We are designing a new building that will be used as an office and shop. To meet the owner's tight construction schedule, the building will have precast wall panels and precast roof panels. Treating the roof as a rigid diaphragm, we have determined the shear generated in each lateral

Question

We are designing a new building that will be used as an office and shop. To meet the owner's tight construction schedule, the building will have precast wall panels and precast roof panels.

Treating the roof as a rigid diaphragm, we have determined the shear generated in each lateral load-resisting element if there is seismic loading in the north-south direction (Example 11.2).

Our assignment is to determine the chord forces that will be carried on the perimeter of the diaphragm due to this loading. The plan is to use full-depth collectors for each of the lateral elements.

Accepted Answer

Approximate Analysis:
Treat each diaphragm segment between lateral elements as an independent diaphragm.

[latex]M_{1max} = \frac{0.465 klf (60 ft)^2}{8} = 209 k[/latex]

[latex]T_1 = \frac{209 k}{40 ft} = 5.22 k[/latex]

[latex]M_{2max} = \frac{0.837 klf(32 ft)^2}{8} = 107 k[/latex]

[latex]T_2 = \frac{107 k}{72 ft} = 1.49 k[/latex]

Corrected Equivalent Beam:
Find the equivalent linear distributed load.

[latex]+\downarrow \sum{V_y} = 17.6 k + 12.2 k + 25.0 k = 54.8 k[/latex]

[latex]+\circlearrowright \sum{V_y}\cdot x = 17.6 k(0) + 12.2 k (60 ft) + 25.0 k (92 ft) = 3032 k\cdot ft[/latex]

[latex]w_1 = \frac{4(\sum{V_{yi}})}{l} - \frac{6\sum(V_{yi}•x_i)}{l^2}[/latex]

[latex]w_1 = \frac{4(54.8 k)}{92 ft} - \frac{6 (3032 k • ft)}{(92 ft)^2} = 2.383 klf - 2.149 klf = 0.234 klf[/latex]

[latex]w_2 = -\frac{2(\sum{V_{yi}})}{l} + \frac{6 \sum(V_{yi} • x_i)}{l^2}[/latex]

[latex]w_2 = -\frac{2(54.8 k)}{92 ft} + \frac{6(3032 k • ft)}{(92 ft)^2} = -1.191 klf + 2.149 klf = 0.958 klf[/latex]

Values at End of Each Segment:

At [latex]x_1 = 0 ft[/latex]:

[latex]w_{a1} = 0.234 klf[/latex]

[latex]V_{a1} = 17.6 k[/latex]

[latex]M_{a1} = 0[/latex]

At [latex]x_2 = 0 ft[/latex]:
Linearly interpolate to find w:

[latex]w_{a2} = \frac{(0.958 klf - 0.234 klf)}{92 ft}(60 ft) + 0.234 klf = 0.706 klf[/latex]

[latex]+\uparrow \sum{F_y} = 0 = -17.6 k - 12.2 k + \frac{1}{2}(0.234 klf + 0.706 klf)(60 ft) - V_{a2}[/latex]

[latex]V_{a2} = -1.6 k[/latex]

[latex]+\circlearrowright M_{a2} = 0 = 17.6 k (60 ft) - 0.234 klf(60 ft)(30 ft) - \frac{1}{2}(0.706 klf - 0.234 klf)(60 ft)(20 ft) + M_{a2}[/latex]

[latex]M_{a2} = -352 k • ft[/latex]

Note that because the free body diagram is upside down compared with our derived formulas, we must reverse the signs on [latex]V_{a2}[/latex] and [latex]M_{a2}[/latex] before using them to find M(x) and [latex]M_{max}[/latex].

