CIVE 210 Chapter 6, Spring 2017

1/11/2013 Chapter 6 - Structural Analysis 1 STATICS: CE201 Chapter 6 Structural Analysis Notes are prepared based on: Engineering Mechanics, Statics by R. C. Hibbeler, 12E Pearson Dr M. Touahmia & Dr M. Boukendakdji Civil Engineering Department, University of Hail (2012/2013) 1 Chapter 6 -Structural Analysis 6. Structural Analysis ________________________________________________________________________________________________________________________________________________ Main Goal: To show how to determine the forces in the members of a truss using the method of joints and the method of section. Contents: 6.1 Simple Trusses 6.2 The Method of Joints 6.3 Zero Force Members 6.4 The Method of Sections 2 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 2 6.1 Simple Trusses  A truss is a structure composed of slender members joined together at their end points.  The members are usually wooden struts or metal bars.  Planar trusses lie in a single plane and are often used to support roofs and bridges.  The roof load is transmitted to the truss at the joints by means of a series of purlins. 3 Chapter 6 -Structural Analysis 6.1 Simple Trusses  The analysis of the forces developed in the truss members will be two-dimensional.  In the case of a bridge, the load on the deck is first transmitted to stringers, then to floor beams, and finally to the joints of the two supporting side trusses.  The bridge truss loading is also coplanar. 4 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 3 6.1 Simple Trusses Assumption for Design:  To design both the members and the connections of a truss, it is necessary first to determine the force developed in each member when the truss is subjected to a given loading.  Each truss member is considered as a two-force member so that the forces at the ends of the member must be directed along the axis of the member. All loadings are applied at the joints. The members are joined together by smooth pins. 5 Chapter 6 -Structural Analysis 6.1 Simple Trusses Assumption for Design:  The joint connections are usually formed by bolting or welding the ends of the members to a common plate, called a gusset plate, or by simply passing a large bolt or pin through each of the members. 6 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 4 6.1 Simple Trusses Assumption for Design:  The force acting at each end of the member is directed along the axis of the member.  If this tends to elongate the member, it is a tensile force (T).  If it tends to shorten the member, it is a compressive force (C). 7 Chapter 6 -Structural Analysis 6.1 Simple Trusses  To prevent collapse, the form of a truss must be rigid. The simplest form which is rigid or stable is a triangle.  Consequently, a simple truss is constructed by starting with a basic triangular element.  Additional elements consisting of two members and a joint are added to the triangular element to form a simple truss. 8 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 5 6.2 The Method of Joints  The method of joint is based on the fact that if the entire truss is in equilibrium, then each of its joints is also in equilibrium.  Therefore, if the free-body diagram of each joint is drawn, the force equilibrium equations can then be used to obtain the member forces acting on each joint.  Since the members of a plane truss are straight two-force members lying in a single plane, each joint is subjected to a force system that is coplanar and concurrent.  As a result, and need to be satisfied for equilibrium. 9 Chapter 6 -Structural Analysis  0  Fx  0  y F 6.2 The Method of Joints For example consider the pin at joint B of the truss:  Three forces act on the pin: the 500 N force and the forces exerted by members BA and BC.  is pulling on the pin, which means that member BA is in tension.  is pushing on the pin, and consequently member BC is in compression. 10 Chapter 6 -Structural Analysis BA F FBC 1/11/2013 Chapter 6 - Structural Analysis 6 6.2 The Method of Joints  When using the method of joints, always start at a joint having at least one known force and at most two unknown forces.  and can be solved to determine the two unknowns.  The correct sense of an unknown member force can be determined using one of two possible methods:  The correct sense of direction of an unknown member force can be determined by inspection.  Assume the unknown member forces acting on the joint’s free-body diagram to be in tension. 11 Chapter 6 -Structural Analysis  0  Fx  0  y F Example 1  Determine the force in each member of the truss and indicate whether the members are in tension or compression. 12 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 7 Solution 1  Since we should have no more than two unknown forces at the joint and at least one known force acting there, we will begin our analysis at joint B. Joint B:  Applying the equations of equilibrium, we have  Since the force in member BC has been calculated, we can proceed to analyze joint C 13 Chapter 6 -Structural Analysis  0  Fx 0 sin 45 500N   o FBC BC  707 1. N (C) F  0  Fy 0 cos45   BA o BC F F (T) BA  500N F Solution 1 Joint C: 14 Chapter 6 -Structural Analysis  0  Fx CA  500N (T) F  0  Fy y  500N C 0 707 1. cos45 N    o CA F 0 707 1. sin 45 N   o Cy 1/11/2013 Chapter 6 - Structural Analysis 8 Solution 1 Joint A: 15 Chapter 6 -Structural Analysis 0 500   Ax N  0  x F  0  y F N Ax 500  0 500   Ay N N Ay 500  Solution 1 Note: The results of the analysis are summarized below, which shows the free-body diagram for each joint and member: 16 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 9 Example 2  Determine the force in each member of the truss and indicate whether the members are in tension or compression. 