Chapter 8, Problem 8.4.19P

A W 12 X 14 wide-flange beam (see Table F-l(a), Appendix F) is simply supported with a span length of 120 in. (see figure). The beam supports two anti-symmetrically placed concentrated loads of 7,5 kips each.

At a cross section located 20 in. from the right-hand support, determine the principal stresses (7_]and (7\ and the maximum shear stress T_maw at each of the following locations: (a) the top of the beam, (b) the top of the web, and (c) the neutral axis,

  

To determine

To find: Values of principal stresses and maximum shear stress at top of beam.

Answer to Problem 8.4.19P

Values of principal stress :

${\sigma }_1=3.46\text{ ksi}$ ,${\sigma }_2=0\text{ ksi}$

Maximum shear stress${\tau }_{\max }=1.73\text{ ksi}$

Explanation of Solution

Given Information:

Beam length$L=10\text{ ft}=120\text{ in}$

Point load$P=7.5\text{ kips}$

Dimensions of beam,

$\begin{array}{l}h=11.91\\ b=3.97\\ t=0.2\\ h_1=11.46\end{array}$

Concept Used:

Bending stress${\sigma }_x=\frac{My}{I}$

Shear stress${\tau }_{xy}=\frac{VQ}{It}$

Principal normal stresses${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

Maximum shear stress${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

From equilibrium:

$\begin{array}{l}{R}_B\times 120-7.5\times 80+7.5\times 40=0\\ R_B=2.5\text{ kips}\end{array}$

So, bending moment at point$C\text{ is}$ :

  $M=R_B\times \text{20=2}\text{.5}\times \text{20=50 kip-in}$

Shear force at point$C\text{ is}$

  $\text{V}=R_B\text{=2}\text{.5 kip}$

Moment of inertia:

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{3.97\times {11.91}^3}{12}-\frac{3.97\times {11.46}^3}{12}+\frac{0.2\times {11.46}^3}{12}\\ I=86.074{\text{ in}}^{\text{4}}\end{array}$

First moment of area at the top of beam shall be zero,$\text{Q}=0{\text{ m}}^3$

So, bending stress at top:

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{50\times 5.955}{86.1}\\ {\sigma }_x=3.46\text{ ksi}\end{array}$

And shear stress at that point:

$\begin{array}{l}\tau =\frac{VQ}{It}\\ \tau =0\text{ ksi}\end{array}$

For this situation no stress in$y$ direction so${\sigma }_y=0$

Values of normal stress is given by following equation:

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{3.46+0}{2}\pm \sqrt{{\left(\frac{3.46-0}{2}\right)}^2+0^2}\\ {\sigma }_{1,2}=1.73\pm 1.73\\ \\ {\sigma }_1=1.73+1.73=3.46\text{ ksi}\\ {\sigma }_2=1.73-1.73=0\text{ ksi}\end{array}$

Maximum shear stress:

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{3.46-0}{2}\right)}^2+0^2}\\ {\tau }_{\max }=1.73\text{ ksi}\end{array}$

Conclusion:

Hence, we get:

Values of principal stress:

${\sigma }_1=3.46\text{ ksi}$ ,${\sigma }_2=0\text{ ksi}$

Maximum shear stress${\tau }_{\max }=1.73\text{ ksi}$

To determine

To find: Values of principal stresses and maximum shear stress at top of web.

Answer to Problem 8.4.19P

Values of principal stress:

${\sigma }_1=3.5\text{ ksi}$ ,${\sigma }_2=-0.165\text{ ksi}$

Maximum shear stress${\tau }_{\max }=1.83\text{ ksi}$

Explanation of Solution

Given Information:

Beam length$L=10\text{ ft}=120\text{ in}$

Point load$P=7.5\text{ kips}$

Dimensions of beam,

$\begin{array}{l}h=11.91\\ b=3.97\\ t=0.2\\ h_1=11.46\end{array}$

Concept Used:

Bending stress${\sigma }_x=\frac{My}{I}$

Shear stress${\tau }_{xy}=\frac{VQ}{It}$

Principal normal stresses${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

Maximum shear stress${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

From equilibrium,

$\begin{array}{l}{R}_B\times 120-7.5\times 80+7.5\times 40=0\\ R_B=2.5\text{ kips}\end{array}$

So, bending moment at point$C\text{ is}$ :

  $M=R_B\times \text{20=2}\text{.5}\times \text{20=50 kip-in}$

Shear force at point$C\text{ is}$

  $\text{V}=R_B\text{=2}\text{.5 kip}$

Moment of inertia:

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{3.97\times {11.91}^3}{12}-\frac{3.97\times {11.46}^3}{12}+\frac{0.2\times {11.46}^3}{12}\\ I=86.074{\text{ in}}^{\text{4}}\end{array}$

