Chapter 10, Problem 10.7.7P

To determine

(a)

The girder cross section and the spacing of intermediate stiffeners that is required.

Answer to Problem 10.7.7P

Web dimensions $\frac{1}{2}\text{in}\times 83\text{in}$

Flange $1\frac{1}{2}\text{in}\times 24\text{in}$

Explanation of Solution

Given information:

Span length $L=66ft$

Uniform dead load $w_D=1.3\text{kip/ft}$

Uniform live load $w_L=2.3\text{kip/ft}$

Dead load $P_D=28\text{kip}$

Live load $P_L=49\text{kip}$

Depth $d=80\text{in}$

$F_y=50$

Calculation:

Determine girder cross section and girder spacing of intermediate stiffeners.

$\begin{array}{l}{w}_u=1.2w_D+1.6w_L\\ w_u=1.2\left(1.3\right)+1.6\left(2.3\right)\\ w_u=5.24\text{kips/ft}\\ P_u=1.2P_D+1.6P_L\\ P_u=1.2\times 28+1.6\times 49\\ P_u=112\text{kips}\end{array}$

Calculate design strength.

$\begin{array}{l}{V}_u=\frac{w_{u}L}{2}+P_u\\ V_u=\frac{5.24\times 66}{2}+112\\ V_u=284.92\text{kips}\\ M_u=284.92\times 33-\frac{5.24\times {33}^2}{2}-112\frac{66}{6}\\ M_u=5317.18\text{ft-kips}\end{array}$

Try $t_f=1.5\text{in}$

Determine the trial web size.

$\begin{array}{l}h=d-2t_f\\ h=80-2\left(1.5\right)\\ h=83\text{in}\end{array}$

Consider the condition for the web to slender.

$\begin{array}{l}\frac{h}{t_w}>5.70\sqrt{\frac{E}{F_y}}\\ 45\ge 5.70\sqrt{\frac{29,000}{50}}\\ t_w\le \frac{83}{137.3}\\ t_w\le 0.6045in\end{array}$

Estimate the web thickness from the following limitations:

$\begin{array}{l}\text{For }\frac{a}{h}\le 1.5\\ \frac{h}{t_w}\le 12.0\sqrt{\left(\frac{E}{F_y}\right)}\\ \frac{h}{t_w}\le 12\sqrt{\left(\frac{29,000}{50}\right)}\\ \frac{h}{t_w}=289\end{array}$

Estimate the web thickness from the following limitations:

$\begin{array}{l}\text{For }\frac{a}{h}\ge 1.5\\ \frac{h}{t_w}\le 0.4\left(\frac{E}{F_y}\right)\\ \frac{83}{t_w}\le 0.4\left(\frac{29,000}{50}\right)\\ t_w\ge \frac{83}{232}\\ t_w\ge 0.3578\text{in}\end{array}$

From the estimated limit of ${\text{t}}_{\text{w}}$ try a web size of $\frac{1}{2}\text{in}\times 83\text{in}$

Determine area of web.

$\begin{array}{l}{\text{A}}_{\text{w}}\text{=h×}{\text{t}}_{\text{w}}\\ {\text{A}}_{\text{w}}\text{=83}\times \text{0}\text{.5}\\ {\text{A}}_{\text{w}}\text{=}41.5{\text{n}}^2\end{array}$

Determine area of flange.

$\begin{array}{l}{A}_f=\frac{M_n}{h{F}_y}-\frac{A_w}{6}\\ A_f=\frac{5317.18\times 12}{0.9\times 83\times 50}-\frac{41.5}{6}\\ A_f=10.167\text{i}{\text{n}}^2\end{array}$

Determine the breadth of flange.

$\begin{array}{l}{b}_f\ge \frac{A_f}{t_f}\\ b_f\ge \frac{10.167}{1.5}\\ b_f\ge 6.778in\\ A_f=1.5\times 24\\ A_f=36\text{i}{\text{n}}^{\text{2}}\end{array}$

Conclusion:

Therefore, girder size of $\frac{1}{2}\text{in}\times 83\text{in}$ web and $1\frac{1}{2}\text{in}\times 24\text{in}$ flanges should be used.