17 Chapter 6 -Structural Analysis Solution 2 Support Reactions: No joint can be analyzed until the support reactions are determined, because each joint has more than three unknown forces acting on it. Free-body diagram of the truss: 18 Chapter 6 -Structural Analysis  0  x F 0 600   Cx N y  600 N A  0  y F N Cx 600   0  C M       0 600N 4 m 400N 3m 6 m     y A 0 400N 600N    Cy Cy  200N 1/11/2013 Chapter 6 - Structural Analysis 10 Solution 2  The analysis can now start at either joint A or C: Joint A: FAB is assumed to be compressive and FAD is tensile  Applying the equations of equilibrium, we have: 19 Chapter 6 -Structural Analysis  0  Fy 750 N( ) C FAB   0  Fx AD  450 N(T) F 0 5 4 600N   AB F   0 5 750 3  AD  F Solution 2 Joint D: Use the result for FAD  The negative sign indicates that FDB acts in the opposite sense to that shown in the figure.  To determine FDC, we can correct the sense of FDB on the free body diagram. 20 Chapter 6 -Structural Analysis  0  x F DB  250N F  0  y F DB  250N (T) F 0 600N 5 3 450N     DB F   0 250N 5 4     FDC FDC  200N (C) 1/11/2013 Chapter 6 - Structural Analysis 11 Solution 2 Joint C: (Check) 21 Chapter 6 -Structural Analysis  0  x F  0  y F FCB  600N (C) 0 600N  CB  F 0 200N 200N   Solution 2 Note: The analysis is summarized below, which shows the free-body diagram for each joint and member: 22 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 12 6.3 Zero-Force Members  The zero-force members of a truss can generally be found by inspection of each of the joints.  For example, consider the truss below.  At Joint A, it is seen that members AB and AF are zero- force members: 23 Chapter 6 -Structural Analysis  0  x F  0  y F AB  0 F AF  0 F 6.3 Zero-Force Members  At Joint D, it is seen that members DC and DE are zero-force members.  From these observations, we can conclude that if only two members form a truss joint and no external load or support reaction is applied to the joint, the two members must be zero-force members. 24 Chapter 6 -Structural Analysis  0  x F  0  Fy 0 sin   FDC DC  0 F FDE  0 1/11/2013 Chapter 6 - Structural Analysis 13 6.3 Zero-Force Members  Now consider this truss:  At Joint D, it is seen that DA is zero-force member: 25 Chapter 6 -Structural Analysis  0  x F  0  y F DA  0 F DE DC F F  6.3 Zero-Force Members  At Joint D, it is seen that DA is zero-force member.  In general then, if three members form a truss joint for which two of the members are collinear, the third member is a zero-force member provided no external force or support reaction is applied to the joint. 26 Chapter 6 -Structural Analysis  0  x F  0  y F CA  0 F CD CB F F  0 sin   FCA 1/11/2013 Chapter 6 - Structural Analysis 14 6.4 The Method of Sections  When we need to find the force in only a few members of a truss we can analyze the truss using the method of sections.  This method is based on the principle that if the truss is in equilibrium then any segment of the truss is also in equilibrium.  If the forces within the members are to be determined, then an imaginary section can be used to cut each member into two parts. 27 Chapter 6 -Structural Analysis 6.4 The Method of Sections  The method of sections can also be used to cut or section the members of an entire truss. 28 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 15 6.4 The Method of Sections  For example, consider the truss below:  If the forces in members BC, GC, and GF are to be determined, then section a-a would be appropriate. 29 Chapter 6 -Structural Analysis 6.4 The Method of Sections  The free-body diagrams of the two segments are then:  Members BC and GC are assumed to be in tension since they are subjected to a pull, whereas GF in compression since it is subjected to a push. 30 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 16 6.4 The Method of Sections  The Three unknown member forces FBC, FGC, and FGF can be obtained by applying the three equilibrium equations of the segments.  The three support reactions Dx, Dy and Ex can be obtained by considering a free-body diagram of the entire truss. 31 Chapter 6 -Structural Analysis Example 2  Determine the force in members GE, GC, and BC of the truss. Indicate whether the members are in tension or compression. 32 Chapter 6 -Structural Analysis 1/11/2013 Chapter 6 - Structural Analysis 17 Solution 2  Section a-a has been chosen since it cuts through the three members whose forces are to be determined.  It is first necessary to determine the external reactions at A and D. 33 Chapter 6 -Structural Analysis  0  Fx 0 400N   x A x  400N A  0  MA       0 12m 400N 3m 1200N 8m     y D y  900N D  0  Fy 0 900N 1200N   Ay  y  300N A Solution 2 Free-Body Diagram: For the analysis the free-body diagram of the left portion of the sectioned truss will be used. Equations of Equilibrium: 34 Chapter 6 -Structural Analysis  0  G M BC  800N (T) F  0  C M       0 3m 400N 3m 300N 4m     BC F GE  800N (C) F  0  Fy 0 5 3 300N   FGC GC  500N (T) F     0 3m 300N 8m    FGE