Chapter 9, Problem 31P

9-30 to 9-31 A steel bar of thickness h is subjected to a bending force F. The vertical support is stepped such that the horizontal welds are b₁ and b₂ long. Determine F if the maximum allowable shear stress is τ_allow.

Problem Number	b1	b2	c	d	h	τallow
9–30	2 in	4 in	6 in	4 in	$\frac{5}{16}$ in	25 kpsi
9–31	30 mm	50 mm	150 mm	50 mm	5 mm	140 MPa

Problems 9–30 to 9–31

To determine

The bending force $F$.

Answer to Problem 31P

The bending force $F$ is $9.932\text{kN}$.

Explanation of Solution

The below figure shows the dimension of the weld.

Figure-(1)

Write the expression for throat area of weld-1.

$A_1=0.707h{b}_1$ . (I)

Here, the thickness of weld is $h$ and length of weld is $b_1$.

Write the expression for throat area of weld-2.

$A_2=0.707h{b}_2$ (II)

Here, the length of weld is $b_2$.

Write the expression for throat area of weld-3.

$A_3=0.707hd$ (III)

Here, the length of weld is $d$.

Write the expression for total throat area.

$A=A_1+A_2+A_3$ . (IV)

Write the expression for total length of the weld.

$l=b_1+b_2+d$ (V)

Write the expression for location of centroid along x-direction.

$\overline{X}=\frac{x_1{b}_1+x_2{b}_2+x_3{b}_3}{l}$ . (VI)

Here, the centroidal length of weld-1 is $x_1$, the centriodal distance of length of weld-2 is $x_2$ and the centriodal distance of weld-3 is $x_3$.

Write the expression for centroid along y-direction.

$\overline{Y}=\frac{y_1{b}_1+y_2{b}_2+y_{3}d}{l}$ (VII)

Here, the centroidal distance of weld-1 is $y_1$ , the centroidal distance of weld-2 is $y_2$ and the centroidal distance of weld-3 is $y_3$.

Write the expression for the length between total weld and weld-1.

$r_1=\sqrt{{\left(\frac{b_1}{2}-\overline{X}\right)}^2+{\left(d-\overline{Y}\right)}^2}$ (VIII)

Write the expression for length between total weld and weld-2.

$r_2=\sqrt{{\left(\frac{b_2}{2}-\overline{X}\right)}^2+{\left(\overline{Y}\right)}^2}$ . (IX)

Write the expression for length of total weld and weld-3.

$r_3=\sqrt{{\left(\overline{X}\right)}^2+{\left(\frac{d}{2}-\overline{Y}\right)}^2}$ . (X)

Write the expression for polar moment of inertia of weld-1 about its centroid.

$J_1=\frac{0.707h{b}_1{}^3}{12}$ (XI)

Write the expression for moment of inertia of weld-2 about its centroid.

$J_2=\frac{0.707h{b}_2{}^3}{12}$ (XII)

Write the expression for polar moment of inertia of weld-3 about its centroid.

$J_3=\frac{0.707h{d}^3}{12}$ . (XIII)

Write the expression for polar moment of inertia of the total weld.

$J=\left(J_1+A_1{r}_1{}^2\right)+\left(J_2+A_2{r}_2{}^2\right)+\left(J_3+A_3{r}_3{}^2\right)$ (XIV)

Write the expression for length where maximum secondary shear stress occurs.

$r=\sqrt{{\overline{Y}}^2+{\left(b_2-\overline{X}\right)}^2}$ . (XV)

Write the expression for angle of $r$ with $x$- axis.

$\alpha ={\tan }^{-1}\left(\frac{\overline{Y}}{b_2-\overline{X}}\right)$ (XVI)

Write the expression for moment about weld center.

$M=F\left(c+b_2-\overline{X}\right)$ (XVII)

Here, the load on the weld is $F$.

Write the expression for primary shear stress.

${\tau }^{\prime }=\frac{F}{A}$ . (XVIII)

Write the expression for secondary shear.

${\tau }^{″}=\frac{Mr}{J}$ (XIX)

Write the expression for maximum shear stresss.

