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CHAPTER FOUR
ANALYSIS OF STRUCTURES
4.1 Introduction
The problems considered in the preceding chapters concerned the equilibrium of a single rigid body, and all forces involved were external to the rigid body. We now consider problems dealing with the equilibrium of structures made of several connected parts. These problems call for the determination not only of the external forces acting on the structure but also of the forces which hold together the various parts of the structure. From the point of view of the structure as a whole, these forces are internal forces.
In this chapter, three broad categories of engineering structures will be considered: Trusses:- which are designed to support loads and are usually stationary, fully constrained structures. Trusses consist exclusively of straight members connected at joints located at the ends of each member. Members of a truss, therefore, are two-force members , i.e., members acted upon by two equal and opposite forces directed along the member. Frames:- which are also designed to support loads and are also usually stationary, fully constrained structures. However, frames always contain at least one multiforce member , i.e., a member acted upon by three or more forces which, in general, are not directed along the member. Machines:- which are designed to transmit and modify forces and are structures containing moving parts. Machines, like frames, always contain at least one multiforce member.
The truss is one of the major types of engineering structures. It provides both a practical and an economical solution to many engineering situations, especially in the design of bridges and buildings. A typical truss is shown in Fig. 4.1a. A truss consists of straight members connected at joints. Truss members are connected at their extremities only; thus no member is continuous through a joint. In Fig. 4.1a , for example, there is no member AB ; there are instead two distinct members AD and DB. Most actual structures are made of several trusses joined together to form a space framework. Each truss is designed to carry those loads which act in its plane and thus may be treated as a two-dimensional structure. Figure 4.
The weights of the members of the truss are also assumed to be applied to the joints, half of the weight of each member being applied to each of the two joints the member connects. Although the members are actually joined together by means of welded, bolted, or riveted connections, it is customary to assume that the members are pinned together; therefore, the forces acting at each end of a member reduce to a single force and no couple. Thus, the only forces assumed to be applied to a truss member are a single force at each end of the member. Each member can then be treated as a two-force member, and the entire truss can be considered as a group of pins and two-force members (Fig. 4.1b). An individual member can be acted upon as shown in either of the two sketches of Fig. 4.2. In Fig. 4.2a , the forces tend to pull the member apart, and the member is in tension; in Fig. 4.2b , the forces tend to compress the member, and the member is in compression. A number of typical trusses are shown in Fig. 4.3.
Figure 4.
Figure 4.
4.3 Analysis of Trusses
4.3.1 Method of Joints
The truss of Fig. 4.1 , whose free-body diagram is shown in Fig. 4.5a , can thus be dismembered, and a free-body diagram can be drawn for each pin and each member (Fig. 4.5b). Each member is acted upon by two forces, one at each end; these forces have the same magnitude, same line of action, and opposite sense. Furthermore, Newton’s third law indicates that the forces of action and reaction between a member and a pin are equal and opposite. Therefore, the forces exerted by a member on the two pins it connects must be directed along that member and be equal and opposite. The common magnitude of the forces exerted by a member on the two pins it connects is commonly referred to as the force in the member considered, even though this quantity is actually a scalar. Since the lines of action of all the internal forces in a truss are known, the analysis of a truss reduces to computing the forces in its various members and to determining whether each of its members is in tension or in compression. Figure 4.
Since the entire truss is in equilibrium, each pin must be in equilibrium. The fact that a pin is in equilibrium can be expressed by drawing its free-body diagram and writing two equilibrium equations. If the truss contains n pins, there will, therefore, be 2n equations available, which can be solved for 2n unknowns. In the case of a simple truss, we have m = 2n - 3 , that is, 2n = m + 3 , and the number of unknowns which can be determined from the free-body diagrams of the pins is thus m + 3. This means that the forces in all the members, the two components of the reaction RA , and the reaction RB can be found by considering the free-body diagrams of the pins.
The fact that the entire truss is a rigid body in equilibrium can be used to write three more equations involving the forces shown in the free-body diagram of Fig. 4.5a. Since they do not contain any new information, these equations are not independent of the equations associated with the free-body diagrams of the pins. Nevertheless, they can be used to determine the components of the reactions at the supports.
The arrangement of pins and members in a simple truss is such that it will then always be possible to find a joint involving only two unknown forces. These forces can be determined and their values transferred to the adjacent joints and treated as known quantities at these joints. This procedure can be repeated until all unknown forces have been determined.
