Chapter 2, Problem 18P

(a)

To determine

Calculate the design wind load, base shear, and overturning moment.

Answer to Problem 18P

The resultant force acting on roof slab is $\underset{\_}{F_{\text{Roof}}=13.4\text{kips}}$.

The resultant force acting on second floor is $\underset{\_}{F_{\text{2nd}}=26.81\text{kips}}$.

The seismic base shear is $\underset{\_}{V_{\text{Base}}=40.21\text{kips}}$.

The overturning moment is $\underset{\_}{M_{OT}=804.15\text{kip-ft}}$.

Explanation of Solution

Given information:

The importance factor (I) is 1.15.

The value of $K_z$ is 1.

The mean roof height is 30 ft.

The height of the single floor (h) is 15 ft.

The basic wind speed is $90\text{mi}/\text{h}$.

The exposure is D.

Calculation:

Consider the exposure of the D.

Refer Table 2.9, “Adjustment factor $\lambda$ for building height and exposure” in the textbook.

The value of $\lambda$ is 1.66 for the height of 30 ft and the exposure D.

Calculate the design wind pressure as shown below.

$\begin{array}{c}{P}_S=\lambda K_{z}I{p}_{s30}\\ =1.66\times 1\times 1.15p_{s30}\\ =1.909p_{s30}\end{array}$        (1)

Refer Table 2.8 “Simplified horizontal design wind pressure $p_{s30}$” for zone A and zone C as shown in Table 1

Calculate the design wind pressures for zone A and zone C as shown in Table 1.

Zone	$p_{s30}\left(\text{psf}\right)$	$P_S=1.909p_{s30}\left(\text{psf}\right)$
A	12.8	24.44
C	8.5	16.23

Calculate the resultant force for each level as shown below.

The distance at which the load acts for zone A is $\begin{array}{c}2a=2\times 0.1\times 100\\ =20\text{ft}\end{array}$

The distance at which the load acts for zone C is $\begin{array}{c}{a}^{\prime }=100-2a\\ =100-20\\ =80\text{ft}\end{array}$

Calculate the resultant force acting on roof slab as shown below.

$\begin{array}{c}{F}_{\text{Roof}}={\left(P_s\times \text{Area}\right)}_A+{\left(P_s\times \text{Area}\right)}_C\\ ={\left(P_s\times \frac{h}{2}\times 2a\right)}_A+{\left(P_s\times \frac{h}{2}\times a^{\prime }\right)}_C\\ =\left(24.44\text{psf}\times \frac{15\text{ft}}{2}\times 20\text{ft}+16.23\text{psf}\times \frac{15\text{ft}}{2}\times 80\text{ft}\right)\times \frac{1\text{kip}}{1,000\text{lb}}\\ =13.4\text{kips}\end{array}$

Hence, the resultant force acting on roof slab $\underset{\_}{F_{\text{Roof}}=13.4\text{kips}}$.

Calculate the resultant force acting on second floor as shown below.

$\begin{array}{c}{F}_{\text{2nd}}={\left(P_s\times \text{Area}\right)}_A+{\left(P_s\times \text{Area}\right)}_C\\ =\left(24.44\text{psf}\times 15\text{ft}\times 20\text{ft}+16.23\text{psf}\times 15\text{ft}\times 80\text{ft}\right)\times \frac{1\text{kip}}{1,000\text{lb}}\\ =26.81\text{kips}\end{array}$

Hence, the resultant force acting on second floor $\underset{\_}{F_{\text{2nd}}=26.81\text{kips}}$.

Calculate the base shear force as shown below.

$\begin{array}{c}{V}_{\text{Base}}=F_{\text{Roof}}+F_{\text{2nd}}\\ =13.4+26.81\\ =40.21\text{kips}\end{array}$

Hence, the seismic base shear is $\underset{\_}{V_{\text{Base}}=40.21\text{kips}}$.

Calculate the overturning moment as shown below.

$\begin{array}{c}{M}_{OT}=F_{\text{Roof}}\times h_{\text{Roof}}+F_{\text{2nd}}\times h_{\text{2nd}}\\ =13.4\times 30+26.81\times 15\\ =804.15\text{kip-ft}\end{array}$

Therefore, the overturning moment is $\underset{\_}{M_{OT}=804.15\text{kip-ft}}$.

(b)

To determine

Calculate the base shear and overturning moment using the equivalent lateral force procedure.

Answer to Problem 18P

The seismic base shear is $\underset{\_}{V_{\text{Base}}=59.2\text{kips}}$.

The overturning moment is $\underset{\_}{M_{OT}=1,480.5\text{kip-ft}}$.

Explanation of Solution

Given information:

The average weight of the floor and roof is $90\text{lb}/{\text{ft}}^{\text{2}}$.

