Truss Analysis Method of Sections Steven Vukazich San ose State University
General Procedure for the Analysis of Simple Trusses using the Method of Sections 1. Draw a ree Body Diagram (BD) of the entire truss cut loose from its supports and find the support reactions using the euations of euilibrium (we will see that for some truss structures this step is not always necessary); 2. Make a cut through the members of the truss that are of interest. The cut must define two separate sections of the truss; 3. Draw a BD of the section of the truss that is to be analyzed. There are three euations of euilibrium available to find unknown truss member forces; 4. Note that due to the geometry of simple trusses, several forces often intersect at a point. These points are often good points to take moment euilibrium about. Often one can isolate one unknown member force with a moment euilibrium euation.
Analysis Example Using the Method of Sections A E B C 5 kn D G I 5 kn 5 kn 5 m K 5 m Consider the idealized truss structure with a pin support at A and a roller support at. The truss is subjected to applied loads shown. ind the truss member forces, G, and GI
1. Draw a ree Body Diagram (BD) of the entire truss cut loose from its supports and find the support reactions using the euations of euilibrium (we will see that for some truss structures this step is not always necessary) A x A A y E B C 5 kn D G I 5 kn 5 kn 5 m K 5 m E y
A x Use Euilibrium to ind Support Reactions E B A A y C 5 kn D G I K 5 kn 5 kn E y! M # = 0 y = 7.5 kn
A x Use Euilibrium to ind Support Reactions E B A A y C 5 kn D G I K 5 kn 5 kn E y! ) = 0! * = 0 A y = 12.5 kn
BD Showing Support Reactions A 12.5 kn E B C 5 kn D G I 5 kn 5 kn 5 m K 7.5 kn 5 m 2. Make a cut through the members of the truss that are of interest. The cut must define two separate sections of the truss; Can use a BD of either section to find unknown member forces
BD of the Section to the Right of Cut - G G GI 5 m I K 7.5 kn Notes: Unknown truss member forces are assumed to act in tension (pulling away form the joint); Members G and GI intersect at G; Members and GI intersect at ; Members G and intersect at. 3. Draw a BD of the section of the truss that is to be analyzed. There are three euations of euilibrium available to find unknown truss member forces;
The Principle of Transmissibility The Principle of Transmissibility states: The condition of euilibrium (or motion) of a body remain unchanged if force acting at a given point on a rigid body is replaced by a force that has the same magnitude, line of action, and sense but acts at a different point. =
ind orce in Member Using the Principle of Transmissibility, slide to point 15 17 12 G 8 17 12 17 8 15! M 4 = 0 G GI 5 m I K 7.5 kn = -13.8125 kn
ind orce in Member Note that we could also have slid to point! M 4 = 0 G 5 m 17 8 G 15 GI I K 15 17 12 8 17 12 7.5 kn = -13.8125 kn
ind orce in Member GI 8 15 = 26 10 26 = 5.333 m! M 2 = 0 G G GI 5 m 5.333 m I K 7.5 kn GI = 13.125 kn
ind orce in Member G Using the Principle of Transmissibility, slide G to point G 5.333 CD θ = tan 5 sin θ = 0.72954 cos θ = 0.68394! M : = 0 = 46.85 0.68394 42 0.72954 42 5 5.333 G 46.85 GI 5 m 5.333 m I K 7.5 kn G = -1.3707 kn
BD of the Section to the eft of Cut - A 12.5 kn E B C D G 5 kn 5 kn 5 kn G GI 5.333 m! M 2 = 0 GI = 13.125 kn
Confirm orce in Member E B A C D G 5 kn 12.5 kn 5 kn 5 kn G GI 8 17 12 15 17 12! M 4 = 0 = -13.8125 kn
Confirm orce in Member G A B E C D G 5 kn 12.5 kn 5 kn 5 kn 0.78295 42 0.68394 42 GI! M : = 0 G = -1.3707 kn
Use orce Euilibrium to Check Results 13.8125 kn 1.3707 kn G I K 13.125 kn 7.5 kn 5 m Notes: Show known member forces in their actual directions; Member GI is in tension; Member G is in compression; Member is in compression. Express the inclined member forces in terms of horizontal and vertical components, and examine force euilibrium of the section.
Use orce Euilibrium to Check Results 15 17 13.8125 = 12.1875 kn G 0.68394 1.3707 = 0.9375 kn 0.72954 1.3707 =! ) = 0 8 17 13.8125 = 6.5 kn I K 13.125 kn 7.5 kn 5 m! * = 0