To determine

(b)

The size of intermediate and bearing stiffeners.

Answer to Problem 10.7.7P

Web size $\frac{1}{2}\text{in}\times 83\text{in}$

Bearing stiffeners $1\frac{1}{2}\text{in}\times 24\text{in}$

Explanation of Solution

Given information:

Span length $L=66ft$

Uniform dead load $w_D=1.3\text{kip/ft}$

Uniform live load $w_L=2.3\text{kip/ft}$

Dead load $P_D=28\text{kip}$

Live load $P_L=49\text{kip}$

Depth $d=80\text{in}$

$F_y=50$

Calculation:

Calculate the elastic section modulus by using the formula.

${\text{S}}_{\text{x}}\text{=}\frac{{\text{I}}_{\text{x}}}{\text{c}}$

Moment of inertia is given by,

$\begin{array}{l}{\text{I}}_{\text{x}}\text{=}\frac{\text{1}}{\text{12}}{\text{t}}_{\text{w}}{\text{h}}^{\text{3}}\text{+2}{\text{A}}_{\text{f}}{\left(\frac{\text{h+}{\text{t}}_{\text{f}}}{\text{2}}\right)}^{\text{2}}\\ {\text{I}}_{\text{x}}\text{=}\frac{\text{1}}{\text{12}}\left(0.5\right){\left(\text{83}\right)}^{\text{3}}\text{+2}\left(36\right){\left(\frac{\text{83+}\left(1.5\right)}{\text{2}}\right)}^{\text{2}}\\ {\text{I}}_{\text{x}}\text{=1}.523\times \text{1}{\text{0}}^4\text{i}{\text{n}}^4\end{array}$

Calculate the maximum distance whichis given by,

$\begin{array}{l}\text{c=}\frac{\text{h}}{\text{2}}\text{+}{\text{t}}_{\text{f}}\\ \text{c=}\frac{83}{\text{2}}\text{+}1.5\\ \text{c=43in}\end{array}$

Elastic section modulus about the axis $S_x$ is calculated using the values of ${\text{I}}_{\text{x}}$ and $\text{c}$.

$\begin{array}{l}{\text{S}}_{\text{x}}\text{=}\frac{{\text{I}}_{\text{x}}}{\text{c}}\\ {\text{S}}_{\text{x}}\text{=}\frac{1523\times {10}^4}{\text{43}}\\ {\text{S}}_{\text{x}}\text{=3541}\text{.86i}{\text{n}}^{\text{3}}\end{array}$

From $AISCB4$, $TableB4.1$ the relevant slenderness parameters for local buckling are,

$\begin{array}{l}\text{λ=}\frac{{\text{b}}_{\text{f}}}{\text{2}{\text{t}}_{\text{f}}}\\ \text{λ=}\frac{24}{\text{2}\left(1.5\right)}\\ \text{λ=8}\text{.0}\\ {\text{λ}}_{\text{p}}\text{=0}\text{.38}\sqrt{\frac{\text{E}}{{\text{F}}_y}}\\ {\text{λ}}_{\text{p}}\text{=0}\text{.38}\sqrt{\frac{29,000}{\text{50}}}\\ {\text{λ}}_{\text{p}}\text{=}9.152\end{array}$

$\begin{array}{l}\text{λ<}{\text{λ}}_p\\ F_{cr}=F_y=50ksi\end{array}$

Radius of gyration is given by,

$r_t\text{=}\sqrt{\frac{I_{\text{y}}}{A}}$

Calculating the moment of inertia.

$\begin{array}{l}{\text{I}}_y\text{=}\frac{\text{1}}{\text{12}}\left(\frac{h}{6}\right){\left({\text{t}}_w\right)}^{\text{3}}\text{+}\frac{\text{1}}{\text{12}}\left({\text{t}}_f\right){\left(b_f\right)}^{\text{3}}\\ {\text{I}}_y\text{=}\frac{\text{1}}{\text{12}}\left(13.83\right){\left(0.5\right)}^{\text{3}}\text{+}\frac{\text{1}}{\text{12}}\left(1.5\right){\left(24\right)}^{\text{3}}\\ {\text{I}}_y\text{=}1782.14\text{i}{\text{n}}^{\text{4}}\end{array}$

Determine area of cross-section.

