types of flanges possible and a beam with cross section through web at holes. This type of reinforcement has advantage when compared to donut doubler type. Since donut doubler provides bending moments, which have bending stress in reinforcement. By providing flanges bending stresses can be avoided. Lightning holes with flanges is checked for buckling.
2
⎛t⎞ Average Shear Stress at which Plate will buckle f s = KE ⎜ ⎟ in N/mm2. ⎝b⎠
Where K = Elastic Buckling Stress Coefficient. E = Modulus of Elasticity for Plate Material. t = Thickness of Plate in mm. b = width of the plate in mm Actual shear Stress in the section τ =
P q in N/mm2. (Or ) A t
P = Ultimate Applied Load in n. A = Area of cross Section in N/mm2. q = Shear flow in the section in N/mm. t = Thickness of plate in mm. Reserve Factor = fs/τ. To find K and for further reference refer ESDU data sheets 75035 and 71005
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12.5 Structures with Non-Circular and Non-Rectangular Cutouts:
Even though standard circular and rectangular cutouts are available, in order to keep cutouts size as minimum as possible above shown cutouts are also used in both civil and military aircraft structures. These types of flanged holes are simple and low cost method for fabricating access holes. Generally elliptical flanged holes are used in lightly and moderately loaded rib webs in wing and empennage applications.
Maximum stress in cutout due to applied load f = K '
P in N/mm2. (w − 2h )t
Where W = width if the plate in mm. 2h = width of the cutout in mm. t = thickness of sheet in mm. P = applied load in N. K’ = stress concentration factor For the values of K’ and for further reference refer ESDU data sheets 69020, 70005
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12.6 Cutouts in Fuselage:
Apart from bending and skin shear, fuselage structures are subjected to Pressurization also. Moreover fuselage structures are curved in nature, which is to be analysed by a different procedure. Fuselage cutout analysis is important since size of the cutout is large as compared to cutouts in other structures and magnitude of load acting on these structures is also high. Since fuselage structures are pressurized it has to check for burst pressure. Thus in an aircraft load distributions are due to 1. Fuselage skin shear. 2. Fuselage bending. 3. Fuselage cabin pressurization. Diagram for a large cutout in fuselage in an aircraft with reinforcements is shown below
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12.6.1 Load Distribution Due to Fuselage Skin shear:
The redistribution of shear flow, which occurs in the vicinity of the cutout due to the cutout on aircraft fuselage, is self-balancing like as in the case of cutouts in flat panels. The redistribution occurs in the vicinity of the cutout. It is shown below A Redistribution of Shear Flow
The shear flows h ⎞ ⎛ q1 = ⎜1 + ⎟q 0 ⎝ a +b⎠ l ⎞ ⎛ q 2 = ⎜1 + ⎟q 0 ⎝ c+d⎠ ⎡ ⎛ l ⎞⎛ h ⎞ ⎤ q3 = ⎢1 − ⎜ ⎟⎜ ⎟ − 1⎥ q0 ⎣ ⎝ c + d ⎠⎝ a + b ⎠ ⎦ Equal and opposite shear flows qa, qb, qc is applied
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B Variable Shear Flows:
C Final Shear Flow Distribution:
New shear flows q1a = q a + Δq1a , q1b = qb + Δq1b q 2 ha = q ha + Δq 2 ha , q 2 hb = q hb + Δq 2 hb q3a = Δq3a − q a , q3b = Δq3b − qb
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Where Δq1a =
q a hb q h , Δq1b = b c a b
Δq 2 ha =
qa l ql , Δq 2 hb = b c+d c+d
Δq 3 a =
q 2 ha hb q h , Δq3b = 2 hb c a b
hb =
ah , hc = h − hb a+b
Thickness of the sheet = t in mm Actual Ultimate Shear Stress τ = q/t in N/mm2. Allowable Ultimate Shear Stress = fs in N/mm2. Reserve factor = fs/ τ 12.6.2 Load Distribution due to Fuselage Bending:
The cut stringer loads are assumed to be generally diffused in to the main sills over one frame bay on either side of the cutout as shown in the diagram. This approach results in somewhat higher panel shear loads and frame axial loads than act5ually occur in practise. If the
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cut stringer loads are not the same on either side of the cutout, use average stringer loads at the centre line of the cutout for this method. Stresses formed due to fuselage bending PL and PU is obtained from moment and force balance PL =
P2 h2 + P3 h3 + P4 h4 + P5 h5 in N h6
PU = (P2 + P3 + P4 + P5 ) − PL in N Skin shear flows q h1 =
PU in N/mm. c
q h 2 = q h1 −
P2 in N/mm. c
q h3 = q h 2 −
P3 in N/mm. c
Where c is the distance between two frames as shown in the diagram. Actual Ultimate Shear Stress τ =
qh in N/mm2. t
Where t is thickness of the sheet in mm. Allowable Ultimate Shear Stress = fs in N/mm2. Reserve Factor = fs/τ. 12.6.3 Load Distribution due to Fuselage Cabin Pressurization:
Due to cabin pressurization Hoop stress and Longitudinal stress will set up in the fuselage structure due to that both tension loads and stresses will be redistributed in to the edge frames and main sills.
