Topic: Distribution of force in framed structure Introductions
In the Lecture 1, it gives the general view about the nature of the force. In this Lecture, such nature of force will be applied in analyzing framed (truss) structure. The analysis of the framed structure involves determining how forces are transmitted, from their points of application, through the various members of structures to their external supports. What is frame structure?
A frame is a structure built up of three or more members, which are normally considered as being pinned or hinged at the various joints. Any loads, which are applied to the frame, are usually transmitted to it at the joints, so that the individual members are in pure tension or compression. A very simple frame is shown in Figure 1. It consists of three individual members hinged at the ends to form a triangle, and the only applied loading consists of a vertical load of W at the apex. There are also, of course, reactions at the lower corners.
Figure 1:- Frame structure
Under the action of the loads the frame tends to take the form shown in broken lines, i.e. the bottom joints move outward putting the member C in tension, and the member A and B in compression. Members A and B are termed struts and member C is termed a tie. Assumption in analyzing the frame (or ‘ truss’) structure
Figure 2:- Axial compression and Axial tension Prepared by Dr. Thomas Tong 1
In order to simplify the analysis and design of trusses, the following assumptions are commonly made: In plane trusses, all members are connected at their ends by frictionless hinged connections; whilst the members of space structures are connected by ball-and-socket joints; All external loads and support reactions are applied only at the joints of t he truss; and The centroidal axis of each member coincides with the line connecting the centers of the adjacent joints. The effect of these assumptions is that all members of the truss act as two-force members; thus each member will be either in axial tension (being pulled apart, as shown in Figure 2(b)) or in axial compression (being pushed in as shown in Figure 2(a)). The axial member forces determined from analyzing such an idealized truss are called primary forces. Internal Stability of Frame (T (Truss) russ) structure
An internally stable truss (or ‘frame structure) has members whose number and geometry is such that the truss does not change shape and remains a rigid body when subjected to a general system of forces at its joints. The truss must maintain its shape and remain rigid even when it is detached from the supports. The term internal is used here to refer to the number and arrangement of members contained within the truss. Any instability arising due to insufficient and/or improper arrangement of external supports is referred to as external.
Figure 3:- Simple Frame Structure
As the number of members in a simple frame structure consists of the 3 basic truss element members plus two additional members for each additional joint, then the total number of members, m, in an internally stable simple truss is given by the expression: m = 3 + 2( j – 3) =2 j – 3………………….Eqn.(2)
in which j equals the number of joints (including those attached to the support system. Hence according to Eqn. 2, for any plane truss having m members and j joints, if m < 2 j – 3, then the truss Prepared by Dr. Thomas Tong 2
concerned is internally unstable whilst if m ≥ 2 j − 3 , it is internally stable. Statically Determine Simple Plane frame (truss) structure
A simple plane truss is considered to be statically determined if all the external reactions, as well as all the forces in all its members, can be determined using the three laws of equilibrium. The analysis of a simple truss or frame structure having m members and j joints, and supported by r reaction components, requires the determination of m member forces and r reaction components; a total of (m + r) forces. If the simple plane truss is in equilibrium then each of its j joints must also be in equilibrium each under the action of a concurrent coplanar plane system. It is hence possible to formulate two equilibrium equation,
∑ F x
=0 and
∑ F y
=0 , for each joint giving a total number of 2j
equilibrium equations. If the number of unknowns forces for a simple plane truss is equal to the number of equilibrium equation (i.e. m + r = 2j), then all of the unknowns can be determined by solving the equilibrium equations and the said truss is statically determinate. If a simple plane truss has more unknowns than the available equilibrium equations (i.e. m +r > 2j), then not all unknowns can be determined by solving the available equilibrium equations. Such a truss is considered statically indeterminate and has more members than the minimum required for internal stability, and/or more reaction components than the minimum required for external stability. The excess members and reaction components are called redundants, and the number of excess members and reaction components is referred to as the degree of indeterminacy i, which can be determined using the expression (i = (m + r) – 2j) . The analysis of statically indeterminate structures is beyond the scope of this course and will not be considered any further. If a simple plane truss has fewer unknowns than the number of joint equilibrium equations (i.e. m + r < 2j), then the truss is considered statically instable. This static instability may be due to the said
truss having fewer members than the minimum required for internal stability, and/or fewer reaction components than the minimum required for external stability. Summarizing the above relationships: If (m + r = 2j), simple frame structure is statically determinate; If (m + r > 2j), simple frame structure is statically indeterminate; If (m + r < 2j), simple frame structure is statically unstable. The condition for static instability is both necessary and conclusive. However, the conditions for Prepared by Dr. Thomas Tong 3
both static determinacy and indeterminacy are necessary but not conclusive as a truss may have sufficient number of members and reactions, but may still be unstable due to improper arrangement of its members and/or supports. Figure 4 contains examples of statically determinate, indeterminate, and unstable trusses.