$\begin{array}{l}\text{A=}\left(\frac{h}{6}\right)\left({\text{t}}_w\right)\text{+}\left({\text{t}}_f\right)\left(b_f\right)\\ \text{A=}\left(13.83\right)\left(0.5\right)\text{+}\left(1.5\right)\left(24\right)\\ \text{A=}42.92\text{i}{\text{n}}^2\end{array}$

Radius of gyration is given by,

$\begin{array}{l}{r}_t\text{=}\sqrt{\frac{I_{\text{y}}}{A}}\\ r_t\text{=}\sqrt{\frac{1728.14}{42.92}}\\ r_t\text{=}6.35\text{in}\end{array}$

Unbraced length $L_b=7\text{ft}\text{.}$

Determine the $L_P$ as shown below,

$\begin{array}{l}{L}_p=1.1r_1\sqrt{\frac{E}{F_y}}\\ L_p=1.1\times 6.35\sqrt{\frac{29,000}{50}}\\ L_p=88.38\times \left(\frac{\text{ft}}{\text{12in}}\right)\\ L_p=14.02\text{ft}\end{array}$

Determine the $L_r$ as shown below:

$\begin{array}{l}{L}_r=\pi r_1\sqrt{\frac{E}{0.7F_y}}\\ L_p=\pi \times 6.35\sqrt{\frac{29,000}{0.7\times 50}}\\ L_p=47.85\text{ft}\end{array}$

Since $L_p\le L_b<L_r$

$\begin{array}{l}{F}_{cr}=C_b\left[F_y-0.3F_y\left(\frac{L_b-L_p}{L_r-L_p}\right)\right]\le F_y\\ F_{cr}=1.046\left[50-0.3\times 50\left(\frac{23.33-14.02}{47.85-14.02}\right)\right]\le 50\\ F_{cr}=45.87\text{ksi<50ksi}\end{array}$

Calculate compression flange strength.

$\begin{array}{l}{a}_w=\frac{h_c{t}_w}{b_f{t}_f}\\ a_w=\frac{83\times \left(0.5\right)}{24\times 1.5}\\ a_w=1.15<10\end{array}$

Calculate bending strength reduction factor.

$\begin{array}{l}{\text{R}}_{\text{pg}}\text{=1-}\frac{{\text{a}}_{\text{w}}}{\text{1200+300}{\text{a}}_{\text{w}}}\left(\frac{{\text{h}}_{\text{c}}}{{\text{t}}_{\text{w}}}\text{-5}\text{.7}\sqrt{\frac{\text{E}}{{\text{F}}_{\text{y}}}}\right)\le \text{1}\\ {\text{R}}_{\text{pg}}\text{=1-}\frac{1.15}{\text{1200+300}\left(1.15\right)}\left(\frac{83}{\frac{3}{16}}\text{-5}\text{.7}\sqrt{\frac{29,000}{\text{50}}}\right)\\ {\text{R}}_{\text{pg}}\text{=0}\text{.9786<1}\end{array}$

Determine compression flange strength.

$\begin{array}{l}{\text{M}}_{\text{n}}\text{=}{\text{R}}_{\text{pg}}{\text{F}}_{\text{cr}}{\text{S}}_{\text{x}}\\ {\text{M}}_{\text{n}}\text{=0}\text{.9786×45}\text{.87×3541}\text{.86}\\ {\text{M}}_{\text{n}}\text{=}1.5899\times {10}^5\text{kips-in}\end{array}$

For proper flexural strength,

$\begin{array}{l}{\varphi }_b{M}_n=0.90\times \frac{1.5899\times {10}^5}{12}\\ {\varphi }_b{M}_n=11924.128\text{kips-ft>5317}\text{.18kips-ft}\end{array}$

Conclusion:

Therefore, stiffeners are not needed in the middle and use girder size of $\frac{1}{2}\text{in}\times 83\text{in}$ web and $1\frac{1}{2}\text{in}\times 24\text{in}$ flanges should be used.

