Chapter 8, Problem 8.4.16P

A W 200 x 41.7 wide-flange beam (see Table F-l(b), Appendix F) is simply supported with a span length of 2.5 m (see figure). The beam supports a concentrated load of 100 kN at 0.9 m from support B. At a cross section located 0,7 m from the left-hand support, determine the principal stresses tr, and 2 and the maximum shear stress r _{n M J t} 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 maximum stress and principal shear stress at top of beam.

Answer to Problem 8.4.16P

Values of principal stress :

${\sigma }_1=151.6\text{ MPa}$ ,${\sigma }_2=0\text{ MPa}$

Maximum shear stress${\tau }_{\max }=75.8\text{ MPa}$

Explanation of Solution

Given Information:

Beam length$L=2.5\text{ m}$

Point load$P=100\text{ kN}$

Dimensions of beam,

$\begin{array}{l}h=200\\ b=165\\ t=7.24\\ h_1=176.4\end{array}$

Concept Used:

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

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

Values of principal normal stress :${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

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

From equilibrium:

$\begin{array}{l}{R}_B\times 2.5-100\times 1.6=0\\ R_B=64\text{ kN}\end{array}$

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

  $M=R_B\times 0.9\text{=57}\text{.6 kN-m}$

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

  $\text{V}=R_B=64\text{ kN}$

Moment of inertia:

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{165\times {200}^3}{12}-\frac{165\times {176.4}^3}{12}+\frac{7.24\times {176.4}^3}{12}\\ I=37837529.34{\text{ mm}}^{\text{4}}\\ I=3.8\times {10}^{-5}{\text{ m}}^{\text{4}}\end{array}$

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

So, bending stress at top:

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{57.6\times 0.1}{3.8\times {10}^{-5}}\\ {\sigma }_x=151579\text{ kPa =151}\text{.6 MPa}\end{array}$

And shear stress at that point:

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

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

Values of Principal and normal stress are given by following equation:

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{151.6+0}{2}\pm \sqrt{{\left(\frac{151.6-0}{2}\right)}^2+0^2}\\ {\sigma }_{1,2}=75.8\pm 75.8\\ \\ {\sigma }_1=75.8+75.8=151.6\text{ MPa}\\ {\sigma }_2=75.8-75.8=0\text{ MPa}\end{array}$

Maximum shear stress:

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{151.6-0}{2}\right)}^2+0^2}\\ {\tau }_{\max }=75.8\text{ MPa}\end{array}$

Conclusion:

Hence, we get:

Values of principal stress :

${\sigma }_1=151.6\text{ MPa}$ ,${\sigma }_2=0\text{ MPa}$

Maximum shear stress${\tau }_{\max }=75.8\text{ MPa}$

To determine

To find: The values of principal stress and principal stress at top of web.

Answer to Problem 8.4.16P

Values of principal stress:

${\sigma }_1=146.13\text{ MPa}$ ,${\sigma }_2=-12.43\text{ MPa}$.

Maximum shear stress:${\tau }_{\max }=79.28\text{ MPa}$.

Explanation of Solution

Given Information:

Beam length$L=2.5\text{ m}$

Point load$P=100\text{ kN}$

Dimensions of beam,

$\begin{array}{l}h=200\\ b=165\\ t=7.24\\ h_1=176.4\end{array}$

Concept Used:

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

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

Values of principal normal stress:${\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}$

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

From equilibrium:

$\begin{array}{l}{R}_B\times 2.5-100\times 1.6=0\\ R_B=64\text{ kN}\end{array}$

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

  $M=R_B\times 0.9\text{=57}\text{.6 kN-m}$

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

  $\text{V}=R_B=64\text{ kN}$

Moment of inertia:

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{165\times {200}^3}{12}-\frac{165\times {176.4}^3}{12}+\frac{7.24\times {176.4}^3}{12}\\ I=37837529.34{\text{ mm}}^{\text{4}}\\ I=3.8\times {10}^{-5}{\text{ m}}^{\text{4}}\end{array}$

First, moment of area of flange:

  $\begin{array}{l}Q=A_1\times {\text{y}}_{\text{1}}=165\times \frac{200-176.4}{2}\times (\frac{176.4}{2}+\frac{200-176.4}{4})\\ Q=183212.7{\text{ mm}}^{\text{3}}=1.832\times {10}^{-4}{\text{ m}}^{\text{3}}\end{array}$

