Question 3.22: The two support rods in Fig. 1 are made of structural steel ......
The two support rods in Fig. 1 are made of structural steel that may be assumed to have a stress-strain diagram like Fig. 3.28b, with E = 30(10³) ksi and σ_Y = 36 ksi. (a) Construct a load-displacement diagram that relates the load P to the vertical displacement at D. Make the usual small-angle assumption for the rotation of the “rigid” beam AD. (b) Determine the allowable load if the factor of safety (see Eq. 2.26) for first yielding is FS_Y = 2.0, and the factor of safety for fully-plastic-behavior is FS_P = 2.5.
Plan the Solution The equilibrium analysis and the geometry-of-deformation analysis will be the same as in Example Problem 3.8. As in the above discussion of elastic-plastic behavior, the material behavior here falls under three cases—(1) both support rods are linearly elastic, (2) one rod has yielded, and (3) both rods have yielded. The load-displacement diagram should resemble Fig. 3.29. The allowable load is determined by applying the factors of safety as defined in Eq.
2.26, with the allowable load being the lower of the two loads (P_{\text{allow}})_Y and (P_{\text{allow}})_P.
F_S = \frac{\text{Failure load}} {\text{Allowable load}} (2.26)
(a) Construct a load-displacement diagram relating load P to the vertical displacement at D.
Equilibrium: For the free-body diagram in Fig. 2, the equation of equilibrium is
\left(∑M\right)_A = 0: F_1 + 2F_2 = 3P Equilibrium (1)
Geometry of Deformation: Figure 3 is a deformation diagram for small rotation of the rigid beam AD. From it, the equations of geometric compatibility can be written as
e_1 = \frac{1} {3} δ Geometric Compatibility (2)
e_2 = \frac{2} {3} δwhere δ is the vertical displacement at D.
Material Behavior: There are three cases to be considered: Case 1, both rods are linearly elastic; Case 2, one rod has yielded; and Case 3, both rods have yielded.
Case 1—Both rods are linearly elastic. From Eq. 3.15 we have
F = ke, where k = \frac{AE} {L} (3.15)
k_1 = \left(\frac{AE} {L} \right)_1 = 375 \frac{kips} {in.}, k_2 = \left(\frac{AE} {L} \right)_2 = 600 \frac{kips} {in.},and
F_1 = k_1e_1 = 375e_1 Material Behavior—Case 1 (3)
F_2 = k_2e_2 = 600e_2Equations (1) through (3) may be combined to give the following load-displacement equation for Case 1:
δ = \frac{3} {925} P, or P = \frac{925}{3}δ Ans. (a) (4)
The corresponding stresses in the two rods are obtained by combining Eqs. (2) through (4).Thus,
σ_1 = \frac{F_1} {A_1} = \frac{15} {37}P, σ_2 = \frac{F_2}{A_2} = \frac{48}{37}P (5)
Since σ_2 > σ_1, rod 2 will yield before rod 1 does. Therefore, the yield load P_Y is given by setting σ_2 = σ_Y = 36 ksi in Eq. (5b), the equation for σ_2.This gives
P_Y = 27.8 kips Ans. (a) (6)
The corresponding displacement at which first yield occurs is obtained by substituting Eq. (6) into Eq. (4a), the equation for δ, giving
δ_Y = \frac{3} {925} (27.8) = 0.090 in. Ans. (a) (7)
The load-displacement formula for loads up to load P_Y is given by Eqs. (4).
Case 2—One rod has yielded; one is linearly elastic. After rod 2 yields and up until the load at which rod 1 yields, the material-behavior equations for rods 1 and 2 are:
F_1 = 375e_1 kips Material Behavior—Case 2 (8)
F_2 = σ_YA_2 = 36 kips
Combining Eqs. (1) (equilibrium equation), (2a) (deformation-geometry equation for e_1), and (8) (force-deformation equations), we get
375 \left(\frac{1}{3}\right) δ + 2 \left(36\right) = 3P
or
P = \left(\frac{125 }{3} δ + 24\right) kips Ans. (a) (9)
This formula characterizes the load-displacement behavior from the yield load P_Y up to the fully plastic load P_P, which corresponds to the yielding of rod 1
Case 3—Both rods have yielded. Once rod 1 also yields, the material behavior equations become
F_1 = σ_YA_1 = 36 kips Material Behavior—Case 3 (10)
F_2 = σ_YA_2 = 36 kips
These rod forces can be substituted into the equilibrium equation. Eq.(1), to give the plastic load P_P.
P_P = \frac{F_1 + 2F_2 } {3} = \frac{36 + 2(36)} {3}or
P_P = 36 kips Ans. (a) (11)
The corresponding displacement δ_P is obtained by substituting Eq. (11) into Eq. (9) to obtain
δ_P = \frac{3} {125} (12) = 0.288 in. Ans. (a) (12)
We can now use Eqs. (4), (6), (7), (9), (11), and (12) to plot the load -displacement diagram, Fig. 4.
(b) Determine the allowable load. Based on first yield, the allowable load is given by
(P_{\text{allow}})_Y = \frac{P_Y} {FS_Y} (13a)
(P_{\text{allow}})_Y = \frac{27.75 kips} {2.0} = 13.875 kips (13b)
Based on the fully-plastic load, the allowable load is given by
(P_{\text{allow}})_P = \frac{P_P} {FS_P} (14a)
So,
(P_{\text{allow}})_P = \frac{36 kips}{2.5} = 14.40 kips (14b)
In this case, the allowable load is determined by the first-yield criterion.
Therefore,
P_{\text{allow}} = 13.88 kips Ans. (b) (15)
Review the Solution The key results in the preceding analysis are the “break points,” (P_Y, δ_Y) and (P_P, δ_P), in the load -displacement diagram.
Let us check the first of these, starting with δ_Y = 0.090 in. The compatibility equations, Eqs. (2) are easily checked by referring to Fig. 3, and these give
e_{1Y} = 0.030 in., e_{2Y} = 0.060 in.
The strain ε_{2Y} = e_{2Y}/L_2 = 0.0012 is in agreement with the yield strain σ_Y/E, so we have the correct elongations. We can get the rod forces by using Eqs. (3). Thus, F_{1Y} = 11.25 kips and F_{2Y} = 36 kips. When these forces are substituted into the equilibrium equation, Eq. (1), we do get P_Y = 27.75 kips, so our solution appears to be correct.
