Chapter 9, Problem 30P

To determine

The bending force $F$.

Answer to Problem 30P

The bending force $F$ is $1.77\text{kip}$

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 stress.

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

Conclusion:

Substitute $\frac{5}{16}\text{in}$ for $h$ and $2\text{in}$ for $b_1$ in Equation (I).

$\begin{array}{c}{A}_1=0.707\left(\frac{5}{16}\text{in}\right)\left(2\text{in}\right)\\ =0.707\left(0.625{\text{in}}^2\right)\\ =0.4418{\text{in}}^2\end{array}$

Substitute $\frac{5}{16}\text{in}$ for $h$ and $4\text{in}$ for $b_2$ in Equation (II).

$\begin{array}{c}{A}_2=0.707\left(\frac{5}{16}\text{in}\right)\left(4\text{in}\right)\\ =0.707\left(1.25{\text{in}}^2\right)\\ =0.88375{\text{in}}^2\end{array}$

Substitute $\frac{5}{16}\text{in}$ for $h$ and $4\text{in}$ for $d$ in Equation (II).

$\begin{array}{c}{A}_3=0.707\left(\frac{5}{16}\text{in}\right)\left(4\text{in}\right)\\ =0.707\left(1.25{\text{in}}^2\right)\\ =0.88375{\text{in}}^2\end{array}$

Substitute $0.4418{\text{in}}^2$ for $A_1$ , $0.88375{\text{in}}^2$ for $A_2$ and $0.88375{\text{in}}^2$ for $A_3$ in Equation (III).

$\begin{array}{c}A=0.4418{\text{in}}^2+0.88375{\text{in}}^2+0.88375{\text{in}}^2\\ =2.2093{\text{in}}^2\end{array}$

Substitute $2\text{in}$ for $b_1$ , $4\text{in}$ for $b_2$ and $4\text{in}$ for $d$ in Equation (IV).

$\begin{array}{c}l=2\text{in}+4\text{in}+4\text{in}\\ =10\text{in}\end{array}$

Substitute $2\text{in}$ for $b_1$ , $4\text{in}$ for $b_2$ , $4\text{in}$ for $d$ , $10\text{in}$ for $l$ , $1\text{in}$ for $x_1$ , $2\text{in}$ for $x_2$ and $0\text{in}$ for $x_3$ in Equation (V).

$\begin{array}{c}\overline{X}=\frac{\left(1\text{in}\right)\left(2\text{in}\right)+\left(2\text{in}\right)\left(4\text{in}\right)+\left(0\text{in}\times 4\text{in}\right)}{10\text{in}}\\ =\frac{2{\text{in}}^2+8{\text{in}}^2+0{\text{in}}^2}{10\text{in}}\\ =\frac{10{\text{in}}^2}{10\text{in}}\\ =1\text{in}\end{array}$

Substitute $2\text{in}$ for $b_1$ , $4\text{in}$ for $b_2$ , $4\text{in}$ for $d$ , $10\text{in}$ for $l$ , $4\text{in}$ for $y_1$ , $0\text{in}$ for $y_2$ and $2\text{in}$ for $y_3$ in Equation (V).

$\begin{array}{c}\overline{Y}=\frac{\left(4\text{in}\right)\left(2\text{in}\right)+\left(0\text{in}\right)\left(4\text{in}\right)+\left(2\text{in}\times 4\text{in}\right)}{10\text{in}}\\ =\frac{8{\text{in}}^2+0{\text{in}}^2+8{\text{mm}}^2}{10\text{in}}\\ =\frac{16{\text{in}}^2}{10\text{in}}\\ =1.6\text{in}\end{array}$

Substitute $2\text{in}$ for $b_1$ , $4\text{in}$ for $d$ , $1\text{in}$ for $\overline{X}$ and $1.6\text{in}$ for $\overline{Y}$ in Equation (VI).

$\begin{array}{c}{r}_1=\sqrt{{\left(\frac{2\text{in}}{2}-1\text{in}\right)}^2+{\left(4\text{in}-1.6\text{in}\right)}^2}\\ =\sqrt{{\left(1\text{in}-1\text{in}\right)}^2+{\left(4\text{in}-1.6\text{in}\right)}^2}\\ =\sqrt{0{\text{in}}^2+5.76{\text{in}}^2}\\ =2.4\text{in}\end{array}$

Substitute $4\text{in}$ for $b_2$ , $1\text{in}$ for $\overline{X}$ and $1.6\text{in}$ for $\overline{Y}$ in Equation (VII).

