Chapter 9, Problem 9.8.7P

To determine

(a)

To select the $\text{W}?X?$ satisfactory shape using LFRD method.

Answer to Problem 9.8.7P

The beam $\text{W}24\times 76$ is safe.

Explanation of Solution

Given:

Thickness of slab, t = 5.0 inches, spacing = 7.0 ft, span length, L = 30 feet, yield stress = 50 Ksi

Construction load = 20 psf, and live load = 800 psf.

The value of $f_{{}_c}^{\text{'}}=4ksi$.

Calculation:

Using LRFD method, we select a suitable shape that will satisfy the given conditions:

Calculate the loads on the beam as follows:

After curing we have,

$W_S=\frac{t}{12}(150)L_p$

Where, $t$ is the thickness of the slab, $W_S$ the self-weight of slab on one beam, $L_p$ is the distance

between two adjacent beams.

$W_S=\frac{t}{12}(150)L_p$

$\begin{array}{l}{W}_S=\frac{5.0}{12}(150)\times 7.0\\ W_S=437.50\frac{lb}{ft}.\end{array}$

The dead load on the beam after the concrete has cured is:

$\begin{array}{l}{W}_D=W_S\\ W_D=W_S=437.50\frac{lb}{ft}\end{array}$

Calculate the live load on the beam using the following equation:

$W_D=W_{{}_{SL}}{L}_P$

Where, $W_{SL}$ is the construction load on the beam.

$\begin{array}{l}{W}_D=W_{{}_{SL}}{L}_P\\ W_D=800.00psf\times 7.0ft\\ W_D=5600.00\frac{lb}{ft}.\end{array}$

Calculate the factored uniformly distributed load after curing has completed by following formula:

$W_u=\left(\left(1.2\times W_D\right)+\left(1.6\times W_L\right)\right)\frac{Kips}{ft}$

Where, $W_D$ is the uniformly distributed dead load on the beam and $W_L$ is the applied live load on the beam.

Substitute the values, we get

$\begin{array}{l}{W}_u=\left(\left(1.2\times W_D\right)+\left(1.6\times W_L\right)\right)\frac{Kips}{ft}\\ W_u=\left(\left(1.2\times 0.4375\right)+\left(1.6\times 5.600\right)\right)\frac{Kips}{ft}\\ W_u=0.525\frac{Kips}{ft}+8.960\frac{Kips}{ft}\\ W_u=9.485\frac{Kips}{ft}\end{array}$

Calculate the bending moment on the beam;

$M_u=\frac{w_u{L}_{AB}^2}{8}$

Where, $M_u$ is the bending moment and $L_{AB}^{}$ is the length of the beam.

$\begin{array}{l}{M}_u=\frac{w_u{L}_{AB}^2}{8}\\ M_u=\frac{\left(9.485\times {30}^2\right)}{8}\frac{Kips}{ft}f{t}^2\\ M_u=1067.06Kips\text{-}ft.\end{array}$

Try for $W24\times 76$, we have from the manual of ASIC

DesignationImperial (in x lb/ft)	Depthh (in)	Widthw (in)	Web Thicknesstw (in)	Flange Thicknesstf (in)	Sectional Area (in2)	Weight (lbf/ft)	Static Parameters
							Moment of Inertia		Elastic Section Modulus
							Ix (in4)	Iy (in4)	Sx (in3)	Sy (in3)
W 24 x 76	23.9	9	0.440	0.680	22.4	76	2100	82.5	176	18.4

${\displaystyle \sum Q_n}=394Kips.ft$

Calculate the distance of the plastic neutral axis from the top of the slab as follows:

$C_C=0.85f_C^{\text{'}}bt$

Where, b is the width of the concrete slab, t is the thickness of the concrete beam, $f_C^{\text{'}}$ is the compressive strength of the concrete and a is the distance of neutral axis from the top of the slab.

