Our team is designing a single-story office building that has a flexible roof. Architectural requirements have led the senior engineer to use rigid frames for lateral load resistance. We are at the point of the design where we are determining the chord forces that the perimeter beams and

Question

Our team is designing a single-story office building that has a flexible roof. Architectural requirements have led the senior engineer to use rigid frames for lateral load resistance. We are at the point of the design where we are determining the chord forces that the perimeter beams and connections must carry. We are using full-depth collectors to transfer shear to each lateral element.
Our task is to determine the chord forces for one of the wind cases: uniform wind from the south.

Accepted Answer

Find Shear in Each Lateral Element:
Equivalent Beam Model:

Apply equilibrium:

[latex]\Sigma M_1 = 0 = 10.0 \mathrm{kN} / \mathrm{m}(16 \mathrm{~m})(7.5 \mathrm{~m}) - V_2(12.5 \mathrm{~m})[/latex]

[latex]V_2 = 96 \mathrm{kN}(+\downarrow)[/latex]

[latex]+\uparrow \Sigma F_y = 0 = 10.0 \mathrm{kN} / \mathrm{m}(16 \mathrm{~m}) - V_1 - 96 \mathrm{kN}[/latex]

[latex]V_1 = 64 \mathrm{kN}(+\downarrow)[/latex]

Expression for Shear and Moment:

[latex]V\left(x_1 ight) = \int(-10 \mathrm{kN} / \mathrm{m}) d x_1 = (-10 \mathrm{kN} / \mathrm{m}) x_1 + C_1[/latex]

At [latex]x_1 = 0, V = 0[/latex]:

[latex] herefore C_1 = 0[/latex]

So,

[latex]V\left(x_1 ight) = (-10 \mathrm{kN} / \mathrm{m}) x_1[/latex]

[latex]V\left(x_2 ight) = \int(-10 \mathrm{kN} / \mathrm{m}) d x_2 = (-10 \mathrm{kN} / \mathrm{m}) x_2 + C_2[/latex]

At [latex]x_2 = 0, x_1 = 0.5 \mathrm{~m}, V\left(x_2 ight) = V\left(x_1 ight) + 64 \mathrm{kN}:[/latex]

[latex]V\left(x_2 ight) = -10 \mathrm{kN} / \mathrm{m}(0.5 \mathrm{~m}) + 64 \mathrm{kN} = 59 \mathrm{kN}[/latex]

[latex] herefore C_2 = 59 \mathrm{kN}[/latex]

So,

[latex]V\left(x_2 ight) = (-10 \mathrm{kN} / \mathrm{m}) x_2 + 59 \mathrm{kN}[/latex]

[latex]V\left(x_3 ight) = \int(-10 \mathrm{kN} / \mathrm{m}) d x_3 = (-10 \mathrm{kN} / \mathrm{m}) x_3 + C_3[/latex]

At [latex]x_3 = 0, x_2 = 12.5 \mathrm{~m}, V\left(x_3 ight) = V\left(x_2 ight) + 96 \mathrm{kN}:[/latex]

[latex]V\left(x_3 ight) = -10 \mathrm{kN} / \mathrm{m}(12.5 \mathrm{~m}) + 59 \mathrm{kN} + 96 \mathrm{kN} = 30 \mathrm{kN}[/latex]

[latex] herefore C_3 = 30 \mathrm{kN}[/latex]

So,

[latex]V\left(x_3 ight) = (-10 \mathrm{kN} / \mathrm{m}) x_3 + 30 \mathrm{kN}[/latex]

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

At [latex]x_3 = 3 \mathrm{~m}:[/latex]

[latex]V\left(x_3 ight) = -10 \mathrm{kN} / \mathrm{m}(3 \mathrm{~m}) + 30 \mathrm{kN} = 0[/latex] ✓

[latex]M\left(x_1 ight) = \int(-10 \mathrm{kN} / \mathrm{m}) x_1 \cdot d x_1 = (-5 \mathrm{kN} / \mathrm{m}) x_1^2 + C_4[/latex]

At [latex]x_1 = 0, M = 0:[/latex]

[latex] herefore C_4 = 0[/latex]

So,

[latex]M\left(x_1 ight) = (-5 \mathrm{kN} / \mathrm{m}) x_1^2[/latex]

[latex]M\left(x_2 ight) = \int\left(-10 \mathrm{kN} / \mathrm{m} \cdot x_2 + 59 \mathrm{kN} ight) d x_2 = (-5 \mathrm{kN} / \mathrm{m}) x_2^2 + (59 \mathrm{kN}) x_2 + C_5[/latex]

At [latex]x_2 = 0, x_1 = 0.5 \mathrm{~m}, M\left(x_2 ight) = M\left(x_1 ight):[/latex]

