Question 2.11: The pin-jointed planar truss shown in Fig. 1a is to be made ......
The pin-jointed planar truss shown in Fig. 1a is to be made of three steel two-force members and support a single vertical load P = 1.2 kips at joint B. The locations of joints A and B are fixed, but the vertical position of joint C can be changed by varying the lengths L_1 and L_3. For the steel truss members, the allowable stress in tension is (σ_T)_{\text{allow}} = 20 ksi, the allowable stress in compression is (σ_C)_{\text{allow}} = 12 ksi, and the weight density is 0.284 lb/in³. You are to consider truss designs for which the vertical member AC has lengths varying from L_1 = 18 in. to L_1 = 50 in.
(a) Show that, if each member has the minimum cross sectional area that meets the strength criteria stated above, the weight W of the truss can be expressed as a function of the length L_1 of member AC by the function that is plotted in Fig. 1b. (b) What value of L_1 gives the minimum-weight truss, and what is the weight of that truss?
Plan the Solution The member forces F_1, F_2, and F_3 can be related to the applied force P by writing equilibrium equations for joints B and C.
The length L_1, which determines the geometry of the triangle ABC, will enter into the expressions for these forces. Next we can use allowable-stress design to determine expressions for the cross-sectional areas of the bars. Finally, we can write an expression for the total weight of the truss as the sum of the weights of its three members, an expression that will contain the design variable L_1; then evaluate this expression over the specified range 18 in. ≤ L_1 ≤ 50 in.
Truss Geometry: The length L_2 is constant, L_2 = 30 in., and, from Fig. 1, the length L_3 is given by
L_3 = \sqrt{(24 in.)^2 + (L_1 – 18 in.)^2} (1)
Equilibrium:
By drawing free-body diagrams of joints B and C and writing equilibrium equations, we can show that the following formulas relate member forces F_1, F_2, and F_3 to the load P:
F_1 = \frac{(L_1 – 18 in.)P} {L_1} , F_2 = \frac{(30 in.)P }{L_1} , F_3 = -\frac{ L_3P} {L_1} (2)
Allowable-Stress Design Constraints: The allowable-stress criterion, Eq. 2.28, leads to the following three equations for the minimum areas that are required to carry the member forces given in Eqs. (2):
σ_{\text{allow}} = \frac{σ_Y} {FS}, or \tau _{\text{allow}} = \frac{\tau Y} {FS} (2.28)
A_1 = \frac{F_1}{(σ_T)_{\text{allow}}}, A_2 = \frac{F_2}{(σ_T)_{\text{allow}}}, A_3 = \frac{|F_3|}{(σ_T)_{\text{allow}}} (3)
Minimum-Weight Optimal Design: The total weight of the truss is the sum of the weights of its three members:
W = γV = γ(A_1L_1 + A_2L_2 + A_3L_3) (4)
Figure 1b is a plot of Eq. (4), evaluated as follows:
1. Select a value for L_1, starting with L_1 = 18 in., and use Eq. (1) to determine L_3.
2. Use Eq. (2) to evaluate the corresponding values of the three member forces.
3. Use Eq. (3) to determine the resulting cross-sectional areas required to meet the stated allowable-stress criteria.
4. Finally, use Eq. (4) to evaluate the weight of the truss that corresponds to the given value of L_1.
5. Repeat Steps (1) through (4) for about fifty values of L_1 in the given range 18 in. ≤ L_1 ≤ 50 in. These values of the function W (L_1) are plotted in Fig. 1b.
A spreadsheet computer program was used to carry out the computations in the above optimal-design solution and to plot the curve of the objective function (weight W) versus the design variable (length L_1).
From Fig. 1b, the optimal design (i.e., the minimum-weight design) is the design for which
L_1 = 30.0 in.
and the corresponding minimum weight is
W_{min} = 1.40 lb
Review the Solution From the plot of the weight function, W(L_1), we can see how the weight depends on the configuration of the truss. The optimum truss configuration is close to that shown in Fig. 1. With L_1 = 30 in., the tension in member (2) and the compression in member (3) each has a significant vertical component that acts to support load P; yet, the lengths L_1 and L_3 are not so long as to make the truss excessively heavy.