The value of $S_{DS}$ is 0.27g.

The value of $S_{D1}$ is 0.06g.

The value of R is 8.

The importance factor (I) is 1.5.

Calculation:

The value of $C_t$ is 0.016 for steel moment frames and $x=0.9$.

Calculate the fundamental period as shown below.

$\begin{array}{c}T=C_t{h}_n^x\\ =0.016\times {\left(30\right)}^{0.9}\\ =0.342\text{sec}\end{array}$

Calculate the total dead load of the building as shown below.

$\begin{array}{c}W=\text{Dead}\text{load}\text{of}\text{the}\text{floors}+\text{Dead}\text{load}\text{of}\text{the}\text{roof}\\ =\left[\left(100\text{ft}\times 100\text{ft}\right)\times 90\text{lb}/{\text{ft}}^{\text{2}}+\left(100\text{ft}\times 100\text{ft}\right)\times 90\text{lb}/{\text{ft}}^{\text{2}}\right]\times \frac{1\text{kip}}{1,000\text{lb}}\\ =1,800\text{kips}\end{array}$

Calculate the magnitude of the base shear as shown below.

$\begin{array}{c}V=\frac{S_{D1}W}{T\left(\frac{R}{I}\right)}\\ =\frac{0.06\times 1,800}{0.342\times \frac{8}{1.5}}\\ =59.2\text{kips}\end{array}$

Calculate the magnitude of the maximum base shear as shown below.

$\begin{array}{c}{V}_{\max }=\frac{S_{DS}W}{\frac{R}{I}}\\ =\frac{0.27\times 1,800}{\frac{8}{1.5}}\\ =91.125\text{kips}\end{array}$

Calculate the magnitude of the minimum base shear as shown below.

$\begin{array}{c}V=0.044S_{DS}IW\\ =0.044\times 0.27\times 1.5\times 1,800\\ =32.1\text{kips}\end{array}$

Hence, take the value of $V=59.2\text{kips}$.

The value of k is 1 for $T=0.342\sec <0.5\sec$.

Calculate the seismic base shear to each floor level as shown below.

$F_x=\frac{w_x{h}_x^k}{{\displaystyle \sum _{i=1}^n{w}_i{h}_i^k}}V$        (1)

Provide the calculated the seismic base shear at each floor levels as shown in Table 1.

Floor	Weight $\left(w_i,\text{kips}\right)$	Height of the floor $\left(h_i,\text{ft}\right)$	$w_i{h}_i^k$	$\frac{w_x{h}_x^k}{{\displaystyle \sum _{i=1}^n{w}_i{h}_i^k}}$	$F_x=\frac{w_x{h}_x^k}{{\displaystyle \sum _{i=1}^n{w}_i{h}_i^k}}V\left(\text{kips}\right)$
Roof	900	30	27, 000	0.667	39.5
2nd	900	15	13,500	0.333	19.7
Sum	1,800		40,500		59.2

Refer to Table 1:

The seismic base shear is $F_x=59.2\text{kips}$.

Hence, the seismic base shear is $\underset{\_}{V_{\text{Base}}=59.2\text{kips}}$.

Calculate the overturning moment as shown below.

$\begin{array}{c}{M}_{OT}=F_{\text{Roof}}\times h_{\text{Roof}}+F_{\text{2nd}}\times h_{\text{2nd}}\\ =39.5\times 30+19.7\times 15\\ =1,480.5\text{kip-ft}\end{array}$

Therefore, the overturning moment is $\underset{\_}{M_{OT}=1,480.5\text{kip-ft}}$.

(c)

To determine

Provide the design strength of the building govern the wind force of seismic force.

Answer to Problem 18P

The design strength of the building governed by the seismic force.

Explanation of Solution

Given information:

The average weight of the floor and roof is $90\text{lb}/{\text{ft}}^{\text{2}}$.

The value of $S_{DS}$ is 0.27g.

The value of $S_{D1}$ is 0.06g.

The value of R is 8.

The importance factor (I) is 1.5.

Calculation:

Refer to part (a).

The seismic base shear and overturning moment due to wind force.

The seismic base shear is $V_{\text{Base}}=40.21\text{kips}$.

The overturning moment is $M_{OT}=804.15\text{kip-ft}$.

Refer to part (b).

The seismic base shear and overturning moment due to seismic force.

The seismic base shear is $V_{\text{Base}}=59.2\text{kips}$.

The overturning moment is $M_{OT}=1,480.5\text{kip-ft}$.

The seismic base shear and overturning moment due to seismic force is greater than to compared with the seismic base shear and overturning moment due to seismic force.

Hence, the design strength of the building governed by the seismic force.