1. Write the following number on the front cover of your exam booklet, top-left corner, above your name:

2. Loads [10 points]. Find the governing load for the 2nd-floor column (in pounds or kips) shown below assuming a wood structure. Consider live load reduction, and assume the following loads: D = 50 psf; L = 75 psf; and S = 30 psf.

Solution: First, find the reduced live load based on the influence area, A_i = 48 x 32 = 1536 sq.ft. The reduction coefficient = 0.25 + 15 / (sq.rt. 1536) = 0.63, so the reduced live load = (.63)(75) = 47.25 psf.

From Table A-5.1 for Allowable Stress Design, the relevant load combinations to consider are as follows:

• D + L = 50(24 x 16 x 2 floors) + 47.25(24 x 16 x 1 floor) = 38,400 + 18,144 = 56,544#
• D + 0.75L + 0.75S = 50(24 x 16 x 2 floors) + 0.75(47.25)(24 x 16 x 1 floor) + 0.75(30)(24 x 16 x 1 floor) = 38,400 + 13,608 + 8,640 = 60,648#

Note that the load selected is the value shown in red, based on the bold-faced value representing the governing condition. In this case, they happen to be the same.

Solution: For Hem-Fir No. 1 (dimension lumber), F_b = 950 psi, F_v = 75 psi, and E = 1,500,000 psi. We can find the maximum bending moment and shear force as follows:

Design for bending: Assuming C_F = 1.0, the adjusted allowable bending stress, F'_b = 950(1.0)(1.0)(1.0) = 950 psi, so the required section modulus, S = M / F'_b = 8400 / 950 = 8.84 in³.

From Table A-4.1, select a 2x8 with S = 13.14 in³. Since C_F = 1.05 (greater than 1.0) for the 2x8, it will work in bending, but a smaller cross-section with a larger C_F might also work: check a 2x6 for bending:

2x6: With C_F = 1.10, the adjusted allowable bending stress, F'_b = 950(1.10)(1.0)(1.0) = 1045 psi, so the required section modulus, S = M / F'_b = 8400 / 1045 = 8.04 in³. This is greater than the actual section modulus for the 2x6 (7.536 in³), so the 2x6 is no good for bending.

Select the 2x8 for bending.

Check for shear: The required area, A = 1.5(V / F'_v) = 1.5(350 / 75) = 7 in². Since the actual area, A = 10.88 in² is greater or equal to the required area, the 2x8 is OK for shear.

Check for live load deflection: The allowable deflection = L/360 = 48"/360 = 0.13".

The actual deflection = CPL³ / (EI) where C = 35.94 (Table A-8.3 for concentrated load at midpoint), P = 400# (live load only), L = 4 feet, E = 1,500,000 psi, and I = 47.63 in⁴ (from Table A-4.1).

Actual deflection = 35.94(400)(4³) / (1,500,000 x 47.63) = 0.01". Since the actual is smaller than the allowable, the 2x8 is OK for deflection.

Solution: The allowable bending stress for A-992 steel (Table 3.12), F_b = 0.66F_y, where F_y = 50 ksi (Table A-3.11). F_b = 0.66(50) = 33 ksi. The maximum moment can be found as follows:

Design for bending: The required section modulus, S = M / F_b = 3600 / 33 = 109 in³.

From Table A-8.4, select a W24x55 with S = 114 in³.

5. Reinforced concrete T-beam design [15 points]: Find As and select 3 bars (if possible) for a positive-moment T-beam with the cross-section shown to the right and Mu = 491 ft-k. Assume 12-foot centerline beam spacing and a 45-foot span. f'c = 5 ksi. Check bar fit (Table A-4.11) and deflection, based on criteria in Table A-8.2 for a continuous beam.
Solution: First, find the effective width of the T-beam (Table A-8.9); it is the smaller of: 1/4 beam span = 1/4(45 x 12) = 135" centerline spacing = 12 x 12 = 144" bw + 16h = 18 + 16(5) = 98" The effective width, b = 98". Mu = φbd2R; so 491(12) = 0.9(98)(262)R; solving for R, we get: R = 0.0988. For f'c = 5 ksi, this corresponds to a steel ratio, ρ = 0.00167 (Table A-8.9). Note that this steel ratio is acceptable as it falls between ρmin and ρmax. To check ρmin: b / bw = 98 / 18 = 5.4 or approximately 5. From Table A-8.9, ρmin corresponding to b / bw = 5 is 0.00067, so the actual steel ratio, ρ = 0.0016, is OK. The required steel area, As = ρbd = 0.00167(98 x 26) = 4.26 in2. From Table A-4.10, select 3 No. 11 bars with As = 4.68 in2. From Table A-4.11, need 12"; have 18", so selected bars fit. From Table A-8.2, need total depth of L/21 = 45x12 / 21 = 25.7" which is less than the actual depth of 29", so beam is OK for deflection control.
6. Reinforced concrete shear (web steel) [15 points]: Find (a) minimum and (b) maximum spacing for web steel (stirrups) in a rectangular reinforced concrete beam, for the value of design shear, Vu, shown in the shear diagram. Do not find location of start of maximum spacing or location where no web steel is required. Use No. 3 bars if possible and f'c = 5 ksi. Round spacing down to 1/2" increment (that is, 4", 4-1/2" 5", 5-1/2" etc.).

Solution: Based on equations in Table A-8.6.
a) Minimum spacing: First, find the critical shear force at a distance "d" from the face of support:

V_u = 107 (267 / 300) = 95.23 k.

V_c = 2(sq.rt.5000)(18 x 33) = 84,004# = 84 k.

V_s = (95.23 / 0.75) - 84 = 43 k.

s = 2(0.11)(60)(33) / 43 = 10.13". Round down to s = 10 inches.

b) Maximum spacing: Use smaller of:

• d/2 = 33 / 2 = 16.5"
• 24"
• 2(0.11)(60,000) / (50 x 18) = 14.66"

Solution: From Table A-4.3, a W21x101 has these properties:

• I_y = 248 in⁴
• A = 29.8 in²

The capacity, P = 22.20(29.8) = 662 k.

Solution:
The equation relating the design (factored) load to the "strength-reduced" resistance of the concrete column is:

P_u = φα(0.85f'_cA_conc. + f_yA_s) where A_conc. = A_gross - A_s. We can therefore write:

675 = (0.65 x 0.80)[0.85(5)(192 - A_s) + 60A_s]; solving for A_s, we get:

A_s = 8.65 in². Select 4 No. 14 bars with A_s = 9.00 in².

Check reinforcement ratio = 9.0 / 192 = 0.47 or 4.7% which is between 1% and 8%, so OK.

9. Wood tension analysis [10 points]. Find the capacity in tension for a Hem-Fir No.1 6x20 with 5/8"-diameter bolts as shown. Remember to find allowable stresses in the correct "size" category.

Solution:
The allowable tensile stress, F_t = 525 psi ("beams and stringers" category for 6x20). There are no adjustments, so F'_t = 525 psi.

The net area = A_g - (bolt hole area) = (5.5 x 19.5) - 3(5.5 x .75) = 107.25 - 12.375 = 94.875 in². In this calculation, the bolt hole diameter is 1/8" larger than the bolt diameter, or 5/8" + 1/8" = 0.75". Three bolt holes are removed from the gross area, as shown in the sketch below:

The capacity, P = 525(94.875) = 49,809#.