${\tau }_{\text{max}}=\sqrt{{\left({\tau }^{″}\sin \alpha \right)}^2+{\left({\tau }^{″}\cos \alpha +{\tau }^{\prime }\right)}^2}$                                            (XX)

Conclusion:

Substitute $5\text{mm}$ for $h$ and $30\text{mm}$ for $b_1$ in Equation (I).

$\begin{array}{c}{A}_1=0.707\left(5\text{mm}\right)\left(30\text{mm}\right)\\ =0.707\left(150{\text{mm}}^2\right)\\ =106.05{\text{mm}}^2\end{array}$

Substitute $5\text{mm}$ for $h$ and $50\text{mm}$ for $b_2$ in Equation (II).

$\begin{array}{c}{A}_2=0.707\left(5\text{mm}\right)\left(50\text{mm}\right)\\ =0.707\left(250{\text{mm}}^2\right)\\ =176.75{\text{mm}}^2\end{array}$

Substitute $5\text{mm}$ for $h$ and $50\text{mm}$ for $d$ in Equation (II).

$\begin{array}{c}{A}_3=0.707\left(5\text{mm}\right)\left(50\text{mm}\right)\\ =0.707\left(250{\text{mm}}^2\right)\\ =176.75{\text{mm}}^2\end{array}$

Substitute $106.05{\text{mm}}^2$ for $A_1$ , $176.75{\text{mm}}^2$ for $A_2$ and $176.75{\text{mm}}^2$ for  $A_3$ in Equation (III).

$\begin{array}{c}A=106.05{\text{mm}}^2+176.75{\text{mm}}^2+176.75{\text{mm}}^2\\ =459.55{\text{mm}}^2\end{array}$

Substitute $30\text{mm}$ for $b_1$ , $50\text{mm}$ for $b_2$ and $50\text{mm}$ for $d$ in Equation (IV).

$\begin{array}{c}l=30\text{mm}+50\text{mm}+50\text{mm}\\ =130\text{mm}\end{array}$

Substitute $30\text{mm}$ for $b_1$ , $50\text{mm}$ for $b_2$ , $50\text{mm}$ for $d$ , $130\text{mm}$ for $l$ , $15\text{mm}$ for $x_1$ , $25\text{mm}$ for $x_2$ and $0\text{mm}$ for $x_3$ in Equation (V).

$\begin{array}{c}\overline{X}=\frac{\left(15\text{mm}\right)\left(30\text{mm}\right)+\left(25\text{mm}\right)\left(50\text{mm}\right)+\left(0\text{mm}\times \text{50}\text{mm}\right)}{130\text{mm}}\\ =\frac{450{\text{mm}}^2+1250{\text{mm}}^2+0{\text{mm}}^2}{130\text{mm}}\\ =\frac{1700{\text{mm}}^2}{130\text{mm}}\\ =13.076\text{mm}\end{array}$

Substitute $30\text{mm}$ for $b_1$ , $50\text{mm}$ for $b_2$ , $50\text{mm}$ for $d$ , $130\text{mm}$ for $l$ , $50\text{mm}$ for $y_1$ , $0\text{mm}$ for $y_2$ and $25\text{mm}$ for $y_3$ in Equation (V).

$\begin{array}{c}\overline{Y}=\frac{\left(50\text{mm}\right)\left(30\text{mm}\right)+\left(0\text{mm}\right)\left(50\text{mm}\right)+\left(25\text{mm}\times \text{50}\text{mm}\right)}{130\text{mm}}\\ =\frac{1500{\text{mm}}^2+0{\text{mm}}^2+1250{\text{mm}}^2}{130\text{mm}}\\ =\frac{2750{\text{mm}}^2}{130\text{mm}}\\ =21.1538\text{mm}\end{array}$

Substitute $30\text{mm}$ for $b_1$ , $50\text{mm}$ for $d$ , $13.076\text{mm}$ for $\overline{X}$ and $21.1538\text{mm}$ for $\overline{Y}$ in Equation (VI).

$\begin{array}{c}{r}_1=\sqrt{{\left(\frac{30\text{mm}}{2}-13.076\text{mm}\right)}^2+{\left(50\text{mm}-21.1538\text{mm}\right)}^2}\\ =\sqrt{{\left(15\text{mm}-13.076\text{mm}\right)}^2+{\left(50\text{mm}-21.1538\text{mm}\right)}^2}\\ =\sqrt{835.8049{\text{mm}}^2}\\ =28.91\text{mm}\end{array}$

Substitute $50\text{mm}$ for $b_2$ , $13.076\text{mm}$ for $\overline{X}$ and $21.1538\text{mm}$ for $\overline{Y}$ in Equation (VII).