Figure 4(a) m = 17, j =10 , r = 3 m + r = 2j it is statically determine
Figure 4(b) m = 17, j = 10, r =2 m + r < 2j Unstable
Figure 4(c) m = 21, j = 10, r =3 m + r > 2j Statically Indeterminate (i = 4)
Analysis of Simple Frame Structur Structuree – Method of Joints
Simple framed structure can be analyzed graphically or analytically. In this course, the analytical method called method of joints (based on the nature of force) are introduced Before considering in detail each of the analytical method named above, it is proposed to look at Prepared by Dr. Thomas Tong 4
zero-force member since the early identification of such members can help expedite the analysis of truss. (a) Method of Joints
In the method of joints, the axial forces in the members of a plane statically determinate truss are determined by considering the equilibrium of its joints. Since the entire truss is in equilibrium, each of its joints (or structural member), or any other part of it, must also be in equilibrium. At each joint of the truss, the forces in the members and any applied loads and reaction components form a coplanar concurrent force system which must satisfy two laws of equilibrium,
∑ F y
∑ F x
=0 and
=0 , in order for the joint to be in equilibrium. Theses two laws of equilibrium must be
satisfied at each joint of the truss. tr uss. As only two equilibrium equations can be setup at each joint, hence no more than two unknown forces can be determined at any one joint using the method of joints. Once the unknown forces have been calculated for one joint, their effects on adjacent joints are known. Successive joints may be then considered until the unknown forces in all the members have been determined. (b) Procedures
The following step-by-step procedure can be used for the analysis of statically determinate simple plane trusses by the method of joints: (a) Check the truss for static determinacy, as discussed in previous section. If truss is statically determinate and stable, proceed to step (b). Otherwise, end the analysis at this stage or modify the layout and arrangement of members and supports accordingly. accordingly. (b) Draw a free-body diagram of the truss, showing all external loads and reaction components. Note: If the truss is subject to inclined loads, think about the possibility of replacing theses by their vertical and horizontal components and showing these instead in the free-body of the whole truss. (c) By inspection, identify any zero-force members and identify these in the free-body diagram of the truss by means of zeros written by the members. (d) Determine the reaction components using the methodology described in Newton third law or by using moment of forces. Note down their values on the free-body diagram of the truss. Note: If there is a suitable joint with two or fewer unknown forces acing on it, it might not be Prepared by Dr. Thomas Tong 5
necessary to carry out step (d). Check before doing so! (e) Examine the free-body diagram of the frame to select a suitable joint with two or fewer unknown forces acting on it. (f)
Draw a free-body diagram of the selected joint, showing tensile forces by arrows pulling away from the joint and compressive forces by arrows pushing into the joint. Assume unknown member forces to be tensile.
(g) Determine the unknown forces by applying the two laws of equilibrium,
∑ F y
∑ F x
=0 and
=0 . A positive answer for a member force means that the member is in tension, as
initially assumed, whereas a negative answer indicates that the member is in compression. (h) If all the desired member forces and reactions have been determined then go to the next step. Otherwise, return to step (e). (i)
It is always good practice to check the results. Possible checks: (1) Select joints not used in calculations and verify that each obeys
∑ Fx =0
and
∑ Fy =0 ;
and (2) Recalculate the
reaction components using an alternative method. (j)
After all member forces and reaction components have been determined, draw a force summary diagram.
(c) Zero-force Members
Simple frame structures are usually designed to support several different loading conditions. Because of this, it is common to find truss members with zero forces in them when one of these trusses is being analyzed for a particular loading condition. The analysis of simple framed structure can be expedited if we can identify the zero-force members by inspection. Two common types of member arrangements that result in zero-force members are: 1.
If two non-collinear members are connected to a joint that has no external loads or reactions applied to it, then the force in both members is zero; and
2.
If three members, two of which are collinear, are connected to a joint that has no external loads or reactions applied to it, then the force in the member that is not collinear is zero.
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(d) Roller and Fixed Point
There are two different method to locate the frame structure on the ground or any fixed point. (1) Roller Roller only restrict the vertical movement but allow the horizontal movement of the frame structure. The free body diagram of the roller can be represented by:
Figure 5:- free body diagram of roller
(2) Fixed Point Fixed point restrict both vertical and horizontal movement of the frame structure. The free body diagram of the fixed point can be represented by the:
Figure 6:- Free body diagram of fixed point
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Example:
Determine the force of all force f orce in the frame structure as shown in Figure 7.