To determine

(c)

Design of all welds.

Answer to Problem 10.7.7P

Web dimensions $\frac{1}{2}\text{in}\times 83\text{in}$

Flanges $1\frac{1}{2}\text{in}\times 24\text{in}$

Explanation of Solution

Given information:

Span length $L=66ft$

Uniform dead load $w_D=1.3\text{kip/ft}$

Uniform live load $w_L=2.3\text{kip/ft}$

Dead load $P_D=28\text{kip}$

Live load $P_L=49\text{kip}$

Depth $d=80\text{in}$

$F_y=50$

Calculation:

Calculate the elastic section modulus by using the formula,

${\text{S}}_{\text{x}}\text{=}\frac{{\text{I}}_{\text{x}}}{\text{c}}$

Moment of inertia is given by,

$\begin{array}{l}{\text{I}}_{\text{x}}\text{=}\frac{\text{1}}{\text{12}}{\text{t}}_{\text{w}}{\text{h}}^{\text{3}}\text{+2}{\text{A}}_{\text{f}}{\left(\frac{\text{h+}{\text{t}}_{\text{f}}}{\text{2}}\right)}^{\text{2}}\\ {\text{I}}_{\text{x}}\text{=}\frac{\text{1}}{\text{12}}\left(0.5\right){\left(\text{83}\right)}^{\text{3}}\text{+2}\left(36\right){\left(\frac{\text{83+}\left(1.5\right)}{\text{2}}\right)}^{\text{2}}\\ {\text{I}}_{\text{x}}\text{=1}.523\times \text{1}{\text{0}}^4\text{i}{\text{n}}^4\end{array}$

Calculate the maximum distance which is given by,

$\begin{array}{l}\text{c=}\frac{\text{h}}{\text{2}}\text{+}{\text{t}}_{\text{f}}\\ \text{c=}\frac{83}{\text{2}}\text{+}1.5\\ \text{c=43in}\end{array}$

Elastic section modulus about the axis $S_x$ is calculated using the values of ${\text{I}}_{\text{x}}$ and $\text{c}$.

$\begin{array}{l}{\text{S}}_{\text{x}}\text{=}\frac{{\text{I}}_{\text{x}}}{\text{c}}\\ {\text{S}}_{\text{x}}\text{=}\frac{1523\times {10}^4}{\text{43}}\\ {\text{S}}_{\text{x}}\text{=3541}\text{.86i}{\text{n}}^{\text{3}}\end{array}$

From $AISCB4$, $TableB4.1$ the relevant slenderness parameters for local buckling are,

$\begin{array}{l}\text{λ=}\frac{{\text{b}}_{\text{f}}}{\text{2}{\text{t}}_{\text{f}}}\\ \text{λ=}\frac{24}{\text{2}\left(1.5\right)}\\ \text{λ=8}\text{.0}\\ {\text{λ}}_{\text{p}}\text{=0}\text{.38}\sqrt{\frac{\text{E}}{{\text{F}}_y}}\\ {\text{λ}}_{\text{p}}\text{=0}\text{.38}\sqrt{\frac{29,000}{\text{50}}}\\ {\text{λ}}_{\text{p}}\text{=}9.152\end{array}$

$\begin{array}{l}\text{λ<}{\text{λ}}_p\\ F_{cr}=F_y=50ksi\end{array}$

Radius of gyration is given by,

$r_t\text{=}\sqrt{\frac{I_{\text{y}}}{A}}$

Calculating the moment of inertia,

$\begin{array}{l}{\text{I}}_y\text{=}\frac{\text{1}}{\text{12}}\left(\frac{h}{6}\right){\left({\text{t}}_w\right)}^{\text{3}}\text{+}\frac{\text{1}}{\text{12}}\left({\text{t}}_f\right){\left(b_f\right)}^{\text{3}}\\ {\text{I}}_y\text{=}\frac{\text{1}}{\text{12}}\left(13.83\right){\left(0.5\right)}^{\text{3}}\text{+}\frac{\text{1}}{\text{12}}\left(1.5\right){\left(24\right)}^{\text{3}}\\ {\text{I}}_y\text{=}1782.14\text{i}{\text{n}}^{\text{4}}\end{array}$

Determine area of cross-section.

