Chapter 9, Problem 39P

To determine

The primary shear stress.

The secondary shear stress.

The maximum shear stress.

Answer to Problem 39P

The value of primary shear stress is $4755.33\text{lbf}/{\text{in}}^2$

The value of secondary shear stress is $2572.68\text{lbf}/{\text{in}}^2$.

The value of maximum shear stress is $2770.88\text{lbf}/{\text{in}}^2$.

The figure of merit of weld is $86.067\text{in}$.

The effectiveness of weld is $121.7\text{in}$.

The primary shear stress for interrupted weld is $4754.65\text{lbf}/{\text{in}}^2$.

The secondary shear stress for interrupted weld is $11886.72\text{lbf}/{\text{in}}^2$.

The maximum shear stress for interrupted weld is $12802.37\text{lbf}/{\text{in}}^2$.

The figure of merit of interrupted weld is $64.542\text{in}$.

The effectiveness of interrupted weld is $91.26\text{in}$.

Explanation of Solution

Write the expression for throat area.

$A=1.414hb$ . (I)

Write the expression for length of weld.

$l=\left(b+d\right)$ (II)

Write the expression for primary shear stress.

$\sigma =\frac{V}{A}$ . (III)

Write the expression for unit second moment of area.

$I_u=\frac{b{d}^2}{2}$ (IV)

Write the expression for second moment of area.

$I=0.707h{I}_u$ (V)

Write the expression for moment.

$M=F\times l$ (VI)

Write the expression for secondary shear.

$\tau =\frac{M{r}_y}{I}$ . (VII)

Write the expression for maximum shear stress.

${\tau }_{\text{max}}=\sqrt{{\left(\sigma \right)}^2+{\left(\tau \right)}^2}$ (VIII)

Write the expression for leg size of weld.

${\tau }_{\max }={\tau }_{\text{all}}$                                                             (IX)

Write the expression for figure of marit.

$fom=\frac{I_u}{hl}$                                                               (X)

Write the expression for effectiveness of weld.

$eff=\frac{I}{\left(\frac{h^2}{2}\right)l}$ . (XI)

Write the expression for throat area for interrupted weld.

$A_I=1.414h_I\left(b-b_1\right)$ . (XII)

Write the expression for length of weld interrupted weld.

$l_I=b+d-2b_1$ (XIII)

Write the expression for primary shear stress interrupted weld.

${\sigma }_I=\frac{V}{A_I}$ . (XIV)

Write the expression for unit second moment of area interrupted weld.

$I_{uI}=\frac{\left(b-b_1\right)d^2}{2}$ (XV)

Write the expression for second moment of area interrupted weld.

$I_I=0.707h{I}_u$ (XVI)

Write the expression for secondary shear interrupted weld.

${\tau }_I=\frac{M{r}_y}{I}$ . (XVII)

Write the expression for maximum shear stress interrupted weld.

${\tau }_{\text{max}\text{I}}=\sqrt{{\left({\sigma }_I\right)}^2+{\left({\tau }_I\right)}^2}$ (XVIII)

Write the expression for leg size of weld interrupted weld.

${\tau }_{\max I}={\tau }_{\text{all}\text{I}}$                                                            (XIX)

Write the expression for figure of marit interrupted weld.

$fo{m}_I=\frac{I_{uI}}{h_I{l}_I}$                                                              (XX)

Conclusion:

Substitute $8\text{in}$ for $b$ and $h\text{in}$ for $h$ in Equation (I).

$\begin{array}{c}A=1.414\left(h\text{in}\right)\left(8\text{in}\right)\\ =11.312h{\text{in}}^2\end{array}$

Substitute $8\text{in}$ for $b$ and $8\text{in}$ for $d$ in Equation (II).

$\begin{array}{c}l=\left(8\text{in}\right)+\left(8\text{in}\right)\\ =16\text{in}\end{array}$

Substitute $10\text{kips}$ for $V$ and $11.312h{\text{in}}^2$ for  $A$ in Equation (III).

$\begin{array}{c}\sigma =\frac{10\text{kips}}{11.312h{\text{in}}^2}\\ =\frac{\left(10\text{kips}\right)\left(\frac{1000\text{lbf}}{1\text{kips}}\right)}{11.312h{\text{in}}^2}\\ =\frac{10000\text{lbf}}{11.312h{\text{in}}^2}\\ =\frac{884.016}{h}\text{lbf}/{\text{in}}^2\end{array}$ (XXI)

Substitute $8\text{in}$ for $b$ and $8\text{in}$ for $d$ in Equation (IV).

