Chapter 8, Problem 8.4.14P

A beam with a wide-flange cross section (see figure) has the following dimensions: h = 120 mm, r = 10 mm, h = 300 mm, and /ij = 260 mm. The beam is simply supported with span length L = 3,0 im A concentrated load P = 120 kN acts at the midpoint of the span.

At across section located 1.0 m from the left-hand support, determine the principal stresses tr, and tr₂and the maximum shear stress T_max 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

Find principal stresses and maximum shear stress at top of beam.

Answer to Problem 8.4.14P

Principal stresses${\sigma }_1=82.64\text{ MPa}$ ,${\sigma }_2=0\text{ MPa}$

Maximum shear stress${\tau }_{\max }=41.32\text{ MPa}$

Explanation of Solution

Given Information:

Beam length$L=3\text{ m}$

Point load$P=120\text{ kN}$

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

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

  $M=\frac{P}{2}\times \frac{L}{3}\text{=}\frac{120}{2}\times \frac{3}{3}\text{=60 kN-m}$

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

  $\text{V}=\frac{P}{2}=60\text{ kN}$

Moment of inertia,

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{120\times {300}^3}{12}-\frac{120\times {260}^3}{12}+\frac{10\times {260}^3}{12}\\ I=108886666.7{\text{ mm}}^{\text{4}}\\ I=1.089\times {10}^{-4}{\text{ m}}^{\text{4}}\end{array}$

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

So bending stress at top,

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{60\times 0.15}{1.089\times {10}^{-4}}\\ {\sigma }_x=82644.63\text{ kPa =82}\text{.64 MPa}\end{array}$

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${\sigma }_y=0$

Principal normal stresses are given by following equation,

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{82.64+0}{2}\pm \sqrt{{\left(\frac{82.64-0}{2}\right)}^2+0^2}\\ {\sigma }_{1,2}=41.32\pm 41.32\\ \\ {\sigma }_1=41.32+41.32=82.64\text{ MPa}\\ {\sigma }_2=41.32-41.32=0\text{ MPa}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{82.64-0}{2}\right)}^2+0^2}\\ {\tau }_{\max }=41.32\text{ MPa}\end{array}$

Conclusion:

Hence we get,

Principal stresses

${\sigma }_1=82.64\text{ MPa}$ ,${\sigma }_2=0\text{ MPa}$

Maximum shear stress${\tau }_{\max }=41.32\text{ MPa}$

To determine

Find principal stresses and maximum shear stress at top of web.

Answer to Problem 8.4.14P

Principal stresses

${\sigma }_1=76.13\text{ MPa}$ ,${\sigma }_2=4.5\text{ MPa}$

Maximum shear stress${\tau }_{\max }=40.32\text{ MPa}$

Explanation of Solution

Given Information:

Beam length$L=3\text{ m}$

Point load$P=120\text{ kN}$

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

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

  $M=\frac{P}{2}\times \frac{L}{3}\text{=}\frac{120}{2}\times \frac{3}{3}\text{=60 kN-m}$

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

  $\text{V}=\frac{P}{2}=60\text{ kN}$

Moment of inertia,

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{120\times {300}^3}{12}-\frac{120\times {260}^3}{12}+\frac{10\times {260}^3}{12}\\ I=108886666.7{\text{ mm}}^{\text{4}}\\ I=1.089\times {10}^{-4}{\text{ m}}^{\text{4}}\end{array}$

First moment of area of flange,

  $\text{Q}=A_1\times {\text{y}}_{\text{1}}=120\times \frac{300-260}{2}\times (\frac{260}{2}+\frac{300-260}{4})=336000{\text{ mm}}^{\text{3}}=3.36\times {10}^{-4}{\text{ m}}^{\text{3}}$

So bending stress at top of web,

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{60\times 0.13}{1.089\times {10}^{-4}}\\ {\sigma }_x=71625.344\text{ kPa =71}\text{.63 MPa}\end{array}$

And shear stress at that point,

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{60\times 3.36\times {10}^{-4}}{1.089\times {10}^{-4}\times 0.01}\\ {\tau }_{xy}=18.51\text{ MPa}\end{array}$