$\begin{array}{l}\text{A=}\left(\frac{h}{6}\right)\left({\text{t}}_w\right)\text{+}\left({\text{t}}_f\right)\left(b_f\right)\\ \text{A=}\left(13.83\right)\left(0.5\right)\text{+}\left(1.5\right)\left(24\right)\\ \text{A=}42.92\text{i}{\text{n}}^2\end{array}$

Radius of gyration is given by,

$\begin{array}{l}{r}_t\text{=}\sqrt{\frac{I_{\text{y}}}{A}}\\ r_t\text{=}\sqrt{\frac{1728.14}{42.92}}\\ r_t\text{=}6.35\text{in}\end{array}$

Unbraced length $L_b=6\text{6ft}$

Determine the $L_P$ as shown below.

$\begin{array}{l}{L}_p=1.1r_1\sqrt{\frac{E}{F_y}}\\ L_p=1.1\times 6.35\sqrt{\frac{29,000}{50}}\\ L_p=88.38\times \left(\frac{\text{ft}}{\text{12in}}\right)\\ L_p=14.02\text{ft}\end{array}$

Determine the $L_r$ as shown below,

$\begin{array}{l}{L}_r=\pi r_1\sqrt{\frac{E}{0.7F_y}}\\ L_r=\pi \times 6.35\sqrt{\frac{29,000}{0.7\times 50}}\\ L_r=47.85\text{ft}\end{array}$

Since, $L_p\le L_b<L_r$

$\begin{array}{l}{F}_{cr}=C_b\left[F_y-0.3F_y\left(\frac{L_b-L_p}{L_r-L_p}\right)\right]\le F_y\\ F_{cr}=1.046\left[50-0.3\times 50\left(\frac{23.33-14.02}{47.85-14.02}\right)\right]\le 50\\ F_{cr}=45.87\text{ksi<50ksi}\end{array}$

To calculate the plate girder strength reduction factor values, ${\text{a}}_{\text{w}}\text{ and }{\text{R}}_{\text{pg}}$ is needed.

$\begin{array}{l}{a}_w=\frac{h_c{t}_w}{b_f{t}_f}\\ a_w=\frac{83\times \left(0.5\right)}{24\times 1.5}\\ a_w=1.15<10\end{array}$

$\begin{array}{l}{\text{R}}_{\text{pg}}\text{=1-}\frac{{\text{a}}_{\text{w}}}{\text{1200+300}{\text{a}}_{\text{w}}}\left(\frac{{\text{h}}_{\text{c}}}{{\text{t}}_{\text{w}}}\text{-5}\text{.7}\sqrt{\frac{\text{E}}{{\text{F}}_{\text{y}}}}\right)\le \text{1}\\ {\text{R}}_{\text{pg}}\text{=1-}\frac{1.15}{\text{1200+300}\left(1.15\right)}\left(\frac{83}{\frac{3}{16}}\text{-5}\text{.7}\sqrt{\frac{29,000}{\text{50}}}\right)\\ {\text{R}}_{\text{pg}}\text{=0}\text{.9786<1}\end{array}$

Calculate compression flange strength.

$\begin{array}{l}{\text{M}}_{\text{n}}\text{=}{\text{R}}_{\text{pg}}{\text{F}}_{\text{cr}}{\text{S}}_{\text{x}}\\ {\text{M}}_{\text{n}}\text{=0}\text{.9786×45}\text{.87×3541}\text{.86}\\ {\text{M}}_{\text{n}}\text{=}1.5899\times {10}^5\text{kips-in}\end{array}$

Calculate proper flexural strength.