The effective flange width is as follows;

$\begin{array}{l}b=\text{Min}\left(\frac{12L_{AB}}{t},12L_p\right)\\ b=\text{Min}\left(\frac{12\times 30}{4},12\times 7.0\right)\\ b=\text{Min}\left(90in,84in\right)\\ b=84in.\end{array}$

Substitute the values, we have

$0.85f_C^{\text{'}}ab=C$

$\begin{array}{l}0.85\times 4Ksi\times a\times 84=394.00kips\\ a=\frac{394.00kips}{0.85\times 4Ksi\times 84}\\ a=1.3795in\\ a=1.380in\end{array}$

Compute the value of Y as shown below:

$\begin{array}{l}y=t-\frac{a}{2}\\ y=5.0in-\frac{1.380in}{2}\\ y=5.0in-0.690in.\\ y=4.310in.\end{array}$

From the manual the value of nominal flexural strength of the beam

${\varphi }_b\times M_n=1102Kips.ft$

Comparing the values of $\left({\varphi }_b{M}_n\right)$ and $M_u$, we have

$\begin{array}{l}1067.00Kips\text{-}ft<1102.00\text{}Kips.ft\\ M_u<\left({\varphi }_b{M}_n\right)\end{array}$

Thus, the beam is satisfactory excluding its self-weight.

Now for including the weight of the beam, we have

Calculate the factored uniformly distributed load after curing has completed by following formula:

Where, $W_S$ is the self-weight of the beam, $W_D$ is the uniformly distributed dead load on the beam and $W_L$ is the applied live load on the beam.

Substitute the values, we get

$\begin{array}{l}{W}_u=\left(1.2\times \left(W_S+W_D\right)+\left(1.6\times W_L\right)\right)\frac{Kips}{ft}\\ W_u=\left(\left(1.2\left(0.4375psf+0.076psf\right)\right)+\left(1.6\left(5.60psf\right)\right)\right)\frac{Kips}{ft}\\ W_u=0.6162\frac{Kips}{ft}+8.960\frac{Kips}{ft}\\ W_u=9.5762\frac{Kips}{ft}\end{array}$

Calculate the maximum bending moment on the beam;

$M_u=\frac{w_u{L}_{}^2}{8}$

Where, $M_u$ is the maximum bending moment and $L_{}^{}$ is the length of the beam.

$\begin{array}{l}{M}_u=\frac{w_u{L}_{}^2}{8}\\ M_u=\frac{\left(9.5762\times {30}^2\right)}{8}\frac{Kips}{ft}f{t}^2\\ M_u=1077.323Kips\text{-}ft.\end{array}$

Check the flexural strength of the beam including its weight.

$\begin{array}{l}1077.00Kips\text{-}ft<1102.00\text{}Kips.ft\\ M_{ul}<\left({\varphi }_b{M}_n\right)\end{array}$

Thus, the beam is satisfactory including its self-weight.

Check for the shear:

Checking the value of nominal value of shear strength of $\text{W}24\times 76$ from the AISC Manual.

${\varphi }_v{V}_n=315.00\text{kips}$

Where, ${\varphi }_v{V}_n$ is the nominal shear strength.

The maximum shear force is as following for the above conditions:

$V_n=\frac{w_{u}L}{2}$

Substitute the values, we have

$\begin{array}{l}{V}_n=\frac{w_{u}L}{2}\\ V_n=\frac{9.576{\text{kips ft}}^{\text{-1}}\times 30\text{ft}}{2}\\ V_n=143.00\text{kips}\text{.}\end{array}$

Now comparing the two we have

$V_n=143.00\text{kips}<{\varphi }_v{V}_n=315.00\text{kips}$

Therefore, the beam is safe in shear and we can use $\text{W}24\times 76$.

Calculate the factored uniformly distributed load before curing has completed by following formula:

$W_u=\left(\left(1.2\times W_D\right)+\left(1.6\times W_L\right)\right)\frac{Kips}{ft}$

Where, $W_D$ is the uniformly distributed dead load on the beam and $W_L$ is the applied live load on the beam.

$W_S=\frac{t}{12}(150)L_p$

Where, $t$ is the thickness of the slab, $W_S$ the self-weight of slab on one beam, $L_p$ is the distance

between two adjacent beams.

$W_S=\frac{t}{12}(150)L_p$

$\begin{array}{l}{W}_S=\frac{5.0}{12}(150)\times 7.0\\ W_S=437.50\frac{lb}{ft}.\end{array}$

The dead load on the beam before the concrete has cured is:

$\begin{array}{l}{W}_D=W_S+W\\ W_D=437.50\frac{lb}{ft}+76.00\frac{lb}{ft}\\ W_D=513.50\frac{lb}{ft}\end{array}$

Calculate the live load on the beam using the following equation:

$W_D=W_{{}_{SL}}{L}_P$

Where, $W_{SL}$ is the construction load on the beam.