As an example, the truss of Fig. 4.5 will be analyzed by considering the equilibrium of each pin successively, starting with a joint at which only two forces are unknown. In the truss considered, all pins are subjected to at least three unknown forces. Therefore, the reactions at the supports must first be determined by considering the entire truss as a free body and using the equations of equilibrium of a rigid body. We find in this way that RA is vertical and determine the magnitudes of RA and RB.
The number of unknown forces at joint A is thus reduced to two, and these forces can be determined by considering the equilibrium of pin A. The reaction RA and the forces FAC and FAD exerted on pin A by members AC and AD , respectively, must form a force triangle. First we draw RA (Fig. 4.6) ; noting that FAC and FAD are directed along AC and AD , respectively, we complete the triangle and determine the magnitude and sense of FAC and FAD. The magnitudes FAC and FAD represent the forces in members AC and AD. Since FAC is directed down and to the left, that is, toward joint A , member AC pushes on pin A and is in compression. Since FAD is directed away from joint A , member AD pulls on pin A and is in tension. Figure 4.
We can now proceed to joint D , where only two forces, FDC and FDB , are still unknown. The other forces are the load P , which is given, and the force FDA exerted on the pin by member AD. As indicated above,
A free-body diagram of the entire truss is drawn; external forces acting on this free body consist of the applied loads and the reactions at C and E. We write the following equilibrium equations. ∑ M 0; 2000 24 + 1000 12 − 6 = 0 = 10,000 ⟹ = , ↑ ∑ F = 0; ⟹ = ∑ F = 0; ! − 2000 − 1000 + 10,000 = 0 ! = −7000 ⟹ # = $ ↓
Free-Body Diagram of Joint A This joint is subjected to only two unknown forces, namely, the forces exerted by members AB and AD. A force triangle is used to determine FAB and FAD. We note that member AB pulls on the joint and thus is in tension and that member AD pushes on the joint and thus is in compression. The magnitudes of the two forces are obtained from the proportion 2000 4 =
Free-Body Diagram of Joint D Since the force exerted by member AD has been determined, only two unknown forces are now involved at this joint. Again, a force triangle is used to determine the unknown forces in members DB and DE. )& = )' ⟹ +0- = 1. , / )2 = 34 )& + 34 )' ⟹ +0 = 5 , Free-Body Diagram of Joint B Since more than three forces act at this joint, we determine the two unknown forces FBC and FBE by solving the equilibrium equations ∑ = 0 and ∑ = 0. We arbitrarily assume that both
unknown forces act away from the joint, i.e., that the members
are in tension. The positive value obtained for FBC indicates that our assumption was correct; member BC is in tension. The negative value of FBE indicates that our assumption was wrong; member BE is in compression.
∑ F = 0; −1000 − 64 2500 − 64 '2 = 0 '2 = −3750 ⟹ +- = 5$. , ∑ F = 0; ' − 1500 − 34 2500 − 34 3750 = 0 ' = +5250 ⟹ +- = .1. , / Free-Body Diagram of Joint E The unknown force FEC is assumed to act away from the joint. Summing x components, we write
∑ F = 0; 34 2 + 3000 + 34 3750 = 0 2 = −8750 ⟹ + = 8$. , Summing y components, we obtain a check of our computations:
∑ F = 0; 10,000 − 64 3750 − 64 8750 = 0 10,000 − 3000 − 7000 = 0 9:;<= Free-Body Diagram of Joint C Using the computed values of FCB and FCE , we can determine the reactions Cx and Cy by considering the equilibrium of this joint. Since these reactions have already been determined from the equilibrium of the entire truss, we will obtain two checks of our computations. We 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:
∑ F = 0; −5250 + 34 8750 = −5250 + 5250 = 0 9:;<= ∑ F = 0; −7000 + 64 8750 = −7000 + 7000 = 0 9:;<=
Example 4. A snow load transfers the forces shown to the upper joints of a Pratt roof truss. Neglect any horizontal reactions at the supports and solve for the forces in all members.
4.3.2 Method of Sections
The method of joints is most effective when the forces in all the members of a truss are to be determined. If, however, the force in only one member or the forces in a very few members are desired, another method, the method of sections, is more efficient.