$\begin{array}{c}{I}_u=\frac{\left(8\text{in}\right){\left(8\text{in}\right)}^2}{2}\\ =\frac{\left(8\text{in}\right)\left(64{\text{in}}^2\right)}{2}\\ =\frac{512{\text{in}}^3}{2}\\ =256{\text{in}}^3\end{array}$

Substitute $256{\text{in}}^3$ for $I_u$ and $h\text{in}$ for $h$ in Equation (V).

$\begin{array}{c}I=0.707\left(h\text{in}\right)\left(256{\text{in}}^3\right)\\ =180.992h{\text{in}}^4\end{array}$

Substitute $10\text{kips}$ for $F$ and $10\text{in}$ for $l$ in Equation (VI).

$\begin{array}{c}M=\left(10\text{kips}\right)\left(10\text{in}\right)\\ =\left(10\text{kips}\right)\left(\frac{1000\text{lbf}}{1\text{kips}}\right)\left(10\text{in}\right)\\ =1\times {10}^5\text{lbf}\cdot \text{in}\end{array}$

Substitute $1\times {10}^5\text{lbf}\cdot \text{in}$ for $M$, $4\text{in}$ for $r_y$ and $181h{\text{in}}^4$ for $I$ in Equation (VII).

$\begin{array}{c}\tau =\frac{\left(1\times {10}^5\text{lbf}\cdot \text{in}\right)\left(4\text{in}\right)}{181h{\text{in}}^4}\\ =\frac{4\times {10}^5\text{lbf}\cdot {\text{in}}^2}{181h{\text{in}}^4}\\ =\frac{2209.94}{h}\text{lbf}/{\text{in}}^2\end{array}$ (XXII)

Substitute $\frac{884.016}{h}\text{lbf}/{\text{in}}^2$ for $\sigma$ and $\frac{2209.94}{h}\text{lbf}/{\text{in}}^2$ for $\tau$ in Equation (VIII).

$\begin{array}{c}{\tau }_{\text{max}}=\sqrt{{\left(\frac{884.016}{h}\text{lbf}/{\text{in}}^2\right)}^2+{\left(\frac{2209.94}{h}\text{lbf}/{\text{in}}^2\right)}^2}\\ =\sqrt{\frac{781484.28}{h^2}{\left(\text{lbf}/{\text{in}}^2\right)}^2+\frac{4883834.804}{h^2}{\left(\text{lbf}/{\text{in}}^2\right)}^2}\\ =\sqrt{\frac{5665319.084}{h^2}{\left(\text{lbf}/{\text{in}}^2\right)}^2}\\ =\frac{2380.1930}{h}\text{lbf}/{\text{in}}^2\end{array}$ (XXIII).

Substitute $\frac{2380.1930}{h}\text{lbf}/{\text{in}}^2$ for ${\tau }_{\text{max}}$ and $12.8\text{kip}/{\text{in}}^2$ for ${\tau }_{\text{all}}$ in Equation (IX).

$\begin{array}{l}\frac{2380.1930}{h}\text{lbf}/{\text{in}}^2=12.8\text{kip}/{\text{in}}^2\\ \frac{2380.1930}{h}\text{lbf}/{\text{in}}^2=\left(12.8\text{kip}/{\text{in}}^2\right)\left(\frac{1000\text{lbf}}{1\text{kip}}\right)\\ \frac{2380.1930}{h}\text{lbf}/{\text{in}}^2=12800\text{lbf}/{\text{in}}^2\\ h=\frac{2380.1930\text{lbf}/{\text{in}}^2}{12800\text{lbf}/{\text{in}}^2}\end{array}$

$h=0.1859\text{in}$

Substitute $0.1859\text{in}$ for $h$ in Equation (XXI).

$\begin{array}{c}\sigma =\frac{884.016}{0.1859}\text{lbf}/{\text{in}}^2\\ =4755.33\text{lbf}/{\text{in}}^2\end{array}$

Thus, the value of primary shear stress is $4755.33\text{lbf}/{\text{in}}^2$

Substitute $0.1859\text{in}$ for $h$ in Equation (XXII).

$\begin{array}{c}\tau =\frac{2209.94}{0.859}\text{lbf}/{\text{in}}^2\\ =2572.68\text{lbf}/{\text{in}}^2\end{array}$

Thus, the value of secondary shear stress is $2572.68\text{lbf}/{\text{in}}^2$.

Substitute $0.1859\text{in}$ for $h$ in Equation (XXIII).

