Question

a circular bar made of steel is to resist an occasional torque of 2.2 knm acting in transverse plane. if the allowable stresses in compression, tension, and shear are 100 mn/m2 , 35 mn/m2 , and 50 mn/m2 respectively, find:

Answer 1

To find the diameter of the circular bar, we can use the formula for torque and consider the allowable stresses. By rearranging the formula and plugging in the values, the diameter of the bar is approximately 16.14 mm.

To find the diameter of the circular bar, we need to consider the torque it can resist and the allowable stresses in compression, tension, and shear. The torque acting on the bar is given as 2.2 kNm.

We can use the formula for torque,

T = (π/16) × τ × d³,

where T is the torque, τ is the shear stress, and d is the diameter of the bar.

First, we need to convert the torque to Nm by multiplying it by 1000. Then, we can rearrange the formula to solve for the diameter,

d = (16 × T) / (π×τ).

Plugging in the values,

d = (16 × 2.2 × 1000) / (π × 50 × 10⁶).

Calculating this, we get a diameter of approximately 16.14 mm for the bar.

Complete Question:

A circular bar made of steel is to resist an occasional torque of 2.2 knm acting in transverse plane. if the allowable stresses in compression, tension, and shear are 100 mn/m² , 35 mn/m² , and 50 mn/m² respectively, find: The diameter of the bar

Answer 2

The necessary diameter of the circular steel bar to resist the occasional torque of 2.2 kN·m while staying within the allowable shear stress is approximately 0.08915 meters, which is approximately 89.15 millimeters.

To find the necessary dimensions (diameter) of the circular steel bar that can resist an occasional torque of 2.2 kN·m while staying within allowable stresses, you'll need to use the formula for torsional stress in a circular shaft:

`\[T = \frac{{\pi d^3 \tau}}{{16}}\]`

Where:

- T is the torque applied (2.2 kN·m, which we need to convert to N·m),

- d is the diameter of the circular bar (in meters),

-
`$\tau$` is the shear stress (in pascals or N/m²).

Given:

- Allowable stresses in compression, tension, and shear:

- Compression
`(\(\sigma_c\)) = 100 MN/m² (convert to N/m², so \(\sigma_c\) = 100 * 10^6 N/m²)`

- Tension
`(\(\sigma_t\)) = 35 MN/m² (convert to N/m², so \(\sigma_t\) = 35 * 10^6 N/m²)`

- Shear
`(\(\tau\)) = 50 MN/m² (convert to N/m², so \(\tau\) = 50 * 10^6 N/m²)`

Step-by-step calculation:

1. Convert torque from kN·m to N·m:

- 1 kN = 1000 N

- 2.2 kN·m = 2.2 * 1000 N·m = 2200 N·m

2. Rearrange the torsional stress formula to solve for the diameter d:

`\[d = \left(\frac{{16T}}{{\pi\tau}}\right)^(1/3)\]`

3. Plug in the values:

`\[d = \left(\frac{{16 * 2200 N·m}}{{\pi * 50 * 10^6 N/m²}}\right)^(1/3)\]`

4. Calculate the diameter d:

`\[d ≈ 0.08915 \, \text{m}\]`