Chapter 9, Problem 53P

For the sake of perspective it is always useful to look at the matter of scale. Double all dimensions in Prob. 9-18 and find the allowable load. By what factor has it increased? First make a guess, then carry out the computation. Would you expect the same ratio if the load had been variable?

To determine

The allowable load.

The factor at which the allowable load is increased.

Answer to Problem 53P

The allowable load is $11.7924\text{kip}$

The factor at which the allowable load is increased is $4.382$.

Explanation of Solution

Write the expression for throat area.

$A_1=1.414h_1{d}_1$ (I)

Here, the weld thickness is $h_1$ and depth of weld is $d_1$.

Write the expression for primary shear stress due to shear force.

${\left({{\tau }^{\prime }}_y\right)}_1=\frac{V}{A_1}$ (II)

Here, the shear force is $V$ and throat area is $A_1$.

Write the expression for unit second polar moment of area.

$J_{u1}=\frac{d_1\left(3b_1{}^2+d_1{}^2\right)}{6}$ (III)

Here, width of weld is $b_1$.

Write the expression for second polar moment of area.

$J_1=0.707h_1{J}_{u1}$ (IV)

Write the expression for length of centroid from force.

$l_1=\left(c_1+\frac{b_1}{2}\right)$ (V)

Here, distance of unfix bar is $c_1$ and distance of welded bar is $b_1$.

Write the expression for bending moment.

$M_1=F_1\times l_1$ (VI)

Here force acting on the bar is $F_1$.

Write the expression for radius of torsion in.

$r_{y1}=\frac{d_1}{2}$ (VII)

Write the expression for radius of torsion in $x-$ direction.

$r_{x1}=\frac{b_1}{2}$ (VIII)

Write the expression for secondary shear in $x-$ direction.

${{\tau }^{″}}_{x1}=\frac{M_1{r}_{y1}}{J_1}$ (IX)

Write the expression for secondary shear in $y-$ direction.

${{\tau }^{″}}_{y1}=\frac{M{r}_{x1}}{J_1}$ (X)

Write the expression for net shear stress.

${\tau }_1=\sqrt{{{\tau }^{″}}_{x1}{}^2+{\left({{\tau }^{\prime }}_{y1}+{{\tau }^{″}}_{y1}\right)}^2}$ (XI)

Write the expression for throat area.

$A=1.414hd$ (XII)

Here, the weld thickness is $h$ and depth of weld is $d$

Write the expression for primary shear stress due to shear force.

${{\tau }^{\prime }}_y=\frac{V}{A}$ (XIII)

Here, the shear force is $V$ and throat area is $A$.

Write the expression for unit second polar moment of area.

$J_u=\frac{d\left(3b^2+d^2\right)}{6}$ (XIV)

Here, width of weld is $b$.

Write the expression for second polar moment of area.

$J=0.707h{J}_u$ (XV)

Write the expression for length of centroid from force.

$l=\left(c+\frac{b}{2}\right)$ (XVI)

Here, distance of unfix bar is $c$ and distance of welded bar is $b$.

Write the expression for bending moment.

$M=F\times l$ (XVII)

Here force acting on the bar is $F$.

Write the expression for radius of torsion in.

$r_y=\frac{d}{2}$ (XVIII)

Write the expression for radius of torsion in $x-$ direction.

$r_x=\frac{b}{2}$ (XIX)

Write the expression for secondary shear in $x-$ direction.

${{\tau }^{″}}_x=\frac{M{r}_y}{J}$ (XX)

Write the expression for secondary shear in $y-$ direction.

${{\tau }^{″}}_y=\frac{M{r}_x}{J}$ (XXI)

Write the expression for net shear stress.

$\tau =\sqrt{{{\tau }^{″}}_x{}^2+{\left({{\tau }^{\prime }}_y+{{\tau }^{″}}_y\right)}^2}$ (XXII)

Write the expression for increased factor.

$C=\frac{F}{F_1}$ . (XXIII)

Conclusion:

Substitute $\frac{5}{16}\text{in}$ for $h_1$ and $2\text{in}$ for $d_1$ in Equation (I).

$\begin{array}{c}{A}_1=1.414\times \left(\frac{5}{16}\text{in}\right)\times \left(2\text{in}\right)\\ =\left(0.441875\text{in}\right)\left(2\text{in}\right)\\ =\text{0}\text{.881}{\text{in}}^2\end{array}$

Substitute $0.881{\text{in}}^2$ for $A_1$ and $F\text{kip}$ for $V$ in Equation (II).