Figure 7:- Frame structure Solution: Step (a)
m = 9, j = 6, and r = 3 ∴ m + r = 2j and m = 2j – 3
Therefore truss is statically determinate, and internally and externally stable. Step (b) Figure 8(a) shows a free-body diagram of frame structure including all external loads and reaction
component Remarks: The red line demonstrate the idea in resolution of force in inclined force for this problem, the same idea is applied in the joint 4 and 5)
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Figure 8(a):- Free body diagram of frame structure Step (c)
Determining the reaction components using the moment and Newton’s Newton’s third law and gives: Take moment about point 5,
(
R1v (8) = 1.273 4
cos 45
)+ 2.545(4 cos 45)
R1v = +1.800 kN By Newton’s third law o
o
o
R5v + R1v = (1.273 cos45 ) + 6 (2.545 cos45 ) + 8 (1.273 cos45 ) R1v = +1.800 kN Similarly, o
o
o
R1H = (1.273 sin45 ) + (2.545 sin45 ) + (1.273 sin45 ) R1H = +3.600 kN
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Step (d), (e), (f)
Three joints with 2 or less unknown forces available, is (1), (5). Assume all unknown force in this stage is being TENSILE FORCE
Figure 8(b) shows a free-body of joint
(1). As the joint is in equilibrium, hence
∑ Fx =0
and
∑ Fy =0 . This enables
the setting up of two linear equations as below:
Figure 8(b) o
+F12 x (sin45 ) + 1.800 = 0 +F12 x (cos45o) +F16 + 3.600 = 0 Solving these two linear equations gives the following values: F12 = -2.546 kN (compression) ( compression) F16 = -1.800 kN (compression) ( compression) Repeat steps (d),(e),(f)
Two joints with two or less unknown forces available, namely: (2) or (5). Select joint (2), ( 2), say, say, Figure 8(c) shows a joint (2). As the
force F12 and force F 23 are collinear and the Force F26 is zero, hence the forces F 12 and F23 must be equal in magnitude and opposite in sense if joint (2) is in equilibrium.
Figure 8(c) F 12 + F 23 = 0 ∴ F23 = -2.546 kN (compression) Prepared by Dr. Thomas Tong 10
Repeat step (d), (e), (f)
Two joints with 2 or less unknown forces available, namely: (3) or (5). Select joint (3). Say. Figure 8(d) shows a free-body diagram
of
joint
(3).
equilibrium,
∑ F y
As hence
the
joint
∑ F x
=0
is
in and
=0 . This enables the setting up of
two linear equations as shown in the following page. Figure 8(d) − F 34 cos 45 − F 36 − 1.273 sin 45 + 2.546 cos 45 = 0 + F 34 sin 45 − 1.273 cos 45 + 2.546 sin 45 = 0
Solving these two linear equation gives the following values:∴ F34 = -1.273 kN(compression) ∴ F36 = +1.800 kN (compression)
Repeat Steps (d), (e) and (f)
Three joints with two or less l ess forces, namely (4), (5), (6). Select joint (4), ( 4), say. say. Figure 8(e) shows a free-body diagram of
joint (4). As the 2.545 kN external force has no effect in a direction at right angle to its line of action and is also collinear with F46, therefore F46 must be equal in magnitude but opposite in sense to it if the joint (4) is in equilibrium. Also, as the forcesF34 and F45 are collinear and the other two forces have no effect in their Figure 8(e)
direction, they must also be equal in magnitude and opposite in sense if joint (4) is in equilibrium.
∴ F46 = -2.545 kN (compression) ∴ F46 = -1.273 kN (compression)
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Repeat Steps (d), (e) and (f)
Two joints with 2 or less unknown force, namely: (5) or (6). Select joint (5), say. Figure 8(f) shows a free-body of joint (5). As
the joint is in equilibrium, hence and
∑ Fx =0
∑ Fy =0 . This enables the setting up of
two linear equations. However only one needs to be set up in order to determine the magnitude and sense of F 56. Figure 8(f) o
o
-F56 + 1.276 x Cos45 – 1.273 x Cos45 = 0 Solving this linear equation gives the following value: ∴ F 56 = 0
Step (g) Check on results obtained Figure 8(g) shows a free-body of joint (6). If
all calculations are correct, then
∑ Fy
for joint (6) must equal zero.
Figure 8(g) o
Horizontal equilibrium: equilibrium: +1.800 – 2.545 x Cos45
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∑ Fx
and
+1.800 – 1.800 1.800 = 0 ok Vertical equilibrium: equilib rium:
+1.800 – 2.545 x Sin45
o
+1.800 – 1.800 1.800 = 0 ok It seems that all calculations are correct.
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Tutorial Exercise Question 1
Determine the following structure whether it is statically determine, unstable or statically indeterminate.
(a)
(b)
(c)
(d)
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Question 2
Determine the force in framed structure (remarks: reaction not required)
Question 3
Determine the force in framed structure.
-END OF LECTURE NOTE 8-
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