$\begin{array}{c}{\tau }_{\text{max}}=\frac{2380.1930}{0.859}\text{lbf}/{\text{in}}^2\\ =2770.88\text{lbf}/{\text{in}}^2\end{array}$

Thus, the value of maximum shear stress is $2770.88\text{lbf}/{\text{in}}^2$.

Substitute $256{\text{in}}^3$ for $I_u$, $0.1859\text{in}$ for $h$ and $16\text{in}$ for $l$ in Equation (X).

$\begin{array}{c}fom=\frac{256{\text{in}}^3}{\left(0.1859\text{in}\right)\left(16\text{in}\right)}\\ =\frac{256{\text{in}}^3}{2.9744{\text{in}}^2}\\ =86.067\text{in}\end{array}$

Thus, the figure of merit of weld is $86.067\text{in}$.

Substitute $180.992h{\text{in}}^4$ for $I$ and $0.1859\text{in}$ for $h$ and $16\text{in}$ for $l$ in Equation (XI).

$\begin{array}{c}eff=\frac{180.992\times 0.1859{\text{in}}^4}{\left(\frac{{\left(0.1859\text{in}\right)}^2}{2}\right)\left(16\text{in}\right)}\\ =\frac{33.6464{\text{in}}^4}{0.2764{\text{in}}^3}\\ =121.7\text{in}\end{array}$

Thus, the effectiveness of weld is $121.7\text{in}$.

Substitute $8\text{in}$ for $b$, $2\text{in}$ for $b_1$ and $h_I\text{in}$ for $h$ in Equation (X).

$\begin{array}{c}A=1.414\left(h_I\text{in}\right)\left(8\text{in}-2\text{in}\right)\\ =1.414\left(h_I\text{in}\right)\left(6\text{in}\right)\\ =8.484h_I{\text{in}}^2\end{array}$

Substitute $8\text{in}$ for $b$, $2\text{in}$ for $b_1$ and $8\text{in}$ for $d$ in Equation (XI).

$\begin{array}{c}l=\left(8\text{in}\right)+\left(8\text{in}\right)-2\left(2\text{in}\right)\\ =16\text{in}-4\text{in}\\ =12\text{in}\end{array}$

Substitute $10\text{kips}$ for $V$ and $8.484h_I{\text{in}}^2$ for  $A$ in Equation (XII).

$\begin{array}{c}{\sigma }_I=\frac{10\text{kips}}{8.484h_I{\text{in}}^2}\\ =\frac{\left(10\text{kips}\right)\left(\frac{1000\text{lbf}}{1\text{kips}}\right)}{8.484h_I{\text{in}}^2}\\ =\frac{10000\text{lbf}}{8.484h_I{\text{in}}^2}\\ =\frac{1178.68}{h_I}\text{lbf}/{\text{in}}^2\end{array}$                                                                    (XXIV)

Substitute $8\text{in}$ for $b$, $2\text{in}$ for $b_1$ and $8\text{in}$ for $d$ in Equation (XIII).

$\begin{array}{c}{I}_u=\frac{\left(8\text{in}-2\text{in}\right){\left(8\text{in}\right)}^2}{2}\\ =\frac{\left(6\text{in}\right)\left(64{\text{in}}^2\right)}{2}\\ =\frac{384{\text{in}}^3}{2}\\ =192{\text{in}}^3\end{array}$

Substitute $192{\text{in}}^3$ for $I_u$ and $h_I\text{in}$ for $h$ in Equation (XIV).

$\begin{array}{c}I=0.707\left(h_I\text{in}\right)\left(192{\text{in}}^3\right)\\ =135.744h_I{\text{in}}^4\end{array}$

Substitute $10\text{kips}$ for $F$ and $10\text{in}$ for $l$ in Equation (XV).

$\begin{array}{c}M=\left(10\text{kips}\right)\left(10\text{in}\right)\\ =\left(10\text{kips}\right)\left(\frac{1000\text{lbf}}{1\text{kips}}\right)\left(10\text{in}\right)\\ =1\times {10}^5\text{lbf}\cdot \text{in}\end{array}$

Substitute $1\times {10}^5\text{lbf}\cdot \text{in}$ for $M$, $4\text{in}$ for $r_y$ and $135.744h_I{\text{in}}^4$ for $I$ in Equation (XVI).