$\begin{array}{l}{W}_D=W_{{}_{SL}}{L}_P\\ W_D=20.00psf\times 7.0ft\\ W_D=140.00\frac{lb}{ft}.\end{array}$

Substitute the values, we get

$\begin{array}{l}{W}_u=\left(\left(1.2\times W_D\right)+\left(1.6\times W_L\right)\right)\frac{Kips}{ft}\\ W_u=\left(\left(1.2\times 0.5135\right)+\left(1.6\times 0.140\right)\right)\frac{Kips}{ft}\\ W_u=0.6162\frac{Kips}{ft}+0.224\frac{Kips}{ft}\\ W_u=0.8402\frac{Kips}{ft}\end{array}$

Calculate the maximum bending moment on the beam;

$M_u=\frac{w_u{L}_{}^2}{8}$

Where, $M_u$ is the maximum bending moment and $L_{}^{}$ is the length of the beam.

$\begin{array}{l}{M}_u=\frac{w_u{L}_{}^2}{8}\\ M_u=\frac{\left(0.8402\times {30}^2\right)}{8}\frac{Kips}{ft}f{t}^2\\ M_u=94.523Kips\text{-}ft.\end{array}$

Check for the value of nominal flexural strength, the flexural strength of the beam before curing is

${\varphi }_b\times M_n={\varphi }_b\times M_P=315.00Kips.ft$

Comparing the values of $\left({\varphi }_b{M}_n\right)$ and $M_u$, We have

$\begin{array}{l}94.00Kips\text{-}ft<315.00\text{}Kips.ft\\ M_u<\left({\varphi }_b{M}_n\right)\end{array}$

Therefore, the beam is satisfactory before the curing has completed.

Now calculating the maximum allowable live load deflection from the given beam using the formula as:

${\Delta }_{Max}=\frac{L}{360}$

Substitute the values, we have

$\begin{array}{l}{\Delta }_{Max}=\frac{L}{360}\\ {\Delta }_{Max}=\frac{30\times 12}{360}\\ {\Delta }_{Max}=\frac{360}{360}in\\ {\Delta }_{Max}=1.00in\end{array}$

We have the value of lower bound moment of inertia for the given condition as follows:

$I=3642i{n}^4$

Calculating the total load on the beam using the following :

$\Delta =\frac{5w_L{L}^4}{384EI}$

Where, $\Delta$ is the maximum deflection, L is the length of the beam, W is the uniformly distributed load on beam, E is the modulus of elasticity and I is the moment of inertia.

$\begin{array}{l}\Delta =\frac{5w_L{L}^4}{384EI}\\ \Delta =\frac{5\times 5.60Kips.ft\times {\left(30\times 12\right)}^4}{384\times 12\times 29000Ksi\times 3642i{n}^4}\\ \Delta =0.9663in.\end{array}$

Now by comparing the values, we have

${\Delta }_{Max}=1.00in>\Delta =0.9663in.$

Conclusion:

Hence, the beam $\text{W}24\times 76$ is safe.

To determine

(b)

Use ASD method to select the $\text{W}?X?$ satisfactory shape.

Answer to Problem 9.8.7P

The beam $\text{W}24\times 76$ is safe.

Explanation of Solution

Calculation:

Now, we will use allowable stress design

Calculate the loads on the beam as follows:

After curing we have,

$W_S=\frac{t}{12}(150)L_p$

Where, $t$ is the thickness of the slab, $W_S$ the self-weight of slab on one beam, $L_p$ is the distance

between two adjacent beams.

