Eleventh E 6 Analysis CHAPTER VECTOR MECHANICS OR ENGINEERS: STATICS erdinand P. Beer E. Russell Johnston, Jr. David. Mazurek of Structures
Contents Application Introduction Definition of a Truss Simple Trusses Method of Joints Joints Under Special Loading Cons Space Trusses Sample Problem 6.1 Method of Sections Trusses Made of Several Simple Trusses Sample Problem 6.3 Analysis of a rame rames That Collapse Without Supports Sample Problem 6.4 Machines 6-2
Application 6-3
Introduction or the equilibrium of structures made of several connected parts, the internal forces as well the external forces are considered. In the interaction between connected parts, Newton s 3 rd Law states that the forces of action and reaction between bodies in contact have the same magnitude, same line of action, and opposite sense. Three categories of engineering structures are considered: a) Trusses: formed from two-force members, i.e., straight members with end point connections and forces that act only at these end points. b) rames: contain at least one multi-force member, i.e., member acted upon by 3 or more forces. c) Machines: structures containing moving parts designed to transmit and modify forces. 6-4
Definition of a Truss A truss consists of straight members connected at joints. No member is continuous through a joint. Most structures are made of several trusses joined together to form a space framework. Each truss carries those loads which act in its plane and may be treated as a two-dimensional structure. Bolted or welded connections are assumed to be pinned together. orces acting at the member ends reduce to a single force and no couple. Only twoforce members are considered. When forces tend to pull the member apart, it is in tension. When the forces tend to compress the member, it is in compression. 6-5
Definition of a Truss Members of a truss are slender and not capable of supporting large lateral loads. Loads must be applied at the joints. 6-6
Definition of a Truss 6-7
Simple Trusses A rigid truss will not collapse under the application of a load. A simple truss is constructed by successively adding two members and one connection to the basic triangular truss. 6-8
Method of Joints Dismember the truss and create a freebody diagram for each member and pin. Cons for equilibrium for the entire truss can be used to solve for 3 support reactions. The two forces exerted on each member are equal, have the same line of action, and opposite sense. orces exerted by a member on the pins or joints at its ends are directed along the member and equal and opposite. Cons of equilibrium are used to solve for 2 unknown forces at each pin (or joint), giving a total of 2n solutions, where n=number of joints. orces are found by solving for unknown forces while moving from joint to joint sequentially. 6-9
Joints Under Special Loading Cons orces in opposite members intersecting in two straight lines at a joint are equal. The forces in two opposite members are equal when a load is aligned with a third member. The third member force is equal to the load (including zero load). The forces in two members connected at a joint are equal if the members are aligned and zero otherwise. Recognition of joints under special loading cons simplifies a truss analysis. 6-10
Space Trusses An elementary space truss consists of 6 members connected at 4 joints to form a tetrahedron. A simple space truss is formed and can be extended when 3 new members and 1 joint are added at the same time. In a simple space truss, m = 3n - 6 where m is the number of members and n is the number of joints. Cons of equilibrium for the joints provide 3n equations. or a simple truss, 3n = m + 6 and the equations can be solved for m member forces and 6 support reactions. Equilibrium for the entire truss provides 6 adal equations which are not independent of the joint equations. 6-11
Sample Problem 6.1 STRATEGY: What s the first step to solving this problem? Think, then discuss this with a neighbor. DRAW THE REE BODY DIAGRAM OR THE ENTIRE TRUSS (always first) and solve for the 3 support reactions Using the method of joints, determine the force in each member of the truss. Draw this BD and compare your sketch with a neighbor. Discuss with each other any differences. 6-12
