Our firm has been selected to design a new manufacturing building in an industrial complex. Because of the sensitivity of the equipment that will be mounted on the roof, we anticipate using a rigid diaphragm. The arrangement of the manufacturing process has dictated the layout of the gravity

Question

Our firm has been selected to design a new manufacturing building in an industrial complex. Because of the sensitivity of the equipment that will be mounted on the roof, we anticipate using a rigid diaphragm. The arrangement of the manufacturing process has dictated the layout of the gravity system and the placement of the diagonal braces for lateral resistance.

Based on the geometry of the diagonal braces, we have been able to estimate the rigidities of the different braced frames. The next step in the process is to design the members of the lateral load resisting system. To do that, we need to know the shear generated in each of the braced frames.

We are using the Directional Procedure to determine the effects of wind on the structure.

The Directional Procedure has four different load cases: (1) maximum uniform wind in one direction at a time, (2) reduced uniform wind in one direction with torsion, (3) reduced uniform wind in two directions simultaneously, and (4) even further reduced uniform wind in two directions simultaneously with torsion. Our task is to determine the shear generated in each of the braced frames for case 2, considering wind from the north.

Accepted Answer

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

x-coordinate:
Braced frames I,II, and III 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{13(7 \mathrm{~m}) + 15(25 \mathrm{~m}) + 19(55 \mathrm{~m})}{13 + 15 + 19} \&= 32.1 \mathrm{~m}\end{aligned}[/latex]

y-coordinate:
Braced frames IV and V contribute because they are oriented in the x-direction.

[latex]\begin{aligned}y_{c . r .} &= \frac{\Sigma R_i y_i}{\Sigma R_i} = \frac{13(0 \mathrm{~m}) + 19(15 \mathrm{~m})}{13 + 19} \&= 8.9 \mathrm{~m}\end{aligned}[/latex]

Translation:
Only braced frames I, II, and III resist translation caused by force in the y-direction.

Net force:

[latex]F = 1.95 \mathrm{kN} / \mathrm{m}(55 \mathrm{~m}) = 107.2 \mathrm{kN}(+\downarrow)[/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{13}{\Sigma(13 + 15 + 19)}(107.2 \mathrm{kN}) = 29.7 \mathrm{kN}(+\uparrow)[/latex]

[latex]V_{I I}^F = \frac{15}{\Sigma(13 + 15 + 19)}(107.2 \mathrm{kN}) = 34.2 \mathrm{kN}(+\uparrow)[/latex]

[latex]V_{I I I}^F = \frac{19}{\sum(13 + 15 + 19)}(107.2 \mathrm{kN}) = 43.3 \mathrm{kN}(+\uparrow)[/latex]

Rotation:
All braced frames resist the rotation.

Element	R	d	Rd	Rd²
I	13	25.1 m	326 m	8190 m²
II	15	7.1 m	107 m	756 m²
III	19	22.9 m	435 m	9964 m²
IV	13	8.9m	116 m	1030 m²
V	19	6.1 m	116 m	707 m²
				Σ = 20,647 m²

Net moment:

[latex]M = 885 \mathrm{kN} \cdot \mathrm{m} + 107.2 \mathrm{kN} \cdot(55 \mathrm{~m} / 2 - 32.1 \mathrm{~m}) = 1378 \mathrm{kN} \cdot \mathrm{m}[/latex]

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{326 \mathrm{~m}}{20,650 \mathrm{~m}^2}(1378 \mathrm{kN} \cdot \mathrm{m}) = 21.8 \mathrm{kN}(+\uparrow)[/latex]

[latex]V_{I I}^M = \frac{107 \mathrm{~m}}{20,650 \mathrm{~m}^2}(1378 \mathrm{kN} \cdot \mathrm{m}) = 7.1 \mathrm{kN}(+\uparrow)[/latex]

[latex]V_{I I I}^M = \frac{435 \mathrm{~m}}{20,650 \mathrm{~m}^2}(1378 \mathrm{kN} \cdot \mathrm{m}) = 29.0 \mathrm{kN}(+\downarrow)[/latex]

[latex]V_{I V}^M = \frac{116 \mathrm{~m}}{20,650 \mathrm{~m}^2}(1378 \mathrm{kN} \cdot \mathrm{m}) = 7.7 \mathrm{kN}(\stackrel{+}{\longleftarrow})[/latex]

[latex]V_V^M = \frac{116 \mathrm{~m}}{20,650 \mathrm{~m}^2}(1378 \mathrm{kN} \cdot \mathrm{m}) = 7.7 \mathrm{kN}(\stackrel{+}{ ightarrow})[/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 = 29.7 \mathrm{kN} + 21.8 \mathrm{kN} = 51.5 \mathrm{kN}(+\uparrow)[/latex]

[latex]V_{I I}^T = 34.2 \mathrm{kN} + 7.1 \mathrm{kN} = 41.3 \mathrm{kN}(+\uparrow)[/latex]

[latex]V_{I I I}^T = 43.3 \mathrm{kN} - 29.0 \mathrm{kN} = 14.3 \mathrm{kN}(+\uparrow)[/latex]

[latex]V_{I V}^T = 0-7.7 \mathrm{kN} \quad = -7.7 \mathrm{kN}(\stackrel{+}{ ightarrow})[/latex]

[latex]V_V^T = 0+7.7 \mathrm{kN} \quad = 7.7 \mathrm{kN}(\stackrel{+}{ ightarrow})[/latex]

Evaluation of Results:
Satisfied Fundamental Principles?

[latex]\stackrel{+}{ ightarrow} \Sigma F_x = -7.7 \mathrm{kN} + 7.7 \mathrm{kN} = 0[/latex] ✓

[latex]+\uparrow \Sigma F_y = -107.2 \mathrm{kN} + 51.5 \mathrm{kN} + 41.3 \mathrm{kN} + 14.3 \mathrm{kN} = -0.1 \mathrm{kN} \approx 0[/latex] ✓

Question 11.1: Our firm has been selected to design a new manufacturing bui...

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