Question 1.1: The plane truss shown in Fig.1-6 has four joints and five me...
The plane truss shown in Fig.1-6 has four joints and five members. Find support reactions at A and B and then use the methods of joints and sections to find all member forces. Let P = 35 kips and c = 10 ft.
Use the following four-step problem-solving approach.
1. Conceptualize [hypothesize, sketch]: First sketch a free-body diagram of the entire truss model (Fig.1-7). Only known applied forces at C and unknown reaction forces at A and B are shown and then used in an equilibrium analysis to find the reactions.
2. Categorize [simplify, classify]: Overall equilibrium requires that the force components in x and y directions and the moment about the z axis must sum to zero; this leads to reaction force components A_{x} , A_{y}, and B_{y}. The truss is statically determinate (unknowns: m + r = 5 + 3 = 8, knowns: 2 j = 8), so all member forces can be obtained using the method of joints. If only a few selected member forces are of interest, the method of sections can be used. Use a statics sign convention when computing external reactions and a deformation sign convention when solving for member forces.
3. Analyze [evaluate; select relevant equations, carry out mathematical solution]:
First find the lengths of members AC and BC needed to compute distances to lines of action of forces.
Law of sines to find member lengths a and b: Use known angles \theta_{A} , \theta_{B}, and \theta_{C} and c = 10 ft to find lengths a and b:
b=c \frac{\sin \left(\theta_{B}\right)}{\sin \left(\theta_{C}\right)}=(10 ft ) \frac{\sin \left(40^{\circ}\right)}{\sin \left(80^{\circ}\right)}=6.527 ft , a=c \frac{\sin \left(\theta_{A}\right)}{\sin \left(\theta_{C}\right)}=(10 ft ) \frac{\sin \left(60^{\circ}\right)}{\sin \left(80^{\circ}\right)}=8.794 ftCheck that computed lengths a and b give length c by using the law of cosines:
c=\sqrt{(6.527 ft )^{2}+(8.794 ft )^{2}-2(6.527 ft )(8.794 ft ) \cos \left(80^{\circ}\right)}=10 ftSupport reactions: Using the truss model free-body diagram in Fig.1-7, sum forces in x and y directions and moments about joint A:
\Sigma M_{A}=0 \quad B_{y}=\frac{1}{c}\left[P\left(b \cos \left(\theta_{A}\right)\right)+2 P\left(b \sin \left(\theta_{A}\right)\right)\right]=51 kips
\Sigma F_{x}=0 \quad A_{x}=-2 P=-70 kips
\Sigma F_{y}=0 \quad A_{y}=P-B_{y}=-16 kips
Reaction force components A_{x} and A_{y} are both negative, so they act in the negative x and y directions, respectively, based on a statics sign convention.
Member forces using method of joints: Begin by drawing free-body diagrams of the pin at each joint (Fig.1-8). Use a deformation sign convention in which each member is assumed to be in tension (so the member force arrows act away from the two joints to which each member is connected). The forces are concurrent at each joint, so use force equilibrium at each location to find the unknown member forces.
First sum forces in the y direction at joint A to find member force AC, and then sum forces in the x direction to get member force AD:
\Sigma F_{y}=0 \quad A C=\frac{-1}{\sin \left(60^{\circ}\right)} A_{y}=18.46 kips
\Sigma F_{x}=0 \quad A D=-A_{x}-A C \cos \left(60^{\circ}\right)=60.8 kips
Summing forces at joint B gives member forces BC and BD as
\Sigma F_{y}=0 \quad B C=\frac{-1}{\sin \left(40^{\circ}\right)} B_{y}=-79.3 kips
\Sigma F_{x}=0 \quad B D=-B C \cos \left(40^{\circ}\right)=60.8 kips
The minus sign means that member BC is in compression, not in tension as assumed. Finally, observe that CD is a zero-force member because forces in the y direction must sum to zero at joint D.
Selected member forces using method of sections: An alternative approach is to make a section cut all the way through the structure to expose member forces of interest, such as AD, CD, and BC in Fig.1-9. Summing moments about joint B confirms that the force in member CD is zero.
Summing moments about joint C (which is not on the free-body diagram but is a convenient point about which to sum moments because forces CD and BC act through joint C) confirms the solution for force AD as
\Sigma M_{C}=0 \quad A D=\frac{1}{b \sin \left(60^{\circ}\right)}\left[B_{y}(a) \cos \left(40^{\circ}\right)\right]=60.8 kips
Finally, summing moments about A in Fig.1-9 confirms member force BC:
\Sigma M_{A}=0 \quad B C=\frac{1}{c \sin \left(40^{\circ}\right)}\left[-B_{y} c\right]=-79.3 kips
4. Finalize [conclude; examine answer—Does it make sense? Are units correct? How does it compare to similar problem solutions?]: There are 2 j = 8 equilibrium equations for the simple plane truss considered, and using the method of joints, these are obtained by applying \Sigma F_{x}=0 and \Sigma F_{y}=0 at each joint in succession. A computer solution of these simultaneous equations leads to the three reaction forces and five member forces. The method of sections is an efficient way to find selected member forces. A key step is the choice of an appropriate section cut, which isolates the member of interest and eliminates as many unknowns as possible. This is followed by construction of a free-body diagram for use in the static equilibrium analysis to compute the member force of interest. The methods of sections and joints were used, a common solution approach in plane and space truss analysis.