Assume, for example, that we want to determine the force in member BD of the truss shown in Fig. 4.9a. To do this, we must determine the force with which member BD acts on either joint B or joint D. If we were to use the method of joints, we would choose either joint B or joint D as a free body. However, we can also choose as a free body a larger portion of the truss, composed of several joints and members, provided that the desired force is one of the external forces acting on that portion. If, in addition, the portion of the truss is chosen so that there is a total of only three unknown forces acting upon it, the desired force can be obtained by solving the equations of equilibrium for this portion of the truss. In practice, the portion of the truss to be utilized is obtained by passing a section through three members of the truss, one of which is the desired member, i.e., by drawing a line which divides the truss into two completely separate parts but does not intersect more than three members. Either of the two portions of the truss obtained after the intersected members have been removed can then be used as a free body. Figure 4.
In Fig. 4.9a , the section nn has been passed through members BD, BE , and CE , and the portion ABC of the truss is chosen as the free body (Fig. 4.9b). The forces acting on the free body are the loads P 1 and P 2 at points A and B and the three unknown forces FBD, FBE , and FCE. Since it is not known whether the members removed were in tension or compression, the three forces have been arbitrarily drawn away from the free body as if the members were in tension.
The fact that the rigid body ABC is in equilibrium can be expressed by writing three equations which can be solved for the three unknown forces. If only the force FBD is desired, we need write only one equation, provided that the equation does not contain the other unknowns. Thus the equation ∑^ M 2 0 yields the value of the magnitude FBD of the force FBD (Fig. 4.9b). A positive sign in the answer will indicate that
our original assumption regarding the sense of FBD was correct and that member BD is in tension; a negative sign will indicate that our assumption was incorrect and that BD is in compression.
On the other hand, if only the force FCE is desired, an equation which does not involve FBD or FBE should be written; the appropriate equation is ∑^ M' 0. Again a positive sign for the magnitude FCE of the desired force indicates a correct assumption, that is, tension; and a negative sign indicates an incorrect assumption, that is, compression. If only the force FBE is desired, the appropriate equation is ∑ = 0.
Whether the member is in tension or compression is again determined from the sign of the answer.
When the force in only one member is determined, no independent check of the computation is available. However, when all the unknown forces acting on the free body are determined, the computations can be checked by writing an additional equation. For instance, if FBD , FBE , and FCE are determined as indicated above, the computation can be checked by verifying that ∑ = 0.
Example 4. Determine the force in members FH,GH , and GI of the roof truss shown. Solution: From the free-body diagram of the entire truss, we find the reactions at A and L : ⟹ , = 12.5 C ↑ ⟹ Z = 7.5 C ↑ We note that
DEF G = HH[ = 158 = 0.53333 ⟹ I = 28.07J
Force in Member GI Section nn is passed through the truss as shown. Using the portion HLI of the truss as a free body, the value of FGI is obtained by writing ∑ (^) M\ 0; 7.5 10 − 1 5 − ]^ 5.33 = 0 ]^ = 13.13 C ⟹ +Y_ = 5. 5 < , / Force in Member FH The value of FFH is obtained from the equation∑^ M] 0. We
∑ M 2 0; j 1.54 + 2 2.4 + 1 4.8 − >?A 2.4 = 0 ⟹ jk = 6.234 C,! j = (^) TUR Gjk = (^) TUR 12.76.234 = l. 5m < , ∑ F = 0; ! − j − j = 0; 7.2 − 6.234 − j = 0 ⟹ jk = 0.966 C,! j = (^) TUR Njk = (^) TUR 22.626.234 = . X$ < , b) Force in Members EG, GH, and HJ Section mm is passed through the truss as shown. Using the portion IJHG of the truss as a free body, the value of force in member EG is obtained by writing ∑ M\ 0; >eA 2.4 − 1 2.4 − H 2.08 = 0 ⟹ Y = 5. Xl < , / ∑ M] 0; >enA 2.4 − 1 2.4 − Oi 2.08 = 0 ⟹ Oik = 3.46 C,! Oi = (^) TUR GOik = (^) TUR 12.73.46 = 5. .. < , ∑ F = 0; en + Oi − 1 − HO = 0; 4 + 3.55 RSF 12.7 − 1 − HO = 0 ⟹ YW = 5. $8 < ,
4.4 Frame and Machines
Under trusses, we have considered structures consisting entirely of pins and straight two-force members. The forces acting on the two-force members were known to be directed along the members themselves. We now consider structures in which at least one of the members is a multi-force member, i.e., a member acted upon by three or more forces. These forces will generally not be directed along the members on which they act; their direction is unknown, and they should be represented therefore by two unknown components.