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12.6.3.1 Panels above and Below The Cutout:
Hoop tension loads above and below the cutout are redistributed in to edge frames via shear flows in the panels between main auxiliary sills. Hoop tension load w = pr in N/mm. Where, Applied Pressure = p in N/mm2. Radius of cross section = in mm. Shear flow in upper centre panel at the edge frame qU =
wl prl = in N/mm. 2a 2a
Shear flow in the upper panel at the centre of the panel qU = 0 Shear flow in lower centre panel at the edge frame q L =
wl prl = in N/mm. 2b 2b
Shear flow in the lower panel at the centre of the panel q L = 0 Hoop tension load in edge frames PHT =
wl prl = in N. 2 2
Where, Length of the centre panel = l in mm. Width of the upper centre panel = a in mm. Width of the lower centre panel = b in mm.
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12.6.3.2 Panels at Side of the Cutout:
Longitudinal Tension load w' =
pr in N/mm. 2
Shear flow at left side centre panel q F =
w' h prh in N/mm. = 2c 4c
Shear flow in left side centre panel at the centre of the panel q F = 0 . Shear flow at right side centre panel q A =
w' h prh in N/mm. = 2d 4d
Shear flow in right side centre panel at the centre of the panel q A = 0 . Longitudinal loads in main sills PLT =
w' h prh in N. = 2 2d
Where Height of side panel = h in mm. Width of the right side panel = d in mm Width of the left side panel = c in mm.
12.6.3.3 Corner Panels of the Cutout:
Since load is acting in the cutout, the frames and panels around the cutout will be stressed. As a result shear flow will be set in the panels. Due to the shear flows in corner panels there will be increment in the axial load in the frame and sill members. This increment is based on corner panel fixity. Distribution factors K1 and K2 are used which are based on the constant moment of inertia of the panels. Distribution factor at panels above and below the cutout K 1 = Distribution factor at panels left and right of the cutout K 2 =
h h+l
l h+l
Shear flow at upper and forward corner panel = qUF in N/mm. Shear flow at lower and forward corner panel = qLF in N/mm. Shear flow at upper and aft corner panel = qUA in N/mm. Shear flow at lower and aft corner panel = qLA in N/mm.
