Question 6.5: From Appendix D.1, select a wide-flange steel beam to suppor......
From Appendix D.1, select a wide-flange steel beam to support the load distribution shown in Fig. 1a. Include the weight of the beam in your calculations.
The moment diagram for the beam, neglecting the weight of the beam, is shown in Fig. 1c. Let σ_{\text{allow}} = 19 ksi.
Plan the Solution We can use Eq. 6.17 to compute the required section modulus. We need to pick the lightest section, with some margin in S so that we can accommodate the maximum moment with the beam weight included.
S_{\text{design}} = \frac{M_{max}} {σ_{\text{allow}}} (6.17)
Preliminary Shape Selection: From Eq. 6.17,
S_{\text{design}} = \frac{M_{max}} {σ_{\text{allow}}} = \frac{(50 kip · ft)(12 in./ft)} {(19 kips/in^2)} = 31.6 in³
From Appendix D.1, there are two candidate sections, a W10×30 with S = 32.4 in³ and a W14×26 with S = 35.3 in^{3.9} The W10×30 weighs 4 lb/ft more than the W14×26, so it seems that the best choice would be the W14×26. However, since this beam is deeper than the W10×30 (13.91 in. vs 10.47 in.), there may be justification for choosing the W10×30 as long as it meets the strength requirement. So, let us see if the W10×30 meets the design requirements when the weight of the beam is added to the loads in Fig. 1a.
Equilibrium Check: We could construct new shear and moment diagrams, but it may be quicker to use free-body diagrams. The maximum positive moment will occur at the point x_m, where V(x_m) = 0 (x_m = 5 ft in Fig. 1b).We first need to compute the reaction at A, which is labeled A_y in Fig. 2.
(4 kips/ft)(12 ft)(6 ft) – (8 kips)(6 ft)
+(w kips/ft)(18 ft)(3 ft) – A_y(12 ft) = 0
A_y = (20 + 4.5w) kips
We can now determine the maximum positive moment using Fig. 3.
V(x_m) = (20 + 4.5w) – (4 + w)x_m = 0
Therefore, the maximum positive moment occurs at
x_m = \frac{20 + 4.5w}{4 + w}\left(\sum{M} \right) _D =0:
M(x_m) = (20 + 4.5w) \left(\frac{20 + 4.5w} {4 + w}\right) – (4 + w) \left(\frac{20 + 4.5w} {4 + w}\right)^2\left(\frac{1}{2}\right)
M(x_m) = \frac{(20 + 4.5w)^2} {2(4 + w)} kip · ft
The maximum negative moment will still occur at B, and its value can be computed using Fig. 4.
\left(\sum{M} \right) _B =0: M_B = (-48 – 18w) kip · ft
For the W10×30 beam, w = 30 lb/ft = 0.030 kips/ft. Therefore,
M(x_m) = 50.3 kip · ft, M_B = -48.5 kip · ft
So M_{max} = 50.3 kip · ft, which would require
S_{\text{design}} = \frac{(50.3 kip · ft)(12 in./ft)} {19 ksi} = 31.8 in³
Final Shape Selection: Since the W14×26 beam is lighter than the W10×30, and since its section modulus is greater (35.3 in³b vs 32.4 in³), the W14×26 would be the best design, unless its added depth is, for some reason, undesirable. In that case, the W10×30 would be a perfectly satisfactory design choice.
Review the Solution It is interesting to note that an 18-ft W14×26 beam, weighing 468 lb is able, in this case, to support a total load of 56,000 lb. (We will again consider the design of the beam in Fig. 1 in Section 9.3, after we have taken up the topics of shear stress in beams and the state of stress at a point.)
