STRENGTH OF MATERIALS (Strenma)
MECHANICS OF DEFORMABLE BODIES
Why do we study Strenma? • Engineers are responsible for design, construction and testing of devices we use from simple to complicated ones • To be familiar with mathematical models to predict system behavior • To learn how to analyze and predict behavior of physical systems of objects • To know the “application to engineering” of “strenma” – in design and safety
What is Strenma? • A branch of applied mechanics that deals with the behavior of solid bodies subjected to various types of loading. • The solid bodies include axially loaded members, shafts in torsion, thin shells, beams, and columns, as well as structures that are assemblies of these components
• The objectives of a strength of materials analysis are the determination of the stresses, strains, and displacements, produced by the loads. • Knowing these quantities for all values of load up to the failure load gives a complete picture of the mechanical behavior of the body.
Stress • Normal stress – strength of a material per unit area or unit strength. It is the force on a member divided by the area which carries the force. Or, ζ = P / A
• Units of stress : Pa or N/m2 , MPa or N/mm2, psi or Lbs/in2 • The maximum stress in tension or compression occurs over a section normal or perpendicular to the load.
• Shearing stress – or tangential stress caused by forces parallel to the area resisting the force. Or, η = V / A • V is the resultant shearing force which passes through the centroid of the area being sheared.
• Single shear or double shear
• Bearing stress – the contact pressure between separate bodies • Differs form compressive stress, it is an internal stress caused by compressive forces • ζ b = P b / Ab
• Stress in thin-walled cylinders A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile forces, which resist bursting, developed across longitudinal and transverse sections. • Tangential stress or circumferential stress, ζt = pD / 2t • Longitudinal stress, ζL = pD / 4t
• If there exist an external pressure po and internal pressure pi, the formula will be, ζt = (pi – po )D / 2t ζL = (pi – po )D / 4t • It can be observed that tangential stress is twice the longitudinal stress, ζt = 2 ζL , thus for a spherical shell, the stress at the wall is, ζL = pD / 4t
Sample Problems in Simple Stress • Determine the largest weight W which can be supported by the two cables. The stress in cables AB and AC are not to exceed 100 MPa and 150 MPA stress respectively. The cross-sectional areas of cable AB is 400mm2 and 200 mm2 for AC.
Sample Problems in Simple Stress • For the truss shown, calculate the stress in members DF, CE and BD. The crosssectional area of each member is 1200mm2. Indicate whether tension or compression.
Sample Problems in Simple Stress • The bars of the pin-connected frame are 30mm by 60mm in section. Determine the maximum load P that can be carried so that the stresses will not exceed 100MN/m2 in tension or 80 MN/m2 in compression.
Sample Problems in Simple Stress •
A timber column 12 x 12 inches in cross section is reinforced on each side by a steel plate 12 inches wide x t inches thick. What is the value of thickness, t if the column can support an axial load of 500 kips without exceeding the allowable timber stress of 1500 psi and allowable steel stress of 30,000 psi. Use Es = 200 GPa and Et = 10 GPa.
STRAIN • •
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Ratio of the change in length caused by an applied force, to the original length. ε = δ / L , δ is elongation, L is original length. ε is dimensionless. Proportional limit (Hooke’s Law) – the stress-strain curve is a straight line. Within this limit, stress is directly proportional to strain or ζ = k ε k is a constant of proportionality called Modulus of Elasticity, E. or σ = E ε
STRAIN •
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Elastic limit – limit beyond which the material will no longer go back to its original shape when the load is removed Elastic and plastic ranges – the region from O to P is the elastic range, from P to R (rupture) is the plastic range. Yield point – the material will have an appreciable elongation or yielding without any increase of load.
STRAIN • •
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Ultimate strength – maximum point in the stress-strain diagram, also tensile strength Rupture strength – strength at rupture or known as breaking strength Working stress – actual stress of a material under a given loading Allowable stress – maximum safe stress that a material can carry, limited to values not exceeding the proportional limit
STRAIN •
Factor of safety – ratio of ultimate or yield strength to the allowable strength, or F.S. = yield stress / allowable stress or = ultimate stress / allowable stress
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Axial deformation, δ = PL / AE, the load must be axial and the bar must have a uniform cross-sectional area, stress must not exceed the proportional limit.
STRAIN • •
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Stiffness – the ratio of steady force acting on elastic body to a resulting displacement, K = P / δ , (N/mm unit)
Shearing deformation, δs – an element subject to shear undergoes change in shape. Shear strain – change in angle at the corner of an original rectangular element, or γ = δs / L
STRAIN •
Modulus of rigidity, G – the ratio of shear stress η and shear strain γ, or G = η / γ ( MPa unit)
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Given an applied shearing force V acting over an area As , δs = V L / As G = η L / G where η = V / As
POISSON’S RATIO •
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When a bar is subjected to a tensile load, there is an increase in length in the direction of the applied load, but there is a decrease in lateral dimension perpendicular to the load. The ratio of sidewise deformation (strain) to longitudinal deformation (strain) is the Poisson’s ratio denoted by μ. For steel, μ lies in the range 0.25 to 0.30 and 0.20 for concrete.