$\begin{array}{l}{\varphi }_b{M}_n=0.90\times \frac{1.5899\times {10}^5}{12}\\ {\varphi }_b{M}_n=11924.128\text{kips-ft>5317}\text{.18kips-ft}\end{array}$

Calculate nominal shear strength.

$\begin{array}{l}{V}_n=\frac{V_u}{\varphi }\\ V_n=\frac{284.92}{0.90}\\ V_n=316.57\text{kips}\end{array}$

Check for design strength.

$\begin{array}{l}{V}_n=0.6A_w{F}_y{C}_v\\ C_v=\frac{V_n}{0.6A_w{F}_y}\\ C_v=\frac{316.57\text{kips}}{0.6\times 8.438\times 50}\\ C_v=1.25\text{kips}\end{array}$

Compute the value of $k_v$.

$\begin{array}{l}{C}_v=\frac{1.51k_{v}E}{{\left(\frac{h}{t_w}\right)}^2{F}_y}\\ k_v=\frac{C_v{\left(\frac{h}{t_w}\right)}^2{F}_y}{1.51E_y}\\ k_v=\frac{1.2{\left(\frac{83}{\frac{3}{16}}\right)}^{2}50}{1.5\times 29,000}\\ k_v=25.1\end{array}$

Assuming the equation $\text{G2-5 }$ controls,

$\begin{array}{l}\text{1}\text{.37}\sqrt{\frac{k_v\text{E}}{{\text{F}}_y}}=\text{1}\text{.37}\sqrt{\frac{25.1\times \text{29,000}}{50}}\\ \text{1}\text{.37}\sqrt{\frac{k_v\text{E}}{{\text{F}}_y}}=165.29<\frac{h}{t_w}=240\end{array}$

Hence, equation $\text{G2-5 }$ controls.

Determine the value of $\text{a}$ from AISC equation $\text{G2-6}$ as shown below:

$\begin{array}{l}\frac{a}{h}=\sqrt{\frac{5}{k_v-5}}\\ \frac{a}{h}=\sqrt{\frac{5}{25.1-5}}\\ a=0.497\times h\\ a=0.497\times 83\\ a=41.33\text{in}\end{array}$

Calculate the value of required shear strength $\text{15}$ in from the left end as shown below:

$\begin{array}{l}{V}_{u1}=V_u-2\left(\frac{15\text{in}}{b_f}\right)\\ V_{u1}=168-2\left(\frac{15\text{in}}{12}\right)\\ V_{u1}=165.5\text{kips}\end{array}$

Compute the value of required shear $\text{15}$ in from the left end as shown below:

$\begin{array}{l}\frac{{\varphi }_v{V}_u}{A_w}=\frac{V_{u1}}{A_w}\\ \frac{{\varphi }_v{V}_u}{A_w}=\frac{165.5}{8.438}\\ \frac{{\varphi }_v{V}_u}{A_w}=19.61\text{ksi}\end{array}$

Using the curves in the table $\text{3-17b}$ with the value of,

$\begin{array}{l}\frac{h}{t_w}=\frac{45}{\frac{3}{16}}\\ \frac{h}{t_w}=240\end{array}$

Values obtained are shown below,

$\begin{array}{l}\text{For},\frac{{\varphi }_v{V}_u}{A_w}=18\text{ksi,}\frac{a}{h}=1.05\\ \text{For},\frac{{\varphi }_v{V}_u}{A_w}=21\text{ksi,}\frac{a}{h}=0.75\end{array}$

By interpolation determine the value of $\text{a}$, $\frac{{\varphi }_v{V}_u}{A_w}=19.61\text{ksi}$

$\begin{array}{l}\frac{a}{h}=1.05-\left(\frac{19.91-18}{21-18}\right)\left(1.05-0.75\right)\\ a=0.859\times h\\ a=0.859\times 45\\ a=38.66\text{in}\end{array}$

Use the value $a=38\text{in}$ which will be added to the distance to the next stiffener.