So, bending stress at top of web:

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{57.6\times 0.0882}{3.8\times {10}^{-5}}\\ {\sigma }_x=133692.6\text{ kPa =133}\text{.7 MPa}\end{array}$

And shear stress at that point ::

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{64\times 1.832\times {10}^{-4}}{3.8\times {10}^{-5}\times 0.00724}\\ {\tau }_{xy}=42.62\text{ MPa}\end{array}$

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

Values of principal normal stress are given by following equation:

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{133.7+0}{2}\pm \sqrt{{\left(\frac{133.7-0}{2}\right)}^2+{42.62}^2}\\ {\sigma }_{1,2}=66.85\pm 79.28\\ \\ {\sigma }_1=66.85+79.28=146.13\text{ MPa}\\ {\sigma }_2=66.85-79.28=-12.43\text{ MPa}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{133.7-0}{2}\right)}^2+{42.62}^2}\\ {\tau }_{\max }=79.28\text{ MPa}\end{array}$

Conclusion:

Hence, we get:

Values of principal stress :${\sigma }_1=146.13\text{ MPa}$ ,${\sigma }_2=-12.43\text{ MPa}$

Maximum shear stress${\tau }_{\max }=79.28\text{ MPa}$

To determine

To find: Values of principal stress and maximum stress at neutral axis.

Answer to Problem 8.4.16P

Values of principal stress:

${\sigma }_1=49.17\text{ MPa}$,${\sigma }_2=-49.17\text{ MPa}$

Maximum shear stress${\tau }_{\max }\text{=}49.17\text{ MPa}\text{.}$

Explanation of Solution

Given Information:

Beam length$L=2.5\text{ m}$

Point load$P=100\text{ kN}$

Dimensions of beam:

$\begin{array}{l}h=200\\ b=165\\ t=7.24\\ h_1=176.4\end{array}$

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:

$\begin{array}{l}{R}_B\times 2.5-100\times 1.6=0\\ R_B=64\text{ kN}\end{array}$

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

  $M=R_B\times 0.9\text{=57}\text{.6 kN-m}$

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

  $\text{V}=R_B=64\text{ kN}$

Moment of inertia,

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{165\times {200}^3}{12}-\frac{165\times {176.4}^3}{12}+\frac{7.24\times {176.4}^3}{12}\\ I=37837529.34{\text{ mm}}^{\text{4}}\\ I=3.8\times {10}^{-5}{\text{ m}}^{\text{4}}\end{array}$

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

  $\begin{array}{l}\text{Q}=A_1\times {\text{y}}_{\text{1}}+A_2\times {\text{y}}_2\\ Q=165\times \frac{200-176.4}{2}\times (\frac{176.4}{2}+\frac{200-176.4}{4})+7.24\times \frac{176.4}{2}\times (\frac{176.4}{4})\\ Q=211373.55{\text{ mm}}^{\text{3}}=2.114\times {10}^{-4}{\text{ m}}^{\text{3}}.\end{array}$

So, bending stress at neutral axis:

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

And shear stress at that point:

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{64\times 2.114\times {10}^{-4}}{3.8\times {10}^{-5}\times 0.00724}\\ {\tau }_{xy}=49.17\text{ MPa}\end{array}$

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

Values of principal stress are given by following equation:

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{0+0}{2}\pm \sqrt{{\left(\frac{0-0}{2}\right)}^2+{49.17}^2}\\ {\sigma }_{1,2}=0\pm 49.17\\ \\ {\sigma }_1=0+49.17=49.17\text{ MPa}\\ {\sigma }_2=0-49.17=-49.17\text{ MPa}\end{array}$

Maximum shear stress:

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{0-0}{2}\right)}^2+{49.17}^2}\\ {\tau }_{\max }=49.17\text{ MPa}\end{array}$

Conclusion:

Hence, we get:

Values of principal stress:${\sigma }_1=49.17\text{ MPa}$ ,${\sigma }_2=-49.17\text{ MPa}$

Maximum shear stress${\tau }_{\max }\text{=}49.17\text{ MPa}$