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

Click here to show or hide the solution

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

Tags:

Add new comment
132627 reads

More Reviewers

Algebra	Engineering Mechanics
Spherical Trigonometry	Strength of Materials
Plane Trigonometry	Structural Analysis
Plane Geometry	General Engineering
Solid Geometry	Derivation of Formulas
Analytic Geometry	Fluid Mechanics and Hydraulics
Integral Calculus	Geotechnical Engineering
Differential Calculus	Reinforced Concrete Design
Elementary Differential Equations	Surveying and Transportation Engineering
Advance Engineering Mathematics	Timber Design
Engineering Economy