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Fixed end moment at upper or lower panels M UL =
wl 2 in N-mm. 12
Fixed end moment at forward or aft panels M FA =
w' h 2 in N-mm. 12
Shear flow in upper forward corner panel qUF
⎡ wl 2 ⎤ − K 1 (M UL + M FA )⎥ ⎢ 12 ⎦ =⎣ 2ac
12.7 Problem:
Problem 1: Shown a sheet frame structure the sheet is mode of material BSL 165T6511 and frame is also of some material. Thickness of sheet is 3mm and cross section of frame is 100 mm2. A cutout is made as shown below. Calculate shear flow and Reserve factor for the setup. All are Ultimate load. For BSL 165T6511 f t = 415 N / mm 2 f s = 249 N / mm 2
Shear flow in cells q =
12000 = 400 N / mm 10 + 10 + 10
For panels 4 and 5 q 4 = q5 = q + q
H2 10 = 400 + 400 = 600 N / mm 10 + 10 H1 + H 2
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for panels 2 and 3 q 2 = q3 = q + q
L2 15 = 400 + 400 = 600 N / mm 15 + 15 L1 + L3
for panels 1,3,6,8 ⎛ L2 ⎞ ⎛ 15 ⎞ 2 ⎟⎟ = 400 − 200⎜ q1 = q3 = q 6 = q8 = q − q 4 ⎜⎜ ⎟ = 300 N / mm ⎝ 15 + 15 ⎠ ⎝ L1 + L2 ⎠
Axial load at upper cap member
at A1 = 30000 N at B1 = 30000 − 15 × 400 = 24000 N at C1 = 24000 − 15 × 300 = 19500 N at D1 = 19500 − 15 × 600 = 10500 N at E1 = 10500 − 15 × 900 = 6000 N at F1 = 6000 − 15 × 400 = 0 Axial load at vertical stiffeners
at B1 = 0
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at B2 = 400 × 10 − 300 × 10 = 1000 N at B3 = 400 × 20 − 300 × 10 − 600 × 10 = −1000 N at B4 = 400 × 30 − 300 × 10 − 600 × 10 − 300 × 10 = 0 at C1 = 0 at C 2 = 300 × 10 − 600 × 10 = −3000 N at C 3 = 600 × 10 − 3000 = 3000 N at C 4 = 300 × 10 + 3000 − 600 × 10 = 0 At sheets: Maximum shear stress due to applied load τ =
q 600 = = 20 N / mm 2 t 3
Allowable shear stress f s = 249 N / mm 2 Reserve factor = 249/200 = 1.245
At Frames: Maximum tensile stress due to applied load σ =
P 30000 = = 300 N / mm 2 A 100
Allowable tensile stress f t = 415 N / mm 2 Reserve factor = 415/300 = 1.38
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Problem 2: Material 2024-T3 clad for both web and ring. Assume the allowable shear flow q is 133.86 N / mm. Determine the bending stress, tensile stress, shear flow due to the ring doublers, rivet load and reserve factors. Rivet material HS 1006 of diameter 4mm.
D = 76.2mm, R = 38.1mm q = 50 N / mm t w + t r = 3.556mm t w = 1.016mm t r = 3.556 − 1.016 = 2.54mm
w = 8.89mm D w = 30mm
R1 = R +
26.67 w = 38.1 + = 51.44mm 2 2
Bending stress
σ be =
3qRR1 3 × 50 × 51.44 × 38.1 = = 321.5 N / mm 2 t w w2 1.016 × 30 2
Tensile stress
σt =
qD 50 × 76.2 = = 50 N / mm 2 wt r 30 × 2.54
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Stress Interaction at failure Applied Ultimate tensile stress, σ = σ be + σ t = 321 .5 + 50 = 371 .5 N / mm 2 Allowable ultimate tensile stress = 465 N/mm2 Reserve factor=
ft
σ
=
465 = 1.252 371.5
Analysis for rivets ⎛ ⎞ tr 2.54 ⎛ ⎞ ⎟⎟ p max = 2 × 50⎜ Load on rivet Pr = 2q⎜⎜ ⎟15 = 1136.4 N ⎝ 2.54 + 0.8 × 1.016 ⎠ ⎝ t r + 0.8t w ⎠
Allowable load for HS 1006 for rivet diameter of 4mm and sheet thickness of 1mm is 1974N. Reserve factor = 1974/1136.4 = 1.74
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13. TENSION CLIPS:
These are also quite numerous in military airplanes, being used to splice relatively light tension loads from one member to another. It should be used only when the load is small, since it has poor fatigue life. It is usually restored to when some structural member such as a bulkhead, web of flange or fitting cannot be efficiently opened up to let an axially loaded member pass through. There is an axial tension load to be transferred and since bulkhead cannot be cut a tension clip arrangement must be shown. Simple tension clips are used for relatively light loads and are usually seen in three forms a) Single angles b) Double angles (back to back) c) Clips cut from the extruded T-section. To obtain maximum strength and stiffness bolts should be used for the attachment purposes. It is noted that for larger clip thickness the bolt may become critical and become to yield. This is because of “Prying action” on the bolt; Because of the prying action the load on the bolt is always greater than the applied load.