$W_S=\frac{t}{12}(150)L_p$

$\begin{array}{l}{W}_S=\frac{5.0}{12}(150)\times 7.0\\ W_S=437.50\frac{lb}{ft}.\end{array}$

The dead load on the beam after the concrete has cured is:

$\begin{array}{l}{W}_D=W_S\\ W_D=W_S=437.50\frac{lb}{ft}\end{array}$

Calculate the live load on the beam using the following equation:

$W_L=W_{{}_{SL}}{L}_P$

Where, $W_{SL}$ is the construction load on the beam

$\begin{array}{l}{W}_L=W_{{}_{SL}}{L}_P\\ W_L=800.00psf\times 7.0ft\\ W_L=5600.00\frac{lb}{ft}.\end{array}$

Calculate the allowable uniformly distributed load after curing has completed by following formula:

$W_a=\left(W_D+W_L\right)\frac{Kips}{ft}$

Where, $W_D$ is the uniformly distributed dead load on the beam and $W_L$ is the applied live load on the beam.

Substitute the values, we get

$\begin{array}{l}{W}_a=\left(W_D+W_L\right)\frac{Kips}{ft}\\ W_u=0.4375\frac{Kips}{ft}+5.60\frac{Kips}{ft}\\ W_u=6.0375\frac{Kips}{ft}\end{array}$

Calculate the bending moment on the beam;

$M_u=\frac{w_u{L}_{AB}^2}{8}$

Where, $M_u$ is the bending moment and $L_{AB}^{}$ is the length of the beam.

$\begin{array}{l}{M}_u=\frac{w_u{L}_{AB}^2}{8}\\ M_u=\frac{\left(6.0375\times {30}^2\right)}{8}\frac{Kips}{ft}f{t}^2\\ M_u=679.22Kips\text{-}ft.\end{array}$

Compute the value of Y, which is the distance from the top of steel shape to compressive force

in concrete and is shown below:

$\begin{array}{l}y=t-\frac{a}{2}\\ y=5.0in-\frac{1.00in}{2}\\ y=5.0in-0.50in.\\ y=4.50in.\end{array}$

Try for $W24\times 76$, we have from the manual of ASIC

${\displaystyle \sum Q_n}=335.00Kips.ft$

Calculate the distance of the plastic neutral axis from the top of the slab as follows:

$C_C=0.85f_C^{\text{'}}bt$

Where, b is the width of the concrete slab, t is the thickness of the concrete beam, $f_C^{\text{'}}$ is the compressive strength of the concrete and a is the distance of neutral axis from the top of the slab.

The effective flange width is as follows;

$\begin{array}{l}b=\text{Min}\left(\frac{12L_{AB}}{t},12L_p\right)\\ b=\text{Min}\left(\frac{12\times 30}{4},12\times 7.0\right)\\ b=\text{Min}\left(90in,84in\right)\\ b=84in.\end{array}$

Substitute the values, we have

$0.85f_C^{\text{'}}ab=C$

$\begin{array}{l}0.85\times 4Ksi\times a\times 84=335.00kips\\ a=\frac{335.00kips}{0.85\times 4Ksi\times 84}\\ a=1.173in\end{array}$

Compute the value of Y as shown below:

$\begin{array}{l}y=t-\frac{a}{2}\\ y=5.0in-\frac{1.173in}{2}\\ y=5.0in-0.5865in.\\ y=4.414in.\end{array}$

From the table 3-19 of the ASIC manual:

Trying for $\text{W}21\times 48$

Check whether the section is safe in flexure if the self -weight is excluded

$\frac{M_n}{{\Omega }_b}=732.00\text{ft-kips}$

Where, $\frac{M_n}{{\Omega }_b}$ is the allowable flexural strength.

Now comparing the values of $M_a$ and $\frac{M_n}{{\Omega }_b}$

Substitute the values, we have

$\frac{M_n}{{\Omega }_b}=732.00\text{ft-kips > }{M}_u=679.00\text{ft-kips}$

Therefore, the section is safe in flexure if the self -weight is excluded.

Let’s check for the beam weight :

Calculation of the maximum bending moment as follows:

$M_{al}=M_a+\frac{w_s{L}^2}{8}$

Substitute the values, we get

$\begin{array}{l}{M}_{al}=M_a+\frac{w_s{L}^2}{8}\\ M_{al}=679.00\text{ft-kips}+\frac{0.076{\text{kips ft}}^{\text{-1}}\times {\left(30\text{ft}\right)}^{\text{2}}}{8}\\ M_{al}=679.00\text{ft-kips}\text{+}8.55\text{ft-kips}\\ M_{al}=687.55\text{ft-kips}\end{array}$

Now check for the flexural strength including the beam weight:

Comparing the values of maximum bending moment and the nominal flexural strength as follows:

$M_{al}=687.55\text{ft-kips}\text{<}\frac{M_n}{{\Omega }_b}=732.00\text{ft-kips}$

Therefore, the section is safe in flexure including the self-weight of the beam.

