The entire perimeter is made of precast double-t panels that serve as shear walls. We have designed the connections such that the panels act coupled. Therefore, the rigidity

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.

The entire perimeter is made of precast double-t panels that serve as shear walls. We have designed the connections such that the panels act coupled. Therefore, the rigidity of each element is proportional to the number of panels. A support beam runs from B2 to B3.

The building is in a zone with moderate seismic activity. A team member has calculated that the equivalent static seismic load will be 0.15 of the dead weight of the roof (seismic response coefficient [latex]C_S[/latex] = 0.15). The resulting distributed lateral loads have been calculated.

Our job is to determine the shear generated in each wall due to the design seismic event in the north-south direction.

Accepted Answer

Center of Rigidity:
Choose corner B1 as the reference location.

x-coordinate:
Walls I, III, and V contribute because they are oriented in the y-direction.

[latex]\begin{aligned}x_{c . r .} & = \frac{\Sigma R_i x_i}{\Sigma R_i} = \frac{10(0 \mathrm{ft}) + 18(92 \mathrm{ft}) + 8(60 \mathrm{ft})}{10 + 18 + 8} \& = 59.3 \mathrm{ft}\end{aligned}[/latex]

y-coordinate:
Walls II, IV, and VI contribute because they are oriented in the x-direction. Note that
wall IV has a negative y-coordinate.

[latex]\begin{aligned}y_{c, r .} & = \frac{\Sigma R_i y_i}{\Sigma R_i} = \frac{23(40 \mathrm{ft}) + 8(-32 \mathrm{ft}) + 15(0 \mathrm{ft})}{23 + 8 + 15} \& = 14.4 \mathrm{ft}\end{aligned}[/latex]

Trunslution:
Only walls I, III, and V resist translation caused by force in the y-direction.
Net force:

[latex]\begin{aligned}F & = 0.465 \mathrm{k} / \mathrm{ft}(60 \mathrm{ft}) + 0.837 \mathrm{k} / \mathrm{ft}(32 \mathrm{ft}) \& = 27.9 \mathrm{k} + 26.8 \mathrm{k} = 54.7 \mathrm{k}(+\uparrow)\end{aligned}[/latex]

Force generated in each lateral element due to translation:

[latex]V_i^F = \frac{R_i}{\Sigma R_j} F[/latex]

[latex]V_I^F = \frac{10}{\Sigma(10 + 18 + 8)}(54.7 \mathrm{k}) = 15.2 \mathrm{k}(+\downarrow)[/latex]

[latex]V_{I I I}^F = \frac{18}{\Sigma(10 + 18 + 8)}(54.7 \mathrm{k}) = 27.4 \mathrm{k}(+\downarrow)[/latex]

[latex]V_V^F = \frac{8}{\Sigma(10 + 18 + 8)}(54.7 \mathrm{k}) = 12.2 \mathrm{k}(+\downarrow)[/latex]

Rotation.:
All walls resist the rotation.

Element	R	d	Rd	Rd²
I	10	59.3 ft	593 ft	35,160 ft²
II	23	25.6 ft	589 ft	15,070 ft²
III	18	32.7 ft	589 ft	19,250 ft²
IV	8	46.4 ft	371 ft	17,220 ft²
V	8	0.7 ft	6 ft	4 ft²
VI	15	14.4 ft	216 ft	3,110 ft²
				Σ = 89,810 ft²

Net moment:

[latex]\begin{aligned}M & = 27.9 \mathrm{k}(30 \mathrm{ft} - 59.3 \mathrm{ft}) + 26.8 \mathrm{k}(76 \mathrm{ft} - 59.3 \mathrm{ft}) \& = -817 \mathrm{k} \cdot \mathrm{ft} + 448 \mathrm{k} \cdot \mathrm{ft} = -369 \mathrm{k} \cdot \mathrm{ft}\end{aligned}[/latex]

Since M is negative, the shear in the walls is opposite the directions we drew. To avoid confusion, let's just carry the negative sign and leave the arrows the way we drew them.

Force generated in each lateral element due to rotation:

[latex]V_i^M = \left(\frac{R_i d_i}{\Sigma R_j d_j^2} ight) M[/latex]

[latex]V_I^M = \frac{593 \mathrm{ft}}{89,810 \mathrm{ft}^2}(-369 \mathrm{k} \cdot \mathrm{ft}) = -2.4 \mathrm{k}(+\uparrow)[/latex]

[latex]V_{I I}^M = \frac{589 \mathrm{ft}}{89,810 \mathrm{ft}^2}(-369 \mathrm{k} \cdot \mathrm{ft}) = -2.4 \mathrm{k}(\stackrel{+}{ ightarrow})[/latex]

[latex]V_{I I I}^M = \frac{589 \mathrm{ft}}{89,810 \mathrm{ft}^2}(-369 \mathrm{k} \cdot \mathrm{ft}) = -2.4 \mathrm{k}(+\downarrow)[/latex]

[latex]V_{I V}^M = \frac{371 \mathrm{ft}}{89,810 \mathrm{ft}^2}(-369 \mathrm{k} \cdot \mathrm{ft}) = -1.5 \mathrm{k}(\stackrel{+}{\longleftarrow})[/latex]

[latex]V_V^M = \frac{6 \mathrm{ft}}{89,810 \mathrm{ft}^2}(-369 \mathrm{k} \cdot \mathrm{ft}) = 0 \mathrm{k}(+\downarrow)[/latex]

[latex]V_{V I}^M = \frac{216 \mathrm{ft}}{89,810 \mathrm{ft}^2}(-369 \mathrm{k} \cdot \mathrm{ft}) = -0.9 \mathrm{k}(\pm)[/latex]

Net Forces:
Let's choose +x and +y as positive directions for our sign convention.

[latex]V_i^T = V_i^F + V_i^M[/latex]

[latex]V_I^T = -15.2 \mathrm{k} - 2.4 \mathrm{k} = -17.6 \mathrm{k}(\stackrel{+}{ ightarrow})[/latex]

[latex]V_{I I}^T = \quad 0 - 2.4 \mathrm{k} = -2.4 \mathrm{k}(\stackrel{+}{ ightarrow})[/latex]

[latex]V_{I I I}^T = -27.4 \mathrm{k} + 2.4 \mathrm{k} = -25.0 \mathrm{k}(+\uparrow)[/latex]

[latex]V_{I V}^T = \quad 0 + 1.5 \mathrm{k} = 1.5 \mathrm{k}(\stackrel{+}{ ightarrow})[/latex]

[latex]V_V^T = -12.2 \mathrm{k} - 0 = -12.2 \mathrm{k}(+\uparrow)[/latex]

[latex]V_{V I}^T = \quad 0 + 0.9 \mathrm{k} = 0.9 \mathrm{k}(+\uparrow)[/latex]

Evaluation of Results:
Satisfied Fundamental Principles?

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = -2.4 \mathrm{k} + 1.5 \mathrm{k} + 0.9 \mathrm{k} = 0[/latex] ✓

[latex]+\uparrow \Sigma F_y = 54.7 \mathrm{k} - 17.6 \mathrm{k} - 25.0 \mathrm{k} - 12.2 \mathrm{k} = -0.1 \mathrm{kN} \approx 0[/latex] ✓

Question 11.2: The entire perimeter is made of precast double-t panels that...

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