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The applied loading and bending moment diagram are shown in the above figure. The internal loads are determined by equating rotations at the angle corner and are checked for both Yield loads and ultimate loads. At rib leg A1 = Wt1 In mm2.
I 1= A1
t12 in mm4. 12
At rib foot A2 = Wt f in mm2. I 2 = A2
t 2f 12
in mm4.
Rib foot at bolt A2b = (W − d )t f in mm2. I 2b = A2b
t 2f 12
in mm4.
13.1 Single Angle:
The single angle is simply supported at the rib attachment 2
⎛ ⎛1⎞ ⎞ 1.5 Papp ⎜⎜ I 2 − ⎜ ⎟t1 ⎟⎟ I 1 ⎝2⎠ ⎠ ⎝ in N-mm. Moment about the angle corner M 0 = ⎛ ⎛ ⎛1⎞ ⎞ ⎛1⎞ ⎞ ⎜⎜ I 1 − ⎜ ⎟t f ⎟⎟ I 2 + 3⎜⎜ I 2 − ⎜ ⎟t1 ⎟⎟ I 1 ⎝2⎠ ⎠ ⎝2⎠ ⎠ ⎝ ⎝
13.2 Double Angle:
The double angle is clamped at the rib attachment. 2
⎛ ⎛1⎞ ⎞ 2 Papp ⎜⎜ I 2 − ⎜ ⎟t1 ⎟⎟ I 1 ⎝2⎠ ⎠ ⎝ in N-mm. Moment about the angle corner M 0 = ⎛ ⎛ ⎛1⎞ ⎞ ⎛1⎞ ⎞ ⎜⎜ I 1 − ⎜ ⎟t f ⎟⎟ I 2 + 4⎜⎜ I 2 − ⎜ ⎟t1 ⎟⎟ I 1 ⎝2⎠ ⎠ ⎝2⎠ ⎠ ⎝ ⎝
1 M1 = − M 0 2
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Yield allowable leg load is Pall =
Reserve factor =
f p2 (1 / 2)t1 M 0 + 1 in N. I1 Papp A1
Pall Papp
Ultimate allowable leg load is Pult =
Reserve factor =
ftu in N. (1 / 2)t1 M 0 1 + I1 Papp A1
Pult Papp
Applied load = Papp in N. ⎛ ⎞ ⎜ I P ⎟ Yield allowable foot load is Pall = f p 2 ⎜ 2b app ⎟ in N ⎜ 1t Mf ⎟ ⎜ f ⎟ ⎝2 ⎠
Reserve factor =
Pall Papp
⎛ ⎞ ⎜ I P ⎟ Ultimate allowable foot load is Pall = f ult ⎜ 2b app ⎟ in N. ⎜ 1t Mf ⎟ ⎜ f ⎟ ⎝2 ⎠
Reserve factor =
Pall Papp
The allowable load based on bolt is Pall =
P 1 π (0.75d )2 ftub app in N. 4 Pb
Where Pb = Papp +
M2 in N. I3
Reserve factor =
Pall Papp
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13.3 Problem:
Sectional properties and load applied on an angle cleat is listed below calculate the proof allowable leg and foot load, ultimate leg and foot load, allowable load on bolt and their corresponding reserve factors. The section properties are A1 = w × t1 = 30 × 2.5 = 75mm 2 A1t12 75 × 2.5 2 l1 = = = 39.0625mm 4 12 12 A2 = w × t f = 30 × 6 = 180mm 4 l2 =
A2 t 2f 12
=
180 × 6 2 = 540mm 4 12
A2b = (w − d ) × t f = (30 − 6.35) × 6 = 141.9mm 4 l 2b =
A2b × t 2f 12
=
141.96 2 = 425.7 mm 4 12
Diameter of nut d0 = 10.57mm The root moment is
(
)
2
1.5Papp × l 2 − 1 t12 × l1 2 M0 = 1 l1 − × t × l 2 + 3 l 2 − 1 × t1 × l1 2 f 2
(
)
(
)
1.5 × 4000 × (20 − 0.5 × 2.5) × 39.1 = = 4061.5 Nmm (36.5 − 0.5 × 6)540 + 3(20 − 0.5 × 2.5) × 39.1 2
The proof allowable leg load is Pall =
f p2 1 t1 2 M0 + 1 l1 Papp A1
=
450 = 9820 N 0.5 × 2.5 4061.5 1 × × 39.1 4000 75
The proof reserve factor for the leg is Reserve factor =
Pall 9820 × 1.5 = = 3.682 Pb 4000
The ultimate allowable leg load, based on an elastic section is