Check for the shear:

Checking the value of nominal value of shear strength of $\text{W}24\text{×76}$ from the AISC Manual.

$\frac{V_n}{{\Omega }_b}=210.00\text{kips}$

Where, $\frac{V_n}{{\Omega }_b}$ is the nominal shear strength.

The maximum shear force is as following for the above conditions:

$V_a=\frac{w_{a}L}{2}$

Substitute the values, we have

$\begin{array}{l}{V}_a=\frac{w_{a}L}{2}\\ V_a=\frac{6.114{\text{kips ft}}^{\text{-1}}\times 30\text{ft}}{2}\\ V_a=91.71\text{kips}\end{array}$

Now comparing the two we have

$V_a=91.71\text{kips}<\frac{V_n}{{\Omega }_b}=210.00\text{kips}$

Therefore, the beam is safe in shear and we can use $\text{W}24\text{×76}$.

Calculate the loads on the beam as follows:

Before curing we have,

$W_S=\frac{t}{12}(150)L_p$

Where, $t$ is the thickness of the slab, $W_S$ the self-weight of slab on one beam, $L_p$ is the distance

between two adjacent beams.

$W_S=\frac{t}{12}(150)L_p$

$\begin{array}{l}{W}_S=\frac{5.0}{12}(150)\times 7.0\\ W_S=437.50\frac{lb}{ft}.\end{array}$

The dead load on the beam before the concrete has cured is:

$\begin{array}{l}{W}_D=W_S+W\\ W_D=437.50\frac{lb}{ft}+76.00\frac{lb}{ft}\\ W_D=513.50\frac{lb}{ft}\end{array}$

Calculate the live load on the beam using the following equation:

$W_D=W_{{}_{SL}}{L}_P$

Where, $W_{SL}$ is the construction load on the beam.

$\begin{array}{l}{W}_D=W_{{}_{SL}}{L}_P\\ W_D=20.00psf\times 7.0ft\\ W_D=140.00\frac{lb}{ft}.\end{array}$

Substitute the values, we get

$\begin{array}{l}{W}_u=\left(W_D+W_L\right)\frac{Kips}{ft}\\ W_u=0.5135\frac{Kips}{ft}+0.140\frac{Kips}{ft}\\ W_u=0.6535\frac{Kips}{ft}\end{array}$

Calculate the maximum bending moment on the beam;

$M_u=\frac{w_u{L}_{}^2}{8}$

Where, $M_u$ is the maximum bending moment and $L_{}^{}$ is the length of the beam.

$\begin{array}{l}{M}_u=\frac{w_u{L}_{}^2}{8}\\ M_u=\frac{\left(0.6535\times {30}^2\right)}{8}\frac{Kips}{ft}f{t}^2\\ M_u=73.52Kips\text{-}ft.\end{array}$

Checking the value of nominal flexural strength of W- section from the Z_X table of the ASIC manual:

Check whether the section is safe in flexure if the self -weight is excluded

$\frac{M_n}{{\Omega }_b}=\frac{M_P}{{\Omega }_b}=499.00\text{ft-kips}$

Where, $\frac{M_n}{{\Omega }_b}$ is the allowable flexural strength and $\frac{M_P}{{\Omega }_b}$ is the plastic moment.

Now comparing the values of $M_a$ and $\frac{M_n}{{\Omega }_b}$

Substitute the values, we have

$\frac{M_n}{{\Omega }_b}=499.00\text{ft-kips > }{M}_a=73.52\text{ft-kips}$

Therefore, the section is safe in before the curing of concrete.

