Question

A horizontal aluminum rod 4.8 cm in diameter projects 5.3 cm from

a wall. A 1200 kg object is suspended from the end of the rod. The
shear modulus of aluminum is 3.0 x 10
10
N/m
2
. Neglecting the
rod’s mass, find (a) the shear stress on the rod and (b) the vertical
deflection of the end of the rod.

Answer 1

To find the shear stress and vertical deflection of the rod, you can use specific formulas and given values. The shear stress can be calculated using the weight of the object and the cross-sectional area of the rod. The vertical deflection can be calculated using the shear force, length of the rod, shear modulus of aluminum, and the cross-sectional area of the rod.

Step-by-step explanation:

To find the shear stress on the rod, we can use the formula:

Shear Stress = Shear Force / Cross-sectional Area

Since the rod is horizontal and supports a 1200 kg object, the shear force is simply the weight of the object, which can be calculated as: Weight = Mass x Gravity.

The cross-sectional area of the rod is given by the formula for the area of a circle: Area = πr^2, where r is the radius of the rod.

Using the given values, we can calculate the shear stress on the rod.

To find the vertical deflection of the end of the rod, we can use the formula:

Vertical Deflection = (Shear Force x Length) / (Shear Modulus x Cross-sectional Area)

Again, using the given values, we can calculate the vertical deflection of the end of the rod.

Answer 2

The shear stress on the aluminum rod is calculated by dividing the force exerted by the suspended object by the rod's cross-sectional area, and the vertical deflection is determined using the shear modulus of aluminum, the force, length of the rod, and its cross-sectional area.

Step-by-step explanation:

To determine the shear stress on the rod and the vertical deflection of its end, we can use two formulas. Shear stress (τ) can be calculated as the force applied divided by the cross-sectional area (A) of the rod. The vertical deflection (\(\Delta y\)) due to the shear force can be found using the formula \(\Delta y = \frac{F \cdot L}{A \cdot G}\) where F is the force, L is the length of the rod, G is the shear modulus, and A is the cross-sectional area.

The cross-sectional area of a rod with a diameter of 4.8 cm (which is 0.048 m) is given as \(A = \pi \cdot (\frac{d}{2})^2\). Inserting the diameter gives \(A = \pi \cdot (0.024 m)^2\).

Using the weight of the object (W), which is the force due to gravity, we have \(W = m \cdot g\) where m is the mass of the object and g is the acceleration due to gravity (\(\approx 9.81 m/s^2\)). So the force from the 1200 kg object is \(F = 1200 kg \cdot 9.81 m/s^2\).

(a) To find the shear stress, \(τ = \frac{F}{A}\).

(b) And to calculate the vertical deflection of the rod, use \(\Delta y = \frac{F \cdot L}{A \cdot G}\).