Sample Problem 6.1 MODELING and ANALYSIS: Based on a free body diagram of the entire truss, solve the 3 equilibrium equations for the reactions at E and C. Looking at the BD, which sum of moments equation could you apply in order to find one of the unknown reactions with just this one equation? M C 0 E 2000 lb24 ft 1000 lb12 ft 6 ft E 10,000 lb Next, apply the remaining equilibrium cons to find the remaining 2 support reactions. 0 C 0 x y x C x 0 2000 lb -1000 lb 10,000 lb C C y y 7000lb 6-13
Sample Problem 6.1 We now solve the problem by moving sequentially from joint to joint and solving the associated BD for the unknown forces. Which joint should you start with, and why? Think, then discuss with a neighbor. Joints A or C are equally good because each has only 2 unknown forces. Use joint A and draw its BD and find the unknown forces. 2000lb AB AD 4 3 5 Which joint should you move to next, and why? Discuss. AB AD 1500 lb 2500 lb Joint D, since it has 2 unknowns remaining (joint B has 3). Draw the BD and solve. T C DB DA DE 2 3 5 DA DB DE 2500lb 3000lb T C 6-14
Sample Problem 6.1 There are now only two unknown member forces at joint B. Assume both are in tension. y BE 0 1000 3750 lb 4 5 2500 4 5 BE BE 3750lb C x BC 0 BC 5250 lb 1500 3 5 2500 3 3750 BC 5 5250 lb T There is one remaining unknown member force at joint E (or C). Use joint E and assume the member is in tension. x EC 0 3 5 EC 8750 lb 3000 3 5 3750 EC 8750lb C 6-15
Sample Problem 6.1 RELECT and THINK: Using the computed values of CB and CE, you can determine the reactions C x and C y by considering the equilibrium of Joint C (ig. 6). Since these reactions have already been determined from the equilibrium of the entire truss, this provides two checks of your computations. You can also simply use the computed values of all forces acting on the joint (forces in members and reactions) and check that the joint is in equilibrium. x y 5250 7000 3 5 4 5 8750 0 checks 8750 0 checks 6-16
Method of Sections When the force in only one member or the forces in a very few members are desired, the method of sections works well. To determine the force in member BD, form a section by cutting the truss at n-n and create a free body diagram for the left side. An BD could have been created for the right side, but why is this a less desirable choice? Think and discuss. Notice that the exposed internal forces are all assumed to be in tension. With only three members cut by the section, the equations for static equilibrium may be applied to determine the unknown member forces, including BD. 6-17
Method of Sections Using the left-side BD, write one equilibrium equation that can be solved to find BD. Check your equation with a neighbor; resolve any differences between your answers if you can. p k p Assume that the initial section cut was made using line k-k. Why would this be a poor choice? Think, then discuss with a neighbor. Notice that any cut may be chosen, so long as the cut creates a separated section. k So, for example, this cut with line p-p is acceptable. 6-18
Trusses Made of Several Simple Trusses Compound trusses are statically determinant, rigid, and completely constrained. m 2n 3 Truss contains a redundant member and is statically indeterminate. m 2n 3 Adal reaction forces may be necessary for a rigid truss. non-rigid m 2n 3 rigid m 2n 4 Necessary but insufficient con for a compound truss to be statically determinant, rigid, and completely constrained, m r 2n 6-19
Sample Problem 6.3 STRATEGY: List the steps for solving this problem. Discuss your list with a neighbor. 1. Draw the BD for the entire truss. Apply the equilibrium cons and solve for the reactions at A and L. Determine the force in members H, GH, and GI. 2. Make a cut through members H, GH, and GI and take the right-hand section as a free body (the left side would also be good). 3. Apply the cons for static equilibrium to determine the desired member forces. 6-20
Sample Problem 6.3 MODELING and ANALYSIS: A x Take the entire truss as a free body. Apply the cons for static equilibrium to solve for the reactions at A and L. A y M A 0 5 m 6 kn 10 m 6 kn 15 m 6 kn L 7.5 kn y 20 m1 kn 25 m1 kn 25 ml 0 20 kn L A y A y 12.5 kn x 0 A x L 6-21