Compute the value required to shear strength from the left end,

$\begin{array}{l}{V}_{u2}=V_u-2\left(\frac{15\text{in+38in}}{b_f}\right)\\ V_{u2}=284.92-2\left(\frac{53}{12}\right)\\ V_{u2}=276.0\text{kips}\end{array}$

Compute the value of required shear $\text{15+38in}$ in from the left end as shown below,

$\begin{array}{l}\frac{{\varphi }_v{V}_u}{A_w}=\frac{V_{u1}}{A_w}\\ \frac{{\varphi }_v{V}_u}{A_w}=\frac{159.2}{8.438}\\ \frac{{\varphi }_v{V}_u}{A_w}=18.87\text{ksi}\end{array}$

Values obtained are shown blow,

$\begin{array}{l}\text{For},\frac{{\varphi }_v{V}_u}{A_w}=18\text{ksi,}\frac{a}{h}=1.05\\ \text{For},\frac{{\varphi }_v{V}_u}{A_w}=21\text{ksi,}\frac{a}{h}=0.75\end{array}$

By interpolation we get the value of $\text{a}$, $\frac{{\varphi }_v{V}_u}{A_w}=18.87\text{ksi}$

$\begin{array}{l}\frac{a}{h}=1.05-\left(\frac{18.87-18}{21-18}\right)\left(1.05-0.75\right)\\ a=0.963\times h\\ a=0.963\times 45\\ a=43.44\text{in}\end{array}$

Use the value $a=43\text{in}$ which we can add to the distance of the next stiffener.

Calculate to shear strength from the left end.

$\begin{array}{l}{V}_{u3}=V_u-2\left(\frac{15\text{in+38in+43}}{b_f}\right)\\ V_{u3}=168-2\left(\frac{96}{12}\right)\\ V_{u3}=152\text{kips}\end{array}$

Calculate shear $\text{96in}$ in from the left end as shown below:

$\begin{array}{l}\frac{{\varphi }_v{V}_u}{A_w}=\frac{V_{u3}}{A_w}\\ \frac{{\varphi }_v{V}_u}{A_w}=\frac{152}{8.438}\\ \frac{{\varphi }_v{V}_u}{A_w}=18\text{ksi}\end{array}$

Values obtained are shown blow,

$\text{For},\frac{{\varphi }_v{V}_u}{A_w}=18\text{ksi,}\frac{a}{h}=1.05$

By interpolation calculate the value of $\text{a}$,

$\frac{{\varphi }_v{V}_u}{A_w}=18\text{ksi}$

$\begin{array}{l}\frac{a}{h}=1.05\\ a=1.05\times h\\ a=1.05\times 83\\ a=87.15\text{in}\end{array}$

Use the value $a=87.15\text{in}$ which we can add to the distance of the next stiffener.

We can use next stiffener will be used at $\text{96+87}\text{.15in=180}\text{.15}$.

Calculate distance remaining concentrated load.

$\begin{array}{l}\text{d=16}\left(b_f\right)\text{-180}\text{.15in}\\ \text{d=16}\left(12\right)\text{-180}\text{.15in}\\ \text{d=}11.85\text{in}\end{array}$

Calculate shear strength $\text{16in}$ from the left end is given by,

$\begin{array}{l}{V}_{u4}=V_u-2\left(66\right)-P_u\\ V_{u4}=284.92-2\left(66\right)-112\\ V_{u4}=40.9\text{kips}\end{array}$

Calculate the required shear $\text{16in}$ from left end.

$\begin{array}{l}\frac{{\varphi }_v{V}_u}{A_w}=\frac{V_{u4}}{A_w}\\ \frac{{\varphi }_v{V}_u}{A_w}=\frac{16}{8.438}\\ \frac{{\varphi }_v{V}_u}{A_w}=1.896\text{ksi}\end{array}$

$\begin{array}{l}\frac{a}{h}=\frac{16\times b_f}{45\text{in}}\\ \frac{a}{h}=\frac{16\times 12\text{in}}{45\text{in}}\\ \frac{a}{h}=4.267>3\end{array}$

Conclusion:

Therefore, girder size of $\frac{1}{2}\text{in}\times 83\text{in}$ web and $1\frac{1}{2}\text{in}\times 24\text{in}$ flanges should be used.