Chapter 10, Problem 10.4.1P

To determine

Nominal flexural strength of the given welded shape.

Answer to Problem 10.4.1P

${\text{M}}_{\text{n}}\text{=2}\text{.381}\text{.5}\times \text{1}{\text{0}}^4\text{ft-kips}$

Explanation of Solution

Given information:

Web thickness $t_w=\frac{3}{8}inch$

Depth of web from inside face of flange to inside face of flange $h=45in$

Thickness of flange $t_f=1in$

Area of flange $A_f=10i{n}^2$

Calculation:

Web width to thickness ratio given by,

$\begin{array}{l}\frac{h}{t_w}=\frac{45in}{\frac{3}{8}in}\\ \frac{h}{t_w}=120in\end{array}$

Determine web is slender or not.

$\begin{array}{l}5.70\sqrt{\frac{E}{F_y}}=5.70\sqrt{\frac{29,000}{50}}\\ =137.27\end{array}$

Now, $\frac{h}{t_w}<5.70\sqrt{\frac{E}{F_y}}$

From the above equation web is non slender.

Calculate the elastic section modulus by using the formula,

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

Calculate ${\text{I}}_{\text{x}}$ that is moment of inertia is given by,

$\begin{array}{l}{\text{I}}_{\text{x}}\text{=}\frac{\text{1}}{\text{12}}{\text{t}}_{\text{w}}{\text{h}}^{\text{3}}\text{+2}{\text{A}}_{\text{f}}{\left(\frac{\text{h+}{\text{t}}_{\text{f}}}{\text{2}}\right)}^{\text{2}}\\ {\text{I}}_{\text{x}}\text{=}\frac{\text{1}}{\text{12}}\left(\frac{\text{3}}{\text{8}}\right){\left(\text{45}\right)}^{\text{3}}\text{+2}\left(\text{10}\right){\left(\frac{\text{48+1}}{\text{2}}\right)}^{\text{2}}\\ {\text{I}}_{\text{x}}\text{=13430i}{\text{n}}^{\text{4}}\end{array}$

Maximum distance is calculated by using the formula,

$\begin{array}{l}\text{c=}\frac{\text{h}}{\text{2}}\text{+}{\text{t}}_{\text{f}}\\ \text{c=}\frac{45}{\text{2}}\text{+1}\\ \text{c=23}\text{.5in}\end{array}$

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

$\begin{array}{l}{\text{S}}_{\text{x}}\text{=}\frac{{\text{I}}_{\text{x}}}{\text{c}}\\ {\text{S}}_{\text{x}}\text{=}\frac{\text{13430}}{\text{23}\text{.5}}\\ {\text{S}}_{\text{x}}\text{=571}\text{.5i}{\text{n}}^{\text{3}}\end{array}$

Tension in flange strength is given by,

$\begin{array}{l}{\text{M}}_{\text{n}}\text{=}{\text{F}}_{\text{y}}{\text{S}}_{\text{x}}\\ {\text{M}}_{\text{n}}\text{=50}\times \text{571}\text{.5}\\ {\text{M}}_{\text{n}}\text{=}28575\text{in-kips}\end{array}$

From AISC table B4-1b than the required slenderness parameters $\text{λ}$ and ${\text{λ}}_{\text{p}}$.

$\begin{array}{l}\text{λ=}\frac{{\text{b}}_{\text{f}}}{\text{2}{\text{t}}_{\text{f}}}\\ \text{λ=}\frac{\text{10}}{\text{2}\left(1\right)}\\ \text{λ=}5\\ {\text{λ}}_{\text{p}}\text{=0}\text{.38}\sqrt{\frac{\text{E}}{{\text{F}}_{\text{f}}}}\\ {\text{λ}}_{\text{p}}\text{=0}\text{.38}\sqrt{\frac{29,000}{\text{50}}}\\ {\text{λ}}_{\text{p}}\text{=}9.152\end{array}$

${\text{F}}_{\text{cr}}\text{=}{\text{F}}_{\text{y}}\text{=50ksi}$

Here $\text{λ<}{\text{λ}}_{\text{p}}$ so the flange is compact and there is no flange local buckling.

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

$\begin{array}{l}{\text{a}}_{\text{w}}\text{=}\frac{{\text{h}}_{\text{c}}{\text{t}}_{\text{w}}}{{\text{b}}_{\text{f}}{\text{t}}_{\text{f}}}\\ {\text{a}}_{\text{w}}\text{=}\frac{\text{45}\left(\frac{3}{8}\right)}{\text{10}\left(1\right)}\\ {\text{a}}_{\text{w}}=1.688<10\end{array}$

From AISC Equation $\text{F5-6}$,

$\begin{array}{l}{\text{R}}_{\text{pg}}\text{=1-}\frac{{\text{a}}_{\text{w}}}{\text{1200+300}{\text{a}}_{\text{w}}}\left(\frac{{\text{h}}_{\text{c}}}{{\text{t}}_{\text{w}}}\text{-5}\text{.7}\sqrt{\frac{\text{E}}{{\text{F}}_{\text{y}}}}\right)\le \text{1}\\ {\text{R}}_{\text{pg}}\text{=1-}\frac{\text{1}\text{.688}}{\text{1200+300}\left(\text{1}\text{.688}\right)}\left(\text{120-5}\text{.7}\sqrt{\frac{29,000}{\text{50}}}\right)\\ {\text{R}}_{\text{pg}}\text{=1}\text{.017=}1.0\end{array}$

From AISC Equation $\text{F5-7}$, nominal flexural strength for the compression flange is:

$\begin{array}{l}{\text{M}}_{\text{n}}\text{=}{\text{R}}_{\text{pg}}{\text{F}}_{\text{cr}}{\text{S}}_{\text{x}}\\ {\text{M}}_{\text{n}}\text{=1×50×571}\text{.5}\\ {\text{M}}_{\text{n}}\text{=28575}\times \frac{1}{12}\\ {\text{M}}_{\text{n}}\text{=2}\text{.381}\text{.5}\times \text{1}{\text{0}}^4\text{ft-kips}\end{array}$

Conclusion:

Therefore, the nominal flexural strength for the given welded plate is $\text{2}\text{.381}\text{.5}\times \text{1}{\text{0}}^4\text{ft-kips}$.