$\begin{array}{c}{r}_2=\sqrt{{\left(\frac{50\text{mm}}{2}-13.076\text{mm}\right)}^2+{\left(21.1538\text{mm}\right)}^2}\\ =\sqrt{{\left(25\text{mm}-13.076\text{mm}\right)}^2+{\left(21.1538\text{mm}\right)}^2}\\ =\sqrt{589.66{\text{mm}}^2}\\ =24.28\text{mm}\end{array}$

Substitute $50\text{mm}$ for $d$ , $13.076\text{mm}$ for $\overline{X}$ and $21.1538\text{mm}$ for $\overline{Y}$ in Equation (VIII).

$\begin{array}{c}{r}_3=\sqrt{{\left(13.076\text{mm}\right)}^2+{\left(\frac{50\text{mm}}{2}-21.1538\text{mm}\right)}^2}\\ =\sqrt{{\left(13.076\text{mm}\right)}^2+{\left(25\text{mm}-21.1538\text{mm}\right)}^2}\\ =\sqrt{185.7732{\text{mm}}^2}\\ =13.62\text{mm}\end{array}$

Substitute $5\text{mm}$ for $h$ and $30\text{mm}$ for $b_1$ in Equation (IX).

$\begin{array}{c}{J}_1=\frac{0.707\left(5\text{mm}\right){\left(30\text{mm}\right)}^3}{12}\\ =\frac{0.707\left(5\text{mm}\right)\left(27000{\text{mm}}^3\right)}{12}\\ =\frac{9544.5{\text{mm}}^4}{12}\\ =795.375{\text{mm}}^4\end{array}$

Substitute $5\text{mm}$ for $h$ and $50\text{mm}$ for $b_2$ in Equation (X).

$\begin{array}{c}{J}_2=\frac{0.707\left(5\text{mm}\right){\left(50\text{mm}\right)}^3}{12}\\ =\frac{0.707\left(5\text{mm}\right)\left(125000{\text{mm}}^3\right)}{12}\\ =\frac{441875{\text{mm}}^4}{12}\\ =36822.91{\text{mm}}^4\end{array}$

Substitute $5\text{mm}$ for $h$ and $50\text{mm}$ for $d$ in Equation (XI).

$\begin{array}{c}{J}_3=\frac{0.707\left(5\text{mm}\right){\left(50\text{mm}\right)}^3}{12}\\ =\frac{0.707\left(5\text{mm}\right)\left(125000{\text{mm}}^3\right)}{12}\\ =\frac{441875{\text{mm}}^4}{12}\\ =36822.91{\text{mm}}^4\end{array}$

Substitute $795.375{\text{mm}}^4$ for $J_1$, $106.05{\text{mm}}^2$ for $A_1$ , $176.75{\text{mm}}^2$ for $A_2$ , $176.75{\text{mm}}^2$ for $A_3$ , $36822.91{\text{mm}}^4$ for $J_2$ , $36822.91{\text{mm}}^4$ for $J_3$ , $28.91\text{mm}$ for $r_1$ , $24.28\text{mm}$ for $r_2$ and $13.62\text{mm}$ for $r_3$ in Equation (XII).

$\begin{array}{c}J=\left[\begin{array}{l}\left(795.375{\text{mm}}^4+\left(106.05{\text{mm}}^2\right){\left(28.91\text{mm}\right)}^2\right)+\\ \left(36822.91{\text{mm}}^4+\left(176.75{\text{mm}}^2\right){\left(24.28\text{mm}\right)}^2\right)+\\ \left(36822.91{\text{mm}}^4+\left(176.75{\text{mm}}^2\right){\left(13.62\text{mm}\right)}^2\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(795.375{\text{mm}}^4+88635.32{\text{mm}}^4\right)+\\ \left(36822.91{\text{mm}}^4+104197.37{\text{mm}}^4\right)+\\ \left(36822.91{\text{mm}}^4+32787.90{\text{mm}}^4\right)\end{array}\right]\\ =89430.695{\text{mm}}^4+141020.28{\text{mm}}^4+69610.81{\text{mm}}^4\\ =300061.785{\text{mm}}^4\end{array}$

Substitute $21.1538\text{mm}$ for $\overline{Y}$ , $50\text{mm}$ for $b_2$ and $13.076\text{mm}$ for $\overline{X}$ in Equation (XIII).