POISSON’S RATIO •
μ = - εy / ε x = - εz / εx where εx is the strain in X direction and εy and εz are strains in the perpendicular direction. The negative sign indicates decrease in transverse dimension when εx is positive.
Biaxial Deformation •
If an element is subjected simultaneously by tensile stresses, σx and σ y in X and Y directions, the strain in X direction is σx / E and the strain in Y direction is σy / E . The stress in Y will produce a lateral contraction on X direction of – μεy or – μ σy / E . The resulting strain in X direction is εx = σ x / E – μ σ y / E or σx = (εx + μ εy ) E / (1 – μ2)
and
Biaxial Deformation εy = σ y / E – μ σ x / E
or
σy = (εy + μ εx ) E / (1 – μ2) •
Triaxial deformation εx , εy , εz εx = 1 / E [ σx – μ (σy + σz ) ]
εy = 1 / E [ σy – μ (σx + σz ) ] εz = 1 / E [ σz – μ (σx + σy ) ]
Sample Problems • A timber column 3 m high has a rectangular cross section, 100mm x 200 mm and is reinforced over its entire length by 2 steel plates each 200mm wide x 10mm thick attached to the 200mm faces. The column is designed to carry a load of 100 KN. The failure stress of timber is 55N/mm2. Check the design using a factor of safety of 3. the allowable stress for steel is 110 MPa and for timber is 5.6 MPa. Et = 15,000 N/mm2 and Es = 200,000N.mm2.
Sample Problems • A block is subjected to a triaxial stress of 16 MPa on the YZ face, -11 MPa on the XZ face, and 52 MPa along the XY face. If these stresses produce a strain in the x direction of 49.2 x 10 -6 m/m, what is the Poisson’s ratio of the block? Assume E of the block is 83,000 MPa.
Torsion • Consider a bar to be rigidly attached at one end and twisted at the other end by a torque or twisting moment T equivalent to F x d, applied perpendicular to the axis of the bar. Such bar is said to be in torsion. • Torsional shearing stress, η For a solid of hollow circular shaft subject to a twisting moment T, the torsional shear stress at a distance ρ from the center of the shaft is, τ = T ρ / J and τmax ( Ss ) = T r / J
Torsion • J is the polar moment of inertia of the section and r is the outer radius. • For solid cylindrical shaft, J = π D4 / 32 Ss = 16 T / π D3 • For hallow cylindrical shaft, J = π(D4 - d4) / 32 Ss = 16 TD / π (D4 - d4)
Angle of Twist • The angle θ through which the bar of length L will twist is, • θ = TL / JG where, T is torque in N-mm, L is the length of shaft in mm, G is shear modulus in MPa, J is polar moment of inertia in mm4, D and d are diameter in mm, and r is radius in mm.
Problems • A solid steel shaft having a diameter of 100 mm has an allowable shearing stress of 80 MPa. Determine the maximum torque in KNm that the shaft can carry without exceeding the allowable shearing stress. Solution : Ss = Tr / J = 16 T / π D3 80 = 16T / π (100)3 T = 15,707,963 N-mm = 15.71 KN-m
Problems • A hallow steel shaft has an OD of 220 mm and an ID of 200 mm. determine the torque in KN-m that will cause a stress of 75.5 MPa in the shaft. Solution : Ss = Tr / J 75.5 = T (220/2) / π(2204 - 2004) / 32 T = 50 KN-m
Exercises • Determine the maximum allowable torque in KN-m for a 50 mm diameter solid steel shaft if the allowable shearing stress is 81.5 MPa.
• Determine the rotation or twist in degrees of a shaft 800 mm long subjected to a torque of 80 N-m, if diameter is 20 mm and shear modulus G is 80,000 MPa.
Exercises • Determine the length of the shortest 2 mm diameter bronze wire which can be twisted through two complete turns without exceeding a stress of 70 MPa. Use G = 35 GPa. • A hallow steel shaft 2540 mm long must transmit torque of 34 KN-m. the total angle of twist must not exceed 3 deg. The maximum shearing stress must not exceed 110 MPa. Find the ID and OD of the shaft. G = 83 GPa.
Thermal Stress • Temperature changes cause a body to expand or contract. The amount of linear deformation, δT, is •
δT = α L (Tf – Ti ) = α L ΔT where α is coefficient of thermal expansion in m/m-0C, L is length in meter, Tf and Ti is final and initial temperatures in 0C. For steel, α = 11.25 x 10-6 / 0C.