$\begin{array}{c}{{\tau }^{\prime }}_{y1}=\frac{\left(F_1\text{kip}\right)}{\text{0}\text{.881}{\text{in}}^2}\\ =1.132F_1\text{kip}/{\text{in}}^2\end{array}$

Substitute $2\text{in}$ for $d_1$ and $2\text{in}$ for $b_1$ in Equation (III).

$\begin{array}{c}{J}_{u1}=\frac{2\text{in}\left(3\times {\left(2\text{in}\right)}^2+{\left(2\text{in}\right)}^2\right)}{6}\\ =\frac{32{\text{in}}^3}{6}\\ =5.33{\text{in}}^3\end{array}$

Substitute $\frac{5}{16}\text{in}$ for $h_1$ and $5.33{\text{in}}^3$ for $J_{u1}$ in Equation (IV).

$\begin{array}{c}{J}_1=0.707\times \left(\frac{5}{16}\text{in}\right)\times \left(5.33{\text{in}}^3\right)\\ =1.17{\text{in}}^4\end{array}$

Substitute $6\text{in}$ for $c_1$ and $2\text{in}$ for $b_1$ in Equation (V).

$\begin{array}{c}{l}_1=\left(\text{6}\text{in+}\frac{2\text{in}}{2}\right)\\ =6\text{in+1}\text{in}\\ =7\text{in}\end{array}$

Substitute $7\text{in}$ for $l_1$ in Equation (VI).

$\begin{array}{c}{M}_1=\left(F_1\text{kip}\right)\times \left(7\text{in}\right)\\ \text{=7}{F}_1\text{kip}\cdot \text{in}\end{array}$

Substitute $2\text{in}$ for $d_1$ in Equation (VII).

$\begin{array}{c}{r}_{y1}=\frac{2\text{in}}{2}\\ =1\text{in}\end{array}$

Substitute $2\text{in}$ for $b_1$ in Equation (VIII).

$\begin{array}{c}{r}_{x1}=\frac{2\text{in}}{2}\\ =1\text{in}\end{array}$

Substitute $1\text{in}$ for $r_{y1}$, $7F_1\text{kip}\cdot \text{in}$ for $M_1$ and $1.17{\text{in}}^4$ for $J_1$ in Equation (IX).

$\begin{array}{c}{{\tau }^{″}}_{x1}=\frac{\left(7F_1\text{kip}\cdot \text{in}\right)\left(1\text{in}\right)}{\left(1.17{\text{in}}^4\right)}\\ =\frac{7F_1\text{kip}\cdot {\text{in}}^2}{1.17{\text{in}}^4}\\ =5.98F_1\text{kip}/{\text{in}}^2\end{array}$

Substitute $1\text{in}$ for $r_{x1}$, $7F_1\text{kip}\cdot \text{in}$ for $M_1$ and $1.17{\text{in}}^4$ for $J_1$ in Equation (X).

$\begin{array}{c}{{\tau }^{″}}_{y1}=\frac{\left(7F_1\text{kip}\cdot \text{in}\right)\left(1\text{in}\right)}{\left(1.17{\text{in}}^4\right)}\\ =\frac{7F_1\text{kip}\cdot {\text{in}}^2}{1.17{\text{in}}^4}\\ =5.98F_1\text{kip}/{\text{in}}^2\end{array}$

Substitute $5.98F_1\text{kip}/{\text{in}}^2$ for ${{\tau }^{″}}_{x1}$, $5.98F\text{kip}/{\text{in}}^2$ for ${{\tau }^{″}}_{y1}$ , $25\text{kpsi}$ for ${\tau }_1$ and $1.132F_1\text{kip}/{\text{in}}^2$ for ${{\tau }^{\prime }}_{y1}$ in Equation (XI).

$\begin{array}{l}25\text{kpsi}=\sqrt{{\left(5.98F_1{\text{kip}/\text{in}}^2\right)}^2+{\left(\left(1.132F_1\text{kip}/{\text{in}}^2\right)+\left(5.98F_1\text{kip}/{\text{in}}^2\right)\right)}^2}\\ 25\text{kpsi}=\sqrt{86.3404F_1{}^2{\left(\text{kip}/{\text{in}}^2\right)}^2}\\ 25\text{kpsi}=9.29F_1\text{kip}/{\text{in}}^2\\ F_1=\frac{25\text{kpsi}}{9.29\text{kpsi}}\end{array}$

$F_1=2.691\text{kip}$

Substitute $\frac{5}{8}\text{in}$ for $h$ and $4\text{in}$ for $d$ in Equation (XI).

$\begin{array}{c}A=1.414\times \left(\frac{5}{8}\text{in}\right)\times \left(4\text{in}\right)\\ =1.414\times \left(2.5{\text{in}}^2\right)\\ =3.535{\text{in}}^2\end{array}$

Substitute $1.7675{\text{in}}^2$ for $A$ and $F\text{kip}$ for $V$ in Equation (XII).