$\begin{array}{c}r=\sqrt{{\left(21.1538\text{mm}\right)}^2+{\left(50\text{mm}-13.076\text{mm}\right)}^2}\\ =\sqrt{447.48{\text{mm}}^2+{\left(36.924\text{mm}\right)}^2}\\ =\sqrt{1810.86{\text{mm}}^2}\\ =42.55\text{mm}\end{array}$

Substitute $21.1538\text{mm}$ for $\overline{Y}$, $50\text{mm}$ for $b_2$ and $13.076\text{mm}$ for $\overline{X}$ in Equation (XIV).

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{21.1538\text{mm}}{50\text{mm}-13.076\text{mm}}\right)\\ ={\tan }^{-1}\left(\frac{21.1538\text{mm}}{36.924\text{mm}}\right)\\ ={\tan }^{-1}\left(0.5729\right)\\ =33.12°\end{array}$

Substitute $150\text{mm}$ for $c$, $50\text{mm}$ for $b_2$, $F\text{kN}$ for $F$ and $13.076\text{m}$ for $\overline{X}$ in Equation (XV).

$\begin{array}{c}M=\left(F\text{kN}\right)\left(150\text{mm}+50\text{mm}-\text{13}\text{.076}\text{mm}\right)\\ =\left(F\text{kN}\right)\left(186.924\text{mm}\right)\\ =186.924F\text{kN}\cdot \text{mm}\end{array}$

Substitute $F\text{kN}$ for $F$ and $459.55{\text{mm}}^2$ for $A$ in Equation (XVI).

$\begin{array}{c}{\tau }^{\prime }=\frac{F\text{kN}}{459.55{\text{mm}}^2}\\ =\frac{\left(F\text{kN}\right)\left(\frac{1000\text{N}}{1\text{kN}}\right)}{459.55{\text{mm}}^2}\\ =\frac{1000F\text{N}}{459.55{\text{mm}}^2}\\ =2.176F\text{MPa}\end{array}$

Substitute $186.924F\text{kN}\cdot \text{mm}$ for $M$ , $42.55\text{mm}$ for $r$ and $300061.785{\text{mm}}^4$ for $J$ in Equation (XVII).

$\begin{array}{c}{\tau }^{″}=\frac{\left(186.924F\text{kN}\cdot \text{mm}\right)\left(42.55\text{mm}\right)}{300061.785{\text{mm}}^4}\\ =\frac{7953.6162F\text{kN}\cdot {\text{mm}}^2}{300061.785{\text{mm}}^4}\\ =\left(0.0265F\text{kPa}\right)\left(\frac{1000\text{Mpa}}{1\text{kPa}}\right)\\ =26.5F\text{MPa}\end{array}$

Substitute $26.5F\text{MPa}$ for ${\tau }^{″}$ , $33.12°$ for $\alpha$ , $140\text{MPa}$ for ${\tau }_{\text{max}}$ and $2.176F\text{MPa}$ for ${\tau }^{\prime }$ in Equation (XVIII).

$\begin{array}{c}140\text{MPa}=\sqrt{{\left\{\begin{array}{l}{\left(26.5F\sin \left(33.12°\right)\text{MPa}\right)}^2\\ +\left(26.5F\cos \left(33.12°\right)\text{MPa}+2.176F\text{MPa}\right)\end{array}\right\}}^2}\\ 140\text{MPa}=\sqrt{{\left(13.1730F\text{MPa}\right)}^2+{\left(22.9939F\text{MPa}+2.176F\text{MPa}\right)}^2}\\ 140\text{MPa}=\sqrt{198.6899F^2{\text{MPa}}^2}\end{array}$

$\begin{array}{l}140\text{MPa}=14.0957F\text{MPa}\\ F=\frac{140\text{MPa}}{14.0957\text{MPa}}\\ F=9.932\text{kN}\end{array}$

Thus, the bending force is $9.932\text{kN}$.