$\begin{array}{c}{r}_2=\sqrt{{\left(\frac{4\text{in}}{2}-1\text{in}\right)}^2+{\left(1.6\text{mm}\right)}^2}\\ =\sqrt{{\left(2\text{in}-1\text{in}\right)}^2+{\left(1.6\text{in}\right)}^2}\\ =\sqrt{2.56{\text{in}}^2}\\ =1.6\text{in}\end{array}$

Substitute $4\text{in}$ for $d$ , $1\text{in}$ for $\overline{X}$ and $1.6\text{in}$ for $\overline{Y}$ in Equation (VIII).

$\begin{array}{c}{r}_3=\sqrt{{\left(1\text{in}\right)}^2+{\left(\frac{4\text{in}}{2}-1.6\text{in}\right)}^2}\\ =\sqrt{{\left(1\text{in}\right)}^2+{\left(2\text{in}-1.6\text{in}\right)}^2}\\ =\sqrt{1.16{\text{in}}^2}\\ =1.077\text{in}\end{array}$

Substitute $\frac{5}{16}\text{in}$ for $h$ and $2\text{in}$ for $b_1$ in Equation (IX).

$\begin{array}{c}{J}_1=\frac{0.707\left(\frac{5}{16}\text{in}\right){\left(2\text{in}\right)}^3}{12}\\ =\frac{0.707\left(\frac{5}{16}\text{in}\right)\left(8{\text{in}}^3\right)}{12}\\ =\frac{1.7675{\text{in}}^4}{12}\\ =0.1472{\text{in}}^4\end{array}$

Substitute $\frac{5}{16}\text{in}$ for $h$ and $4\text{in}$ for $b_2$ in Equation (X).

$\begin{array}{c}{J}_2=\frac{0.707\left(\frac{5}{16}\text{in}\right){\left(4\text{in}\right)}^3}{12}\\ =\frac{0.707\left(\frac{5}{16}\text{in}\right)\left(64{\text{in}}^3\right)}{12}\\ =\frac{14.14{\text{in}}^4}{12}\\ =1.178{\text{in}}^4\end{array}$

Substitute $\frac{5}{16}\text{in}$ for $h$ and $4\text{in}$ for $d$ in Equation (XI).

$\begin{array}{c}{J}_3=\frac{0.707\left(\frac{5}{16}\text{in}\right){\left(4\text{in}\right)}^3}{12}\\ =\frac{0.707\left(\frac{5}{16}\text{in}\right)\left(64{\text{in}}^3\right)}{12}\\ =\frac{14.14{\text{in}}^4}{12}\\ =1.178{\text{in}}^4\end{array}$

Substitute $0.1472{\text{in}}^4$ for $J_1$, $0.4418{\text{in}}^2$ for $A_1$ , $0.88375{\text{in}}^2$ for $A_2$ , $0.88375{\text{in}}^2$ for $A_3$ , $1.178{\text{in}}^4$ for $J_2$ , $1.178{\text{in}}^4$ for $J_3$ , $2.4\text{in}$ for $r_1$ , $1.6\text{in}$ for $r_2$ and $1.077\text{in}$ for $r_3$ in Equation (XII).

$\begin{array}{c}J=\left[\begin{array}{l}\left(0.1472{\text{in}}^4+\left(0.4418{\text{in}}^2\right){\left(2.4\text{in}\right)}^2\right)+\\ \left(1.178{\text{in}}^4+0.88375{\text{in}}^2{\left(1.6\text{in}\right)}^2\right)+\\ \left(1.178{\text{in}}^4+\left(0.88375{\text{in}}^2\right){\left(1.077\text{in}\right)}^2\right)\end{array}\right]\\ =\left[\begin{array}{l}\left(0.1472{\text{in}}^4+2.544{\text{in}}^4\right)+\\ \left(1.178{\text{in}}^4+2.2624{\text{in}}^4\right)+\\ \left(1.178{\text{in}}^4+1.0250{\text{in}}^4\right)\end{array}\right]\\ =2.6912{\text{in}}^4+3.4404{\text{in}}^4+2.203{\text{in}}^4\\ =8.3346{\text{in}}^4\end{array}$

Substitute $1.6\text{in}$ for $\overline{Y}$ , $4\text{in}$ for $b_2$ and $1\text{in}$ for $\overline{X}$ in Equation (XIII).