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

Principal normal stresses are given by following equation,

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{71.63+0}{2}\pm \sqrt{{\left(\frac{71.63-0}{2}\right)}^2+{18.51}^2}\\ {\sigma }_{1,2}=35.815\pm 40.315\\ \\ {\sigma }_1=35.815+40.315=76.13\text{ MPa}\\ {\sigma }_2=35.815-40.315=4.5\text{ MPa}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{71.63-0}{2}\right)}^2+{18.51}^2}\\ {\tau }_{\max }=40.32\text{ MPa}\end{array}$

Conclusion:

Hence we get,

Principal stresses

${\sigma }_1=76.13\text{ MPa}$ ,${\sigma }_2=4.5\text{ MPa}$

Maximum shear stress${\tau }_{\max }=40.32\text{ MPa}$

To determine

Find principal stresses and maximum shear stress at neutral axis.

Answer to Problem 8.4.14P

Principal stresses

${\sigma }_1=23.17\text{ MPa}$ ,${\sigma }_2=-23.17\text{MPa}$

Maximum shear stress${\tau }_{\max }\text{=23}\text{.17 MPa}$

Explanation of Solution

Given Information:

Beam length$L=3\text{ m}$

Point load$P=120\text{ kN}$

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

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

  $M=\frac{P}{2}\times \frac{L}{3}\text{=}\frac{120}{2}\times \frac{3}{3}\text{=60 kN-m}$

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

  $\text{V}=\frac{P}{2}=60\text{ kN}$

Moment of inertia,

  $\begin{array}{l}I=\frac{b{h}^3}{12}-\frac{b{h}_1{}^3}{12}+\frac{t{h}_1{}^3}{12}\text{ about the neutral z axis}\text{.}\\ I=\frac{120\times {300}^3}{12}-\frac{120\times {260}^3}{12}+\frac{10\times {260}^3}{12}\\ I=108886666.7{\text{ mm}}^{\text{4}}\\ I=1.089\times {10}^{-4}{\text{ m}}^{\text{4}}\end{array}$

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

  $\begin{array}{l}\text{Q}=A_1\times {\text{y}}_{\text{1}}+A_2\times {\text{y}}_2\\ Q=120\times \frac{300-260}{2}\times (\frac{260}{2}+\frac{300-260}{4})+10\times \frac{260}{2}\times (\frac{260}{4})\\ Q=420500{\text{ mm}}^{\text{3}}=4.205\times {10}^{-4}{\text{ m}}^{\text{3}}\end{array}$

So bending stress at neutral axis,

$\begin{array}{l}{\sigma }_x=\frac{My}{I}\\ {\sigma }_x=\frac{60\times 0}{1.089\times {10}^{-4}}\\ {\sigma }_x=0\text{ MPa}\end{array}$

And shear stress at that point,

$\begin{array}{l}{\tau }_{xy}=\frac{VQ}{It}\\ {\tau }_{xy}=\frac{60\times 4.205\times {10}^{-4}}{1.089\times {10}^{-4}\times 0.01}\\ {\tau }_{xy}=23.17\text{ MPa}\end{array}$

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

Principal normal stresses are given by following equation,

$\begin{array}{l}{\sigma }_{1,2}=\frac{{\sigma }_x+{\sigma }_y}{2}\pm \sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\sigma }_{1,2}=\frac{0+0}{2}\pm \sqrt{{\left(\frac{0-0}{2}\right)}^2+{23.17}^2}\\ {\sigma }_{1,2}=0\pm 23.17\\ \\ {\sigma }_1=0+23.17=23.17\text{ MPa}\\ {\sigma }_2=0-23.17=-23.17\text{ MPa}\end{array}$

Maximum shear stress,

$\begin{array}{l}{\tau }_{\max }=\sqrt{{\left(\frac{{\sigma }_x-{\sigma }_y}{2}\right)}^2+{\tau }_{xy}{}^2}\\ {\tau }_{\max }=\sqrt{{\left(\frac{0-0}{2}\right)}^2+{23.17}^2}\\ {\tau }_{\max }=23.17\text{ MPa}\end{array}$

Conclusion:

Hence we get,

Principal stresses

${\sigma }_1=23.17\text{ MPa}$ ,${\sigma }_2=-23.17\text{MPa}$

Maximum shear stress${\tau }_{\max }\text{=23}\text{.17 MPa}$