Chapter 9, Problem 1P

9–1 to 9–4 The figure shows a horizontal steel bar of thickness h loaded in steady tension and welded to a vertical support. Find the load F that will cause an allowable shear stress, τ_allow, in the throats of the welds.

Problem Number	b	d	h	τallow
9–1	50 mm	50 mm	5 mm	140 Mpa
9–2	2 in	2 in	$\frac{5}{16}$ in	25 kspi
9–3	50 mm	30 mm	5 mm	140 Mpa
9–4	4 in	2 in	$\frac{5}{16}$ in	25 kpsi

To determine

The load $F$ that will cause an allowable shear stress,${\tau }_{\text{allow}}$ in the throat of the weld.

Answer to Problem 1P

The load $F$ is $49.5\text{kN}$.

Explanation of Solution

Write the expression for the length of the weld.

$l=2b$ (I)

Here, the width of weld is $b$.

Write the expression for load

$F=0.707hl{\tau }_{\text{allow}}$ (II)

Here, the height of weld is $h$ and allowable shear stress is ${\tau }_{\text{allow}}$.

Conclusion:

Substitute $50\text{mm}$ for $b$ in Equation (I).

$\begin{array}{c}l=2\left(50\text{mm}\right)\\ \text{=100}\text{mm}\end{array}$

Substitute $5\text{mm}$ for $h$, $100\text{mm}$ for $l$ and $140\text{MPa}$ for ${\tau }_{\text{allow}}$ in Equation (II).$\begin{array}{c}F=0.707\left(5\text{mm}\right)\left(100\text{mm}\right)\left(140\text{MPa}\right)\\ =\left(49490\text{MN}\right)\times \left(\frac{1\text{kN}}{1000\text{MPa}}\right)\\ =49.49\text{kN}\\ \approx 49.5\text{kN}\end{array}$

Thus, the load is $49.5\text{kN}$.