First moment of area of flange:

  $\text{Q}=A_1\times {\text{y}}_{\text{1}}=3.97\times \frac{11.91-11.46}{2}\times (\frac{11.46}{2}+\frac{11.91-11.46}{4})=5.22{\text{ in}}^{\text{3}}$

So, bending stress at top of web:

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{50\times 5.73}{86.1}\\ {\sigma }_x=3.33\text{ ksi}\end{array}$

And shear stress at that point:

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{2.5\times 5.22}{86.1\times 0.2}\\ {\tau }_{xy}=0.76\text{ ksi}\end{array}$

For this situation no stress in$y$ direction so${\sigma }_y=0$

Principal normal stresses are given by following equation,

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{3.33+0}{2}\pm \sqrt{{\left(\frac{3.33-0}{2}\right)}^2+{0.76}^2}\\ {\sigma }_{1,2}=1.665\pm 1.83\\ \\ {\sigma }_1=1.665+1.83=3.5\text{ ksi}\\ {\sigma }_2=1.665-1.83=-0.165\text{ ksi}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{3.33-0}{2}\right)}^2+{0.76}^2}\\ {\tau }_{\max }=1.83\text{ ksi}\end{array}$

Conclusion:

Hence we get,

Principal stresses

${\sigma }_1=3.5\text{ ksi}$ ,${\sigma }_2=-0.165\text{ ksi}$

Maximum shear stress${\tau }_{\max }=1.83\text{ ksi}$

To determine

Find principal stresses and maximum shear stress at neutral axis.

Answer to Problem 8.4.19P

Principal stresses

${\sigma }_1=4.7\text{ ksi}$ ,${\sigma }_2=-4.7\text{ ksi}$

Maximum shear stress${\tau }_{\max }\text{=}4.7\text{ ksi}$

Explanation of Solution

Given Information:

Beam length$L=10\text{ ft}=120\text{ in}$

Point load$P=7.5\text{ kips}$

Dimensions of beam,

$\begin{array}{l}h=11.91\\ b=3.97\\ t=0.2\\ h_1=11.46\end{array}$

Concept Used:

Bending stress${\sigma }_x=\frac{My}{I}$

Shear stress${\tau }_{xy}=\frac{VQ}{It}$

Principal normal stresses${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

Maximum shear stress${\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

From equilibrium,

$\begin{array}{l}{R}_B\times 120-7.5\times 80+7.5\times 40=0\\ R_B=2.5\text{ kips}\end{array}$

So bending moment at point$C\text{ is}$

  $M=R_B\times \text{20=2}\text{.5}\times \text{20=50 kip-in}$

Shear force at point$C\text{ is}$

  $\text{V}=R_B\text{=2}\text{.5 kip}$

Moment of inertia,

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{3.97\times {11.91}^3}{12}-\frac{3.97\times {11.46}^3}{12}+\frac{0.2\times {11.46}^3}{12}\\ I=86.074{\text{ in}}^{\text{4}}\end{array}$

First moment of area for the section above the neutral axis,

  $\begin{array}{l}\text{Q}=A_1\times {\text{y}}_{\text{1}}+A_2\times {\text{y}}_2\\ Q=3.97\times \frac{11.91-11.46}{2}\times (\frac{11.46}{2}+\frac{11.91-11.46}{4})+0.2\times \frac{11.46}{2}\times (\frac{11.46}{4})\\ Q=8.5{\text{ in}}^{\text{3}}\end{array}$

So bending stress at neutral axis,

$\begin{array}{l}{\sigma }_x=\frac{M(y=0)}{I}\\ {\sigma }_x=0\text{ ksi}\end{array}$

And shear stress at that point,

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{2.5\times 8.5}{86.1\times 0.2}\\ {\tau }_{xy}=1.234\text{ ksi}\end{array}$

For this situation no stress in$y$ direction so${\sigma }_y=0$

Principal normal stresses are given by following equation,

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{0+0}{2}\pm \sqrt{{\left(\frac{0-0}{2}\right)}^2+{1.234}^2}\\ {\sigma }_{1,2}=0\pm 1.234\\ \\ {\sigma }_1=0+1.234=1.234\text{ ksi}\\ {\sigma }_2=0-1.234=-1.234\text{ ksi}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{0-0}{2}\right)}^2+{1.234}^2}\\ {\tau }_{\max }=1.234\text{ ksi}\end{array}$

Conclusion:

Hence we get,

Principal stresses

${\sigma }_1=1.234\text{ ksi}$ ,${\sigma }_2=-1.234\text{ ksi}$

Maximum shear stress${\tau }_{\max }\text{=}1.234\text{ ksi}$