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Pall =
f tu 1 ×t 2 1 × M0 + 1 l1 Papp A1
=
515 = 11238 N 0.5 × 2.5 4061.5 1 × + 39.1 4000 75
Using Ph/M = Pappt1/M0(=2.462)and t/h = 0.5, a form factor of k = 1.04 is obtained. The ultimate allowable leg load, based on a plastic section is Pall =
f p2 1 ×t 1 × 2 1 × M0 K l1 Papp
=
450 = 14417 N 1 0.5 × 2.5 4061.5 × × 1.04 39.1 4000
The ultimate reserve factor for the leg is Reserve factor = Pall / Pb = 11238/4000 = 2.810 The maximum moment in the foot is 1 1 ⎛ ⎞ M f = Papp ⎜ l 2 − t1 − (d + d 0 )⎟ − M 0 2 4 ⎝ ⎠ = 4000(20 − 0.5 × 2.5 − 0.25(6.35 + 10.57 )) − 4061.5 = 54019 Nmm
The allowable foot load is ⎛ ⎜ l P app Pall = f p 2lt ⎜ 2b 1 ⎜ t Mf ⎜ f ⎝2
⎞ ⎟ ⎟ = 450⎛⎜ 425.7 × 4000 ⎞⎟ = 4728 N ⎟ ⎝ 0.5 × 6 54019 ⎠ ⎟ ⎠
The proof reserve factor for the root is Reserve factor = Pall / Pb = 4728/4000 = 1.773 The ultimate allowable foot load, based on a elastic section is ⎛ ⎜ l P app Pall = f tult ⎜ 2b ⎜ 1t M f ⎜ f ⎝2
⎞ ⎟ ⎟ = 515⎛⎜ 425.7 × 4000 ⎞⎟ = 5411N ⎟ ⎝ 0.5 × 6 54019 ⎠ ⎟ ⎠
a form factor of k = 1.4 is applicable. The ultimate allowable foot load, based on a plastic section is ⎛ ⎜ l Papp Pall = f p 2lt ⎜1.4 2b 1 Mf ⎜ tf ⎜ 2 ⎝
⎞ ⎟ ⎟ = 450⎛⎜1.4 425.7 × 4000 ⎞⎟ = 6620 N ⎟ ⎝ 0.5 × 6 54019 ⎠ ⎟ ⎠
The ultimate reserve factor for the root is
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Reserve factor = Pall / Pb = 5411/4000 = 1.353 The moment at the bolt is 1 ⎞ ⎛ M 2 = Papp ⎜ l 2 − t1 ⎟ − M 0 = 4000(20 − 0.5 × 2.5) − 4061.5 = 70939 2 ⎠ ⎝ The applied bolt load is Pb = Papp +
M2 70939 = 4000 + = 9589 N l3 12.7
The allowable load based on the bolt is Papp 1 1 4000 2 2 Pall = π (0.75d ) f tub = π (0.75 × 6.35) × 1100 × = 8177 N 4 Pb 4 9589 The bolt reserve factor (proof and ultimate) is Reserve factor = Pall / Pb = 8177/4000 = 2.044
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14. PRE-TENSIONED BOLTS:
Application of pre-tension loads to bolted joints improves bolt fatigue life and joint stiffness. This is generally achieved by applying torque to bolt – nut combination. The bolt is put in tension and the contact faces of the structure in compression. Subsequent tensile loading to the joint will relieve the contact face loads and increase the tensile loading to the bolt. This is illustrated in the figures given below. The torque applied to a nut to pre – tension a bolt is absorbed in three ways. Part is used to drive the mating threads up the helix, thus creating tensile stress in the bolt. The reminder is used to over come the friction between the mating threads and also to over come the friction between the thrust surfaces under the load.