$\begin{array}{c}{\tau }_I=\frac{\left(1\times {10}^5\text{lbf}\cdot \text{in}\right)\left(4\text{in}\right)}{135.744h_I{\text{in}}^4}\\ =\frac{4\times {10}^5\text{lbf}\cdot {\text{in}}^2}{135.744h_I{\text{in}}^4}\\ =\frac{2946.72}{h_I}\text{lbf}/{\text{in}}^2\end{array}$ . (XXV)

Substitute $\frac{1178.68}{h_I}\text{lbf}/{\text{in}}^2$ for ${\tau }^{\prime }$ and $\frac{2946.72}{h_I}\text{lbf}/{\text{in}}^2$ for ${\tau }^{″}$ in Equation (XVII).

$\begin{array}{c}{\tau }_{\text{maxI}}=\sqrt{{\left(\frac{1178.68}{h_I}\text{lbf}/{\text{in}}^2\right)}^2+{\left(\frac{2946.72}{h_I}\text{lbf}/{\text{in}}^2\right)}^2}\\ =\sqrt{\frac{1389286.542}{h_I{}^2}{\left(\text{lbf}/{\text{in}}^2\right)}^2+\frac{8683158.758}{h_I{}^2}{\left(\text{lbf}/{\text{in}}^2\right)}^2}\\ =\sqrt{\frac{10072445.3}{h_I{}^2}{\left(\text{lbf}/{\text{in}}^2\right)}^2}\\ =\frac{3173.71}{h_I}\text{lbf}/{\text{in}}^2\end{array}$ (XXV).

Substitute $\frac{3173.71}{h_I}\text{lbf}/{\text{in}}^2$ for ${\tau }_{\text{maxI}}$ and $12.8\text{kip}/{\text{in}}^2$ for ${\tau }_{\text{all}}$ in Equation (XVIII).

$\begin{array}{l}\frac{3173.71}{h_I}\text{lbf}/{\text{in}}^2=12.8\text{kip}/{\text{in}}^2\\ \frac{3173.71}{h_I}\text{lbf}/{\text{in}}^2=\left(12.8\text{kip}/{\text{in}}^2\right)\left(\frac{1000\text{lbf}}{1\text{kip}}\right)\\ \frac{3173.71}{h_I}\text{lbf}/{\text{in}}^2=12800\text{lbf}/{\text{in}}^2\\ h_I=\frac{3173.71\text{lbf}/{\text{in}}^2}{12800\text{lbf}/{\text{in}}^2}\end{array}$

$h_I=0.2479\text{in}$

Substitute $0.2479\text{in}$ for $h_I$ in Equation (XXIV).

$\begin{array}{c}{\sigma }_I=\frac{1178.68}{0.2479}\\ =4754.65\text{lbf}/{\text{in}}^2\end{array}$

Thus, the primary shear stress for interrupted weld is $4754.65\text{lbf}/{\text{in}}^2$.

Substitute $0.2479\text{in}$ for $h_I$ in Equation (XXV).

$\begin{array}{c}{\tau }_I=\frac{2946.72}{0.2479}\\ =11886.72\text{lbf}/{\text{in}}^2\end{array}$

Thus, the secondary shear stress for interrupted weld is $11886.72\text{lbf}/{\text{in}}^2$.

Substitute $0.2479\text{in}$ for $h_I$ in Equation (XXV).

$\begin{array}{c}{\tau }_{\max I}=\frac{3173.71}{0.2479}\\ =12802.37\text{lbf}/{\text{in}}^2\end{array}$

Thus, the maximum shear stress for interrupted weld is $12802.37\text{lbf}/{\text{in}}^2$.

Substitute $192{\text{in}}^3$ for $I_u$, $0.2479\text{in}$ for $h$ and $12\text{in}$ for $l$ in Equation (XIX).

$\begin{array}{c}fo{m}_I=\frac{192{\text{in}}^3}{\left(0.2479\text{in}\right)\left(12\text{in}\right)}\\ =\frac{192{\text{in}}^3}{2.9748{\text{in}}^2}\\ =64.542\text{in}\end{array}$

Thus. the figure of merit of interrupted weld is $64.542\text{in}$.

Substitute $135.744h_I{\text{in}}^4$ for $I_I$ and $0.2479\text{in}$ for $h_I$ and $12\text{in}$ for $l$ in Equation (XX).

$\begin{array}{c}ef{f}_I=\frac{135.744\times 0.2479{\text{in}}^4}{\left(\frac{{\left(0.2479\text{in}\right)}^2}{2}\right)\left(12\text{in}\right)}\\ =\frac{33.6509{\text{in}}^4}{0.3687{\text{in}}^3}\\ =91.26\text{in}\end{array}$

Thus, the effectiveness of interrupted weld is $91.26\text{in}$.