Sample Problem 6.3 Make a cut through members H, GH, and GI and take the right-hand section as a free body. Draw this BD. What is the one equilibrium equation that could be solved to find GI? Confirm your answer with a neighbor. Sum of the moments about point H: M H 0 7.50 kn10 m 1 kn GI 13.13 kn 5 m GI 5.33 m 0 GI 13.13 kn T 6-22
Sample Problem 6.3 H is shown as its components. What one equilibrium equation will determine H? tan G GL 8 m 0.5333 28.07 15 m M G 0 10 m 1 kn 7.5 kn15 m 1 kn H cos 8 m 0 H 13.82 kn 5 m H 13.82 kn C There are many options for finding GH at this point (e.g., S x =0, S y =0). Here is one more: tan GI HI M L 0 5 m 2 8 m 3 1 kn10 m 1 kn GH 1.371 kn 0.9375 43.15 5 m GH cos10 m 0 GH 1.371kN C 6-23
Sample Problem 6.3 RELECT and THINK: Sometimes you should resolve a force into components to include it in the equilibrium equations. By first sliding this force along its line of action to a more strategic point, you might eliminate one of its components from a moment equilibrium equation. 2-24
Analysis of a rame rames and machines are structures with at least one multiforce (>2 forces) member. rames are designed to support loads and are usually stationary. Machines contain moving parts and transmit and modify forces. A free body diagram of the complete frame is used to determine the external forces acting on the frame. Internal forces are determined by dismembering the frame and creating free-body diagrams for each component. orces on two force members have known lines of action but unknown magnitude and sense. orces on multiforce members have unknown magnitude and line of action. They must be represented with two unknown components. orces between connected components are equal, have the same line of action, and opposite sense. 6-25
rames That Collapse Without Supports Some frames may collapse if removed from their supports. Such frames can not be treated as rigid bodies. A free-body diagram of the complete frame indicates four unknown force components which cannot be determined from the three equilibrium cons (statically indeterminate). The frame must be considered as two distinct, but related, rigid bodies. With equal and opposite reactions at the contact point between members, the two free-body diagrams show 6 unknown force components. Equilibrium requirements for the two rigid bodies yield 6 independent equations. Thus, taking the frame apart made the problem solvable. 6-26
Sample Problem 6.4 STRATEGY: ollow the general procedure discussed in this section. irst treat the entire frame as a free body, which will enable you to find the reactions at A and B. Then dismember the frame and treat each member as a free body, which will give you the equations needed to find the force at C. Members ACE and BCD are connected by a pin at C and by the link DE. or the loading shown, determine the force in link DE and the components of the force exerted at C on member BCD. 6-27
Sample Problem 6.4 MODELING and ANALYSIS: Create a free-body diagram for the complete frame and solve for the support reactions. y 0 A y 480 N A y 480 N M A Note: 0 480 N100 mm B160 mm x 0 B A x B 300 N A x 300 N A x 300 N 1 tan 80 150 28.07 6-28
Sample Problem 6.4 Create a free body diagram for member BCD (since the problem asked for forces on this body). Choose the best BD, then discuss your choice with a neighbor. Justify your choice. DE,x DE,y DE DE,x DE DE,y 6-29
Sample Problem 6.4 Using the best BD for member BCD, what is the one equilibrium equation that can directly find DE? Please discuss. M C DE 0 561N sin 250 mm 300 N60 mm 480 N100 mm DE 1 tan 80 150 28.07 DE 561N C Sum of forces in the x and y directions may be used to find the force components at C. x 0 0 C C x x DE cos 300 N 561Ncos 300 N C x 795 N y 0 0 C C y y DE sin 480 N 561Nsin 480 N C y 216 N 6-30
Sample Problem 6.4 RELECT and THINK: Check the computations by considering the free body ACE. M A DE cos 300 mm DE sin 100 mm Cx 220 mm 561cos 300 mm 561sin 100 mm 795220 mm 0 (checks) 6-31
Machines Machines are structures designed to transmit and modify forces. Typically they transform input forces (P) into output forces (Q). Given the magnitude of P, determine the magnitude of Q. Create a free-body diagram of the complete machine, including the reaction that the wire exerts. The machine is a nonrigid structure. Use one of the components as a free-body. Discuss why the forces at A are such. Sum moments about A, 0 ap bq M A Q a b P 6-32