Now calculating the maximum allowable live load deflection from the given beam using the formula as:

${\Delta }_{Max}=\frac{L}{360}$

Substitute the values, we have

$\begin{array}{l}{\Delta }_{Max}=\frac{L}{360}\\ {\Delta }_{Max}=\frac{30\times 12}{360}\\ {\Delta }_{Max}=\frac{360}{360}in\\ {\Delta }_{Max}=1.00in\end{array}$

We have the value of lower bound moment of inertia for the given condition as follows:

$I=3642i{n}^4$

Calculating the total load on the beam using the following :

$\Delta =\frac{5w_L{L}^4}{384EI}$

Where, $\Delta$ is the maximum deflection, L is the length of the beam, W is the uniformly distributed load on beam, E is the modulus of elasticity and I is the moment of inertia.

$\begin{array}{l}\Delta =\frac{5w_L{L}^4}{384EI}\\ \Delta =\frac{5\times 5.60Kips.ft\times {\left(30\times 12\right)}^4}{384\times 12\times 29000Ksi\times 3642i{n}^4}\\ \Delta =0.9663in.\end{array}$

Now by comparing the values, we have

${\Delta }_{Max}=1.00in>\Delta =0.9663in.$

Conclusion:

Hence, the beam $\text{W}24\times 76$ is safe.

To determine

(c)

Selecting the stud anchors.

Answer to Problem 9.8.7P

We will use $54$ Studs with $\frac{5}{8}inch\times 2\frac{1}{2}inch$ dimension, with the spacing of $6.50in$.

Explanation of Solution

Calculation:

From AISC specifications, compute the maximum stud diameter using the equation:

$d_{\max }=2.5{\text{t}}_{\text{f}}$

Where, ${\text{t}}_{\text{f}}$ id the flange thickness and $d_{\max }$ is the maximum diameter.

Substitute the values, we get

$\begin{array}{l}{d}_{\max }=2.5{\text{t}}_{\text{f}}\\ d_{\max }=2.5\times 0.680\text{in}\\ d_{\max }=1.70\text{in}\end{array}$

Try for the studs of size $\frac{5}{8}in\times 2\frac{1}{2}in$ studs.

From table 3-21 for lightweight concrete take one stud at each of the beam position.

$Q_n=15.00Kips$

The number of studs for half beam can be found as follows:

$N_1=\frac{{\displaystyle \sum Q_n}}{Q_n}.$

Substitute the values, we have

$\begin{array}{l}{N}_1=\frac{{\displaystyle \sum Q_n}}{Q_n}.\\ {\displaystyle \sum Q_n}=V\\ N_1=\frac{394.00Kips}{15.00Kips}.\\ N_1=26.27\\ N_1\approx 27\end{array}$

Compute the number of studs as follows:

$N=2N_1.$

Substitute the value of $N_1$, we have

$\begin{array}{l}N=2N_1.\\ N=2\times 27\\ N=54\end{array}$

Calculate the spacing of the studs as follows:

Compute the minimum longitudinal spacing for studs using the equation

$S_{t,\min }=6d$

Where, $S_{t,\min }$ is the minimum longitudinal spacing for studs and $d$ is the diameter of the stud.

Substitute the values

$\begin{array}{l}{S}_{t,\min }=6d\\ S_{t,\min }=6\times \frac{5}{8}in\\ S_{t,\min }=3.75in\end{array}$

Compute the minimum transverse spacing for studs using the equation

$S_{t,\min }=4d$

Where, $S_{t,\min }$ is the minimum transverse spacing for studs and $d$ is the diameter of the stud.

Substitute the values

$\begin{array}{l}{S}_{t,\min }=4d\\ S_{t,\min }=4\times \frac{5}{8}in\\ S_{t,\min }=2.50in\end{array}$

Compute the maximum longitudinal spacing for studs using the equation

$S_{t,\max }=8t$

Where, $S_{t,\max }$ is the minimum transverse spacing for studs and $d$ is the thickness of the concrete slab.

Substitute the values

$\begin{array}{l}{S}_{t,\max }=8t\\ S_{t,\max }=8\times 5in\\ S_{t,\max }=40in\end{array}$

But the upper limit of the spacing is 36 inches.

Calculate the require spacing for one stud at each of the section:

$S=\frac{L}{N}$

Substitute the values, we have

$\begin{array}{l}S=\frac{30\times 12}{54}\\ S=6.670in.\end{array}$

Conclusion:

Therefore, we will use $54$ Studs with $\frac{5}{8}inch\times 2\frac{1}{2}inch$ dimension, with the spacing of $6.50in$.