$\begin{array}{c}r=\sqrt{{\left(1.6\text{in}\right)}^2+{\left(4\text{in}-1\text{in}\right)}^2}\\ =\sqrt{2.56{\text{in}}^2+{\left(3\text{in}\right)}^2}\\ =\sqrt{11.56{\text{in}}^2}\\ =3.4\text{in}\end{array}$

Substitute $1.6\text{in}$ for $\overline{Y}$, $4\text{in}$ for $b_2$ and $1\text{in}$ for $\overline{X}$ in Equation (XIV).

$\begin{array}{c}\alpha ={\tan }^{-1}\left(\frac{1.6\text{in}}{4\text{in}-1\text{in}}\right)\\ ={\tan }^{-1}\left(\frac{1.6\text{in}}{3\text{in}}\right)\\ ={\tan }^{-1}\left(0.533\right)\\ =31.19°\end{array}$

Substitute $6\text{in}$ for $c$ , $4\text{in}$ for $b_2$ , $F\text{kip}$ for $F$ and $1\text{in}$ for $\overline{X}$ in Equation (XV).

$\begin{array}{c}M=\left(F\text{kip}\right)\left(6\text{in}+4\text{in}-1\text{in}\right)\\ =\left(F\text{kip}\right)\left(9\text{in}\right)\\ =9F\text{kip}\cdot \text{in}\end{array}$

Substitute $F\text{kip}$ for $F$ and $2.2093{\text{in}}^2$ for $A$ in Equation (XVI).

$\begin{array}{c}{\tau }^{\prime }=\frac{F\text{kip}}{2.2093{\text{in}}^2}\\ =0.4526F\text{kip}/{\text{in}}^2\end{array}$

Substitute $9F\text{kip}\cdot \text{in}$ for $M$ , $3.4\text{in}$ for $r$ and $8.3346{\text{in}}^4$ for $J$ in Equation (XVII).

$\begin{array}{c}{\tau }^{″}=\frac{\left(9F\text{kip}\cdot \text{in}\right)\left(3.4\text{in}\right)}{8.3346{\text{in}}^4}\\ =\frac{30.6F\text{kip}\cdot {\text{in}}^2}{8.3346{\text{in}}^4}\\ =3.67F\text{kip}/{\text{in}}^2\end{array}$

Substitute $3.67F\text{kip}/{\text{in}}^2$ for ${\tau }^{″}$ , $31.19°$ for $\alpha$ , $25\text{kpsi}$ for ${\tau }_{\text{max}}$ and $0.4526F\text{kip}/{\text{in}}^2$ for ${\tau }^{\prime }$ in Equation (XVIII).

$\begin{array}{c}25\text{kpsi}=\sqrt{{\left\{\begin{array}{l}{\left(3.67F\sin \left(31.19°\right)\text{kip}/{\text{in}}^2\right)}^2\\ +\left(3.67F\cos \left(31.19°\right)\text{kip}/{\text{in}}^2+0.4526F\text{kip}/{\text{in}}^2\right)\end{array}\right\}}^2}\\ 25\text{kpsi}=\sqrt{{\left(1.9006F\text{kip}/{\text{in}}^2\right)}^2+{\left(3.136F\text{kip}/{\text{in}}^2+0.4526F\text{kip}/{\text{in}}^2\right)}^2}\\ 25\text{kpsi}=\sqrt{3.6122F^2{\left(\text{kip}/{\text{in}}^2\right)}^2+12.8780F^2{\left(\text{kip}/{\text{in}}^2\right)}^2}\\ 25\text{kpsi}=\sqrt{16.4902F^2{\left(\text{kip}/{\text{in}}^2\right)}^2}\end{array}$

$\begin{array}{l}25\text{kpsi}=14.0957F\text{kpsi}\\ F=\frac{25\text{kpsi}}{14.0957F\text{kpsi}}\\ F=1.77\text{kip}\end{array}$

Thus, the bending force is $1.77\text{kip}$