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With reference to above figure, the initial preload Pi is the start point for all the curves associated with the preloaded joint. If the components being jointed do not have a contact surface then the bolt load will increase with the same magnitude as the applied load - Curve B. Alternatively if the structure has contact surfaces then the bolt load will increase at the lesser rate then the applied while the contact surfaces unload - curve C. once the faces have separated the bolt load will increase at the same rate ac the applied, along the curve D from point A. Separation is indicated by point A. The compression load on the contact surfaces reduces in proportion to applied load Curve E.
14.1 Optimum Preload:
The Optimum preload for the bolts subjected to repeated loading is the preload required such that the tact surfaces of the structure are just at the point of separating under the maximum applied load, point A in the above figure. This is given by ⎡ ⎤ ⎢ ⎥ Pi c 1 ⎢ ⎥ = 1− Am E m ⎥ Pj L⎢ ⎢1 + A E ⎥ b b ⎦ ⎣ Where Preload in bolt = Pi in N. Maximum applied Tension load = Pj in N.
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Distance between bolt head and nut = L in mm. Elastic modulus of bolt = Eb in N/mm2. Elastic modulus of fitting = Em in N/mm2. Shank area of bolt = Ab in mm2. Dearing area of bolt head on fitting Am =
π
(B 4
2
)
− D 2 in mm2.
Diameter of the bolt head = B in mm. Bolt shank diameter = D in mm. C is shown in figures below
An increase in the preload from the optimum value increases the mean stress on the bolt. A decrease from the optimum value will increase the alternating stress. In either case fatigue life will be adversely affected bur more so if alternating stress increases.
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14.2 Total Bolt Load:
The total tensile load on a bolt including preload and applied tensile load to the joint is given by the following cases 1. Mating surfaces not in initial contact 2. Mating surfaces in initial contact.
14.2.1 Mating Surfaces Not in Initial Contact:
Refer curve B Total Tensile Load Pb = Pi + Pj in N. Where, Preload in bolt = Pi in N. Maximum applied Tension load = Pj in N.
14.2.2 Mating Surfaces in the Initial Contact:
Refer curve B Maximum value of (a) or (b) to be taken
(a)
⎡ ⎢ Pb = Pi + Pj ⎢ ⎢1 + ⎢ ⎣
(b)
Pb = Pj in N.
c L Am E m Ab Eb
⎤ ⎥ ⎥ in N. ⎥ ⎥ ⎦
Where Preload in bolt = Pi in N. Maximum applied Tension load = Pj in N. Distance between bolt head and nut = L in mm. Elastic modulus of bolt = Eb in N/mm2. Elastic modulus of fitting = Em in N/mm2. Shank area of bolt = Ab in mm2.
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Dearing area of bolt head on fitting Am =
π 4
(B
2
)
− D 2 in mm2.
Diameter of the bolt head = B in mm. Bolt shank diameter = D in mm.
14.3 Preload for a given Wrench Torque: Pi
TAr R
Where, Wrench torque = T in N-mm. Root area of the bolt = Ar in mm2. Wrench Torque ratio = R. Preload in bolt = Pi in N. Torque T = R*Initial Tension Stress on Root area in N/mm2. Pi = Yield Stress * root area * 0.76 for Tension nuts in N. Pi = Yield Stress * root area * 0.28 for Shear nuts in N. Refer MIL-S-7742, MIL-S-8879 for further references.
14.4 Bolt Strength Requirements:
1. The Ultimate tension load on the bolt, including the effect of Preload shall not exceed the Ultimate strength of the bolt. 2. The applied Limit tension load on the bolt, including the effect of the Preload, based on the thread root area shall not exceed the yield strength of the bolt. 3. The effect of the Shear loading shall be including when evaluating the above points 1 and 2 (Shear is applied on the bolt shank) 4. In calculating reserve factor interactive equations shown should be applied.
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14.5 Ultimate Stress Analysis:
Reserve Factor =
1 Rt2 + Rs2
Where, Rt =
Applied Ultimate Tension Load Allowable Ultimate Tension Load
Rs =
Applied Ultimate Shear Load Allowable Ultimate Shear Load
When calculating Rs Shear is applied on the bolt shank, this area should be used in calculating the allowable shear load. Refer ESDU DATA Sheet 72022 for Tension in Steel Bolts Resulting from Tightening Torque
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