$\begin{array}{c}{{\tau }^{\prime }}_y=\frac{\left(F\text{kip}\right)}{3.535{\text{in}}^2}\\ =0.2828F\text{kip}/{\text{in}}^2\end{array}$

Substitute $4\text{in}$ for $d$ and $4\text{in}$ for $b$ in Equation (XIII).

$\begin{array}{c}{J}_u=\frac{4\text{in}\left(3\times {\left(4\text{in}\right)}^2+{\left(4\text{in}\right)}^2\right)}{6}\\ =\frac{256{\text{in}}^3}{6}\\ =42.66{\text{in}}^3\end{array}$

Substitute $\frac{5}{8}\text{in}$ for $h$ and $42.66{\text{in}}^3$ for $J_u$ in Equation (XIV).

$\begin{array}{c}J=0.707\times \left(\frac{5}{8}\text{in}\right)\times \left(42.66{\text{in}}^3\right)\\ =0.707\times 26.66{\text{in}}^4\\ =18.84{\text{in}}^4\end{array}$

Substitute $12\text{in}$ for $c$ and $4\text{in}$ for $b$ in Equation (XV).

$\begin{array}{c}l=\left(12\text{in+}\frac{4\text{in}}{2}\right)\\ =12\text{in+2}\text{in}\\ =14\text{in}\end{array}$

Substitute $14\text{in}$ for $l$ in Equation (XVI).

$\begin{array}{c}M=\left(F\text{kip}\right)\times \left(14\text{in}\right)\\ \text{=14F}\text{kip}\cdot \text{in}\end{array}$

Substitute $4\text{in}$ for $d$ in Equation (XVII).

$\begin{array}{c}{r}_y=\frac{4\text{in}}{2}\\ =2\text{in}\end{array}$

Substitute $4\text{in}$ for $b$ in Equation (XVIII).

$\begin{array}{c}{r}_x=\frac{4\text{in}}{2}\\ =2\text{in}\end{array}$

Substitute $2\text{in}$ for $r_y$, $14F\text{kip}\cdot \text{in}$ for $M$ and $18.84{\text{in}}^4$ for $J$ in Equation (XIX).

$\begin{array}{c}{{\tau }^{″}}_x=\frac{\left(14F\text{kip}\cdot \text{in}\right)\left(2\text{in}\right)}{\left(18.84{\text{in}}^4\right)}\\ =\frac{28F\text{kip}\cdot {\text{in}}^2}{18.84{\text{in}}^4}\\ =1.4861F\text{kip}/{\text{in}}^2\end{array}$

Substitute $2\text{in}$ for $r_x$, $14F\text{kip}\cdot \text{in}$ for $M$ and $18.84{\text{in}}^4$ for $J$ in Equation (XX).

$\begin{array}{c}{{\tau }^{″}}_y=\frac{\left(14F\text{kip}\cdot \text{in}\right)\left(2\text{in}\right)}{\left(18.84{\text{in}}^4\right)}\\ =\frac{28F\text{kip}\cdot {\text{in}}^2}{18.84{\text{in}}^4}\\ =1.4861F\text{kip}/{\text{in}}^2\end{array}$

Substitute $1.4861F\text{kip}/{\text{in}}^2$ for ${{\tau }^{″}}_x$, $1.4861F\text{kip}/{\text{in}}^2$ for ${{\tau }^{″}}_y$, $25\text{kpsi}$ for $\tau$ and $0.2828F\text{kip}/{\text{in}}^2$ for ${{\tau }^{\prime }}_y$ in Equation (XXI).

$\begin{array}{l}25\text{kpsi}=\sqrt{{\left(1.4861F{\text{kip}/\text{in}}^2\right)}^2+{\left(\left(1.4861F\text{kip}/{\text{in}}^2\right)+\left(0.2828F\text{kip}/{\text{in}}^2\right)\right)}^2}\\ 25\text{kpsi}=\sqrt{4.49677F^2{\left(\text{kip}/{\text{in}}^2\right)}^2}\\ 25\text{kpsi}=2.120F\text{kip}/{\text{in}}^2\\ F=\frac{25\text{kpsi}}{2.120\text{kpsi}}\end{array}$

$F=11.7924\text{kip}$

Thus, the allowable load is $11.7924\text{kip}$.

Substitute $11.7924\text{kip}$ for $F$ and $2.691\text{kip}$ for $F_1$ in Equation (XXII).

$\begin{array}{c}C=\frac{11.7924\text{kip}}{2.691\text{kip}}\\ =4.382\end{array}$

Thus, the factor at which the allowable load is increased is $4.382$.