Thermal Stress • If temperature deformation is permitted to occur freely, no load or stress will be induced in the structure. Where temperature deformation is not permitted, an internal stress is created. This internal stress is called thermal stress. • For a homogeneous rod mounted between unyielding supports, the thermal stress is, - elongation due to temperature change; δT = α L (Tf – Ti ) = α L ΔT
• Elongation due to equivalent axial stress; δp = PL / AE = ζ L / E δT = δp α L ΔT = ζ L / E ζ = E α ΔT where ζ is thermal stress in MPa, E is modulus of elasticity of the rod in MPa. • If the wall yields a distance “a”, δT = x + δp α L ΔT = a + ζ L / E • If T rises above normal, the rod is in compression, if T drops below normal, the rod is in tension.
Sample Problem 1 • A rod with cross-section area of 100 mm2 is stretched between 2 fixed pts such that the initial tensile load is 10 KN. E of rod is 180 GPa and α is 0.00001 m/m-0C. The initial tension is designed such that the rod will have zero stress when there is a certain change in temp. of the environment. Determine the temperature change.
Sample Problem 1 : Solution • Initial stress in tension is, • ζ = P/A = 10,000 N / 100 mm2 = 100 MPa • For the stress to be zero, • ζT = E α ΔT = 100 MPa in compression, or the rod must be heated (change in T is + ),
• 100 MPa =180x103MPa (0.00001 m/m-0C)ΔT ΔT = 55.56 0C
Exercise : Problem 2 • A 5 m. rod has a uniform cross sectional area of 1000 mm2 is secured between 2 walls 5 meters apart. The load on the rod is zero at 27 0C. E = 200 GPa, α = 11.25 x 10-6 m/m-0C. If T rises to 57 0C, a. what is the stress in the rod assuming the walls are rigid. b. what is the stress in the rod if the walls spring together a total distance of 0.5 mm.
Exercise Problem 2 : Solution • a. if the walls are rigid, ζT = E α ΔT ζT = (200,000MPa) x (11.25x 10-6 m/m-0C) x (57-27) 0C = 67.5 MPa, compression b. if the walls spring together by 0.5 mm, δT = α L ΔT = 11.25x 10-6 m/m-0C x 5000 mm x (57 – 27)0C = 1.6875 mm δp = 1.6875 mm – 0.5 mm = ζ L / E 1.1875 mm = ζ (5000mm) / 200,000 MPa ζ = 47.5 MPa, compression.
Shear and Moment in Beams • Beam is a bar subject to forces or couples that lie in a plane containing the longitudinal of the bar. A beam may be determinate or indeterminate. • Statically determinate beams – the reactions of the supports may be determined by the use of the equations of static equilibrium. Ex. cantilever beam, simple beam, overhanging beam
Shear and Moment in Beams • Statically indeterminate beams – the number of reactions exerted on the beam exceeds the number of equations in static equilibrium. To solve the reactions of the beam, the static equations must be supplemented by equations based on elastic deformations of the beam.
• Degree of indeterminacy is the difference between the number of reactions and the number of equations in static equilibrium that can be applied.
Shear and Moment in Beams • Examples of indeterminate beams – propped beams, fixed or restrained beams, continuous beams.
• Types of loading – – – – –
concentrated load uniform load uniformly varying load applied couple combination of above loading
Shear and Moment Diagram • Consider a simple beam of length L and carries a uniform load w (N/m) over its entire length, is held in equilibrium by R1 and R2. • Assume the beam is cut at C, x distance from 1 and the portion to the right of C is removed. • The portion removed must be replaced by a vertical shear force V and a couple M, to hold the left portion of the bar in equilibrium under the action of force R1 and wx. • The couple M is called the resisting moment and force V is the resisting shear.
Load, Shear and Moment • The vertical shear at C is Vc = R1 – wx where R1 = R2 = wL/2. Thus, Vc = wL/2 – wx • The moment at C is Mc = (wL/2)x – (wx)(x/2) Mc = wLx/2 – wx2/2 • If we differentiate M with respect to x, dM/dx = wL/2 – wx = shear. Thus, dM/dx = V
Load, Shear and Moment • The rate of change of bending moment wrt x is equal to shearing force, or the slope of the moment diagram at a given point is the shear at that point. • Differentiate V wrt x gives dV/dx = 0 – w = load dV/dx = Load • The rate of change of shearing force wrt x is equal to the load or the slope of the shear diagram at a given point equals the load at that point.
Properties of Shear and Moment • The area of the shear diagram to the left or to the right of a section is equal to the moment at that section. • The slope of the moment diagram at a given point is the shear at that point. • The slope of the shear diagram at a given point equals the load at that point. • The maximum moment occurs at points of zero shear. When the shear is zero, the tangent to the moment diagram is horizontal.
Properties of Shear and Moment • When the shear diagram is increasing, the moment diagram is concave upward. • When the shear diagram is decreasing, the moment diagram is concave downward.
• Sign conventions – A force that tends to bend the beam downward is said to produce a positive bending moment. A force that tends to shear the left portion of the beam upward wrt the right portion is said to produce a positive shearing force.