[latex]V_{a2} = 1.6 k\M_{a2} = 352 k • ft[/latex]

Expressions for Moment:

[latex]M ( x ) = \frac{(w_a - w_b)}{6l}x^3 - \frac{w_a}{2}x^2 + V_ax + M_a[/latex]

[latex]M (x_1 ) = \frac{(0.234 klf - 0.706 klf)}{6 (60 ft)}x_1^3 - \frac{0.234 klf}{2}x_1^2 + (17.6 k)x_1[/latex]

[latex]M (x_2) = \frac{(0.706 klf - 0.958 klf)}{6 (32 ft)}x_2^3 - \frac{0.706 klf}{2}x_2^2 + (1.6 k)x_2 + 352 k • ft[/latex]

Satisfied Fundamental Principles?
Check that the moment closes to zero:

At [latex]x_2 = 32 ft[/latex]:

[latex]M (x_2) = \frac{(0.706 klf - 0.958 klf)}{6(32 ft)}(32 ft)^3 - \frac{0.706 klf}{2}(32 ft)^2 + 1.6 k (32 ft) + 352 k • ft[/latex]

[latex]= -1.3 k • ft \ll each term ∴ = 0[/latex] ✓

Peak Moments:

[latex]x = \frac{w_a ± \sqrt{w_{a}^2 + 2(w_b - w_a)V_a/l}}{(w_a - w_b)}(l)[/latex]

Left segment:

[latex]x_1 = \frac{0.234 klf ± \sqrt{(0.234 klf)^2 + 2(0.706 klf - 0.234 klf)(17.6 klf)/(60 ft)}}{(0.234 klf - 0.706 klf)}(60 ft)[/latex]

[latex]= \frac{0.234 klf ± 0.576 klf}{-0.472 klf}(60 ft)[/latex]

[latex]= -103.0 ft or 43.5 ft[/latex]

Only one value is within the range of [latex]x_1: 43.5 ft[/latex].

[latex]M_{max} (x_1) = \frac{(0.234 klf - 0.706 klf)}{6(60 ft)}(43.5 ft)^3 - \frac{0.234 klf}{2}(43.5 ft)^2 + 17.6 k(43.5 ft)[/latex]

[latex]M_{max} (x_1) = 436 k • ft[/latex]

Check the value at the end of the segment, [latex]x_1 = 60 ft[/latex]:

[latex]M (x_1) = \frac{(0.234 klf - 0.706 klf)}{6(60 ft)}(60 ft)^3 - \frac{0.234 klf}{2}(60 ft)^2 + 17.6 k (60 ft)[/latex]

[latex]M_{b1} = 352 k • ft < 436 k • ft ∴ ext{not the maximum}[/latex]

Right segment:

[latex]x_2 = \frac{0.706 klf ± \sqrt{(0.706 klf)^2 + 2(0.958 klf - 0.706 klf)(1.6 k)/(32 ft)}}{(0.706 klf - 0.958 klf)}(32 ft)[/latex]

[latex]= \frac{0.706 klf ± 0.724 klf }{-0.252 klf}(32 ft)[/latex]

[latex]= -181.6 ft or 2.3 ft[/latex]

Only one value is within the range of [latex]x_2: 2.3 ft[/latex].

[latex]M_{max} (x_2) = \frac{(0.706 klf - 0.958 klf)}{6(32 ft)}(2.3 ft)^3 - \frac{0.706 klf}{2}(2.3 ft)^2 + 1.6 k (2.3 ft) + 352 k • ft[/latex]

[latex]M_{max} (x_2) = 354 k • ft[/latex]

Chord Forces:
Left segment:

[latex]M_{max} (x_1) = 436 k • ft[/latex]

[latex]T (x_1) = \frac{436 k • ft}{40 ft} = 10.90 k[/latex]

Right segment:

[latex]M_{max} (x_2) = 354 k • ft[/latex]

[latex]T (x_2) = \frac{354 k • ft}{72 ft} = 4.92 k[/latex]

Evaluation of Results:
Approximation Predicted Outcomes?

Left segment approximate result: [latex]T_1 = 5.22 k (52\% low)[/latex]

Right segment approximate result: [latex]T_2 = 1.49 k (70\% low)[/latex]

We expect these results to be low because the interior lateral element carries much less shear than the outer lateral elements; therefore, the moment stays positive throughout the diaphragm and acts more like a simply supported member without the interior support. ✓

Question 11.6: We are designing a new building that will be used as an offi...

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