Solution to Problem 123 Shear Stress | Strength of Materials Review at MATHalino

Problem 123
A rectangular piece of wood, 50 mm by 100 mm in cross section, is used as a compression block shown in Fig. P-123. Determine the axial force P that can be safely applied to the block if the compressive stress in wood is limited to 20 MN/m² and the shearing stress parallel to the grain is limited to 5MN/m². The grain makes an angle of 20° with the horizontal, as shown. (Hint: Use the results in Problem 122.)

Solution 123

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Based on maximum compressive stress:

Normal force:
$N = P \cos 20^{\circ}$

Normal area:
$A_N = 50 (100 \sec 20^\circ)$

$A_N = 5320.89 \, \text{mm}^2$

$N = \sigma A_N$

$P \cos 20^\circ = 20 (5320.89)$

$P = 113\,247 \, \text{N}$

$P = 133.25 \, \text{kN}$

Based on maximum shearing stress:

Shear force:
$V = P \sin 20^\circ$

Shear area:
$A_V = A_N$

$A_V = 5320.89 \, \text{mm}^2$

$V = \tau A_V$

$P \sin 20^\circ = 5 (5320.89)$

$P = 77\,786 \, \text{N}$

$P = 77.79 \, \text{kN}$

For safe compressive force use
$P = 77.79 \, \text{ kN}$ answer

Good day po, ask ko lang sana

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