A shear force of 3.1 105 N is applied to an aluminum bar of length L = 19 cm, width w = 5.2 cm, and height h = 2.3 cm as shown in the figure below. What is the shear deformation Äx?____m
To find the shear deformation (Δx) of the aluminum bar, we can use the formula:
Δx = (F * L) / (G * A)
where:
Δx is the shear deformation
F is the shear force applied
L is the length of the bar
G is the shear modulus of the material
A is the cross-sectional area of the bar
First, let's calculate the cross-sectional area:
A = w * h
A = 5.2 cm * 2.3 cm
Now, let's convert the dimensions to meters since the result will be in meters:
A = 5.2 cm * 2.3 cm
A = 11.96 cm²
A = 11.96 cm² * (1 m / 100 cm)²
A = 11.96 * (0.01 m / 1 cm)²
A = 11.96 * 0.0001 m²
A = 0.001196 m²
Next, we need to know the shear modulus of aluminum. The shear modulus (G) is a material property and can be found through reference materials or online resources. For aluminum, the typical value of the shear modulus is around 25 GPa (gigapascals) or 25 x 10^9 Pa.
Now, we have all the values for the formula:
F = 3.1 * 10^5 N
L = 19 cm = 0.19 m
G = 25 * 10^9 Pa
A = 0.001196 m²
Δx = (F * L) / (G * A)
Δx = (3.1 * 10^5 N * 0.19 m) / (25 * 10^9 Pa * 0.001196 m²)
Simplifying the equation:
Δx = 5.89 * 10^-7 m
Therefore, the shear deformation (Δx) of the aluminum bar is approximately 5.89 * 10^-7 meters.