[latex]M\left(x_2 ight) = -5 \mathrm{kN} / \mathrm{m}(0.5 \mathrm{~m})^2 = -1.25 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex] herefore C_5 = -1.25 \mathrm{kN} \cdot \mathrm{m}[/latex]

So,

[latex]M\left(x_2 ight) = (-5 \mathrm{kN} / \mathrm{m}) x_2^2 + (59 \mathrm{kN}) x_2 - 1.25 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M\left(x_3 ight) = \int\left(-10 \mathrm{kN} / \mathrm{m} \cdot x_3 + 30 \mathrm{kN} ight) d x_3 = (-5 \mathrm{kN} / \mathrm{m}) x_3^2 + (30 \mathrm{kN}) x_3 + C_6[/latex]

At [latex]x_3 = 0, x_2 = 12.5 \mathrm{~m}, M\left(x_3 ight) = M\left(x_2 ight)[/latex]:

[latex]M\left(x_3 ight) = -5 \mathrm{kN} / \mathrm{m}(12.5 \mathrm{~m})^2 + 59 \mathrm{kN}(12.5 \mathrm{~m}) - 1.25 \mathrm{kN} \cdot \mathrm{m} = -45 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex] herefore C_6 = -45 \mathrm{kN} \cdot \mathrm{m}[/latex]

So,

[latex]M\left(x_3 ight) = (-5 \mathrm{kN} / \mathrm{m}) x_3^2 + (30 \mathrm{kN}) x_3 - 45 \mathrm{kN} \cdot \mathrm{m}[/latex]

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

At [latex]x_3 = 3 m[/latex]:

[latex]M\left(x_3 ight) = -5 \mathrm{kN} / \mathrm{m}(3 \mathrm{~m})^2 + 30 \mathrm{kN}(3 \mathrm{~m}) - 45 \mathrm{kN} \cdot \mathrm{m} = 0[/latex] ✓

Peak Moments:
Since the equivalent beam is cantilevered on the left end, [latex]M_{1max}[/latex] is at the lateral element at [latex]x_1 = 0.5 m[/latex]:

[latex]M\left(x_1 ight) = -5 \mathrm{kN} / \mathrm{m}(0.5 \mathrm{~m})^2[/latex]

[latex]M_{1 \max } = -1.25 \mathrm{kN} \cdot \mathrm{m}[/latex]

Since the equivalent beam is cantilevered on the other end, [latex]M_{3max}[/latex] is at the lateral element at [latex]x_3 = 0 m[/latex]:

[latex]M\left(x_3 ight) = -5 \mathrm{kN} / \mathrm{m}(0)^2 + 30 \mathrm{kN}(0) - 45 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{3 \max } = -45 \mathrm{kN} \cdot \mathrm{m}[/latex]

Between lateral elements, M_{2max} is where V = 0 or at a lateral element:

[latex]V\left(x_2 ight) \equiv 0 = (-10 \mathrm{kN} / \mathrm{m}) x_2 + 59 \mathrm{kN}[/latex]

[latex]x_2 = 5.9 \mathrm{~m}[/latex]

[latex]M\left(x_2 ight) = -5 \mathrm{kN} / \mathrm{m}(5.9 \mathrm{~m})^2 + 59 \mathrm{kN}(5.9 \mathrm{~m}) - 1.25 \mathrm{kN} \cdot \mathrm{m} = 172.8 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M\left(x_2 ight) = 172.8 \mathrm{kN} \cdot \mathrm{m} > 1.25 \mathrm{kN} \cdot \mathrm{m} ext { and } 45 \mathrm{kN} \cdot \mathrm{m}[/latex]

[latex]M_{2 \max } = 172.8 \mathrm{kN} \cdot \mathrm{m}[/latex]

Chord Forces:
The depth between chords is 5 m + 7 m =12 m.

In the range of [latex]x_1[/latex]:

[latex]T_1 = \frac{1.25 \mathrm{kN} \cdot \mathrm{m}}{12 \mathrm{~m}} = 0.10 \mathrm{kN}[/latex]

Note: since this is a very short span and small force, no beam will be added.
In the range of [latex]x_2[/latex]:

[latex]T_2 = \frac{172.8 \mathrm{kN} \cdot \mathrm{m}}{12 \mathrm{~m}} = 14.40 \mathrm{kN}[/latex]

In the range of [latex]x_3[/latex]:

[latex]T_3 = \frac{45 \mathrm{kN} \cdot \mathrm{m}}{12 \mathrm{~m}} = 3.75 \mathrm{kN}[/latex]

Summary:

Question 11.5: Our team is designing a single-story office building that ha...

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