Frames and machines are structures containing multiforce members. Frames are designed to support loads and are usually stationary, fully constrained structures. Machines are designed to transmit and modify forces; they may or may not be stationary and will always contain moving parts.
4.4.1 Analysis of a Frame
As a first example of analysis ofa frame, the crane which carries a given load W (Fig. 4.10a). The free-body diagram of the entire frame is shown in Fig. 4.10b. This diagram can be used to determine the external forces acting on the frame. Summing moments about A , we first determine the force T exerted by the cable; summing x and y components, we then determine the components Ax and Ay of the reaction at the pin A.
In order to determine the internal forces holding the various parts of a frame together, we must dismember the frame and draw a free-body diagram for each of its component parts (Fig. 4.10c). First, the two-force members should be considered. In this frame, member BE is the only two-force member. The forces acting at each end of this member must have the same magnitude, same line of action, and opposite sense. They are therefore directed along BE and will be denoted, respectively, by FBE and - FBE. Their sense will be arbitrarily assumed as shown in Fig. 4.10c ; later the sign obtained for the common magnitude FBE of the two forces will confirm or deny this assumption.
Next, we consider the multiforce members, i.e., the members which are acted upon by three or more forces. According to Newton’s third law, the force exerted at B by member BE on member AD must be equal and opposite to the force FBE exerted by AD on BE. Similarly, the force exerted at E by member BE on member CF must be equal and opposite to the force - FBE exerted by CF on BE. Thus the forces that the two-force member BE exerts on AD and CF are, respectively, equal to - FBE and FBE ; they have the same magnitude FBE and opposite sense, and should be directed as shown in Fig. 4.10c.
At C two multiforce members are connected. Since neither the direction nor the magnitude of the forces acting at C is known, these forces will be represented by their x and y components. The components Cx and Cy of the force acting on member AD will be arbitrarily directed to the right and upward. Since, according to Newton’s third law, the forces exerted by member CF on AD and by member AD on CF are equal and opposite, the components of the force acting on member CF must be directed to the left and
Figure 4.
Members:- The frame is now dismembered; since only two members are connected at each joint, equal and opposite components are shown on each member at each joint. Free Body Diagram of Member BCD ∑ M' 0; o>!A 2.4p − 2400 3.6 = 0 ! = 3600 ⟹ # = 5l ∑ M 0; o>QA 2.4p − 2400 3.6 = 0 Q = 1200 ⟹ - # = 1 ∑ F = 0; −Q + ! = 0 We note that neither Bx nor Cx can be obtained by considering only member BCD. The positive values obtained for By and Cy indicate that the force components By and Cy are directed as assumed. Free Body Diagram of Member ABE ∑ M& 0; Q 2.7 = 0 ⟹ - = ∑ F = 0; Q − ? = 0 ⟹ , = ∑ F = 0; −? + Q + 600 = 0 −? + 1200 + 600 = 0 ⟹ ,# = 8 Free Body Diagram of Member BCD Returning now to member BCD, we write ∑ F = 0; −Q + ! = 0 ⟹ = Free Body Diagram of Member ACF (Check) ∑ (^) M 0; 1800 2.4 − o>?A 2.4p − ? 2.7 = 0 1800 2.4 − 1800 2.4 − 0 2.7 = 0 9:;<
Example 4. Knowing that the pulley has a radius of 50 mm, determine the components of the reactions at B and E. Solution: Free Body Diagram of Entire Frame .∑^ M 2 0; Q 0.15 − 300 0.35 = 0 Q = 700 ⟹ - = $ ←
∑ F = 0; − Q = 0 ⟹ = $ →
∑ F = 0; − Q − 300 = 0
The frame is now dismembered; since only two members are connected at each joint, equal and opposite components are shown on each member at each joint. Free Body Diagram of Member BCD
.∑ M 0; o>QA 0.18p − 300 0.12 = 0
Q = 200 ⟹ - # = 1
Free Body Diagram of Entire Frame
. Returning now to Entire Frame, we write ∑ F = 0; − Q − 300 = 0 = 500 ⟹ # = .
