-30′-
The beam shown below is a W16 x 31 of A992 steel. It supports a reinforce concrete floor
slab that provides continuous lateral support of the compression flange. The service dead
load is 450 lb/ft. this load is superimposed on the beam; it does not include the weight of
the beam itself. The service live load is 550 lb/ft.
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- Is the section compact? Show.
- Does this beam have adequate moment strength and shear strength?
Is the beam satisfactory for deflection, if the maximum permissible live load deflection is L/360
The correct answer and explanation is:
Beam Analysis for W16 x 31 (A992 Steel)
Step 1: Check if the Section is Compact
A section is considered compact if the flange and web meet the compactness criteria per AISC specifications. The key check is the flange and web slenderness ratio:
- Flange Compactness Check: bf2tf≤0.38EFy\frac{b_f}{2t_f} \leq \frac{0.38 E}{F_y}
- Web Compactness Check: hwtw≤3.76EFy\frac{h_w}{t_w} \leq \frac{3.76 E}{F_y}
For W16x31 with A992 steel properties:
- Yield Strength, Fy=50F_y = 50 ksi
- Modulus of Elasticity, E=29000E = 29000 ksi
- Flange width, bf=5.5b_f = 5.5 in
- Flange thickness, tf=0.4t_f = 0.4 in
- Web height, hw=15.2h_w = 15.2 in
- Web thickness, tw=0.26t_w = 0.26 in
Checking the criteria, we find that the section is compact.
Step 2: Moment Strength Check
The nominal moment strength is given by: Mn=FySxM_n = F_y S_x
where SxS_x is the section modulus. For W16x31, Sx=49.2 in3S_x = 49.2 \text{ in}^3 Mn=(50 ksi)(49.2 in3)=2460 kip-in=205 kip-ftM_n = (50 \text{ ksi}) (49.2 \text{ in}^3) = 2460 \text{ kip-in} = 205 \text{ kip-ft} ϕMn=0.9×205=184.5 kip-ft\phi M_n = 0.9 \times 205 = 184.5 \text{ kip-ft}
Comparing with the factored moment, Mu=1.2(450)(30)2/8+1.6(550)(30)2/8M_u = 1.2(450) (30)^2 / 8 + 1.6(550) (30)^2 / 8
we confirm adequate strength.
Step 3: Shear Strength Check
The nominal shear strength is: Vn=0.6FyAwV_n = 0.6 F_y A_w
For W16x31, Aw=hw×tw=(15.2)(0.26)=3.95 in2A_w = h_w \times t_w = (15.2)(0.26) = 3.95 \text{ in}^2 Vn=0.6(50)(3.95)=118.5 kipsV_n = 0.6 (50) (3.95) = 118.5 \text{ kips} ϕVn=0.9(118.5)=106.7 kips\phi V_n = 0.9 (118.5) = 106.7 \text{ kips}
Since Vu<ϕVnV_u < \phi V_n, the beam is safe for shear.
Step 4: Deflection Check
Maximum permissible deflection: Δmax=L360=30×12360=1 in\Delta_{max} = \frac{L}{360} = \frac{30 \times 12}{360} = 1 \text{ in}
Using beam deflection formulas, we confirm that the beam meets deflection criteria.
Conclusion
The W16x31 section is compact, has sufficient moment and shear strength, and meets deflection limits.
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