Question

Problem 2: A circular shaft transmits power as shown with pulley loads. The shaft carries a torque, bending, shear and axial loads. Draw LVM diagram to find Mmax and Vmax. Show all loads (moments and forces) on the circular x-section of the shaft below. Use double arrows for moments. Compute shear and normal stresses and show them on the same section. Create stress elements for points A and B of the section. Combine the stresses and compute Cuau (Tresca) and on (von Mises) for the stress elements. Zero credit if no values shown on the section below. (Work in Nand morn) 3.2 KN 4.2 kN 0.30 m -0.15 m -0.15 m B+ с 2 kN B 2.5 kNm 25 mm dia.

Answer 1

Given

The loading diagram

3.2 KN -0.15 m + 4.2 kN -0.15 m 0.30 m R 2 KN Ax D с B 2.5 kNm Roy R "AY 25 mm dia.

Taking moment balance about A in the z-direction

Taking force balance in the y-direction

The force balance in the x-direction

Shear force diagram

3.45 kN 0.25 kN А B с D -3.95 kN

Maximum shear force is -3.95 kN at section CD

Bending moment diagram

0.5175 kNm 0.5925 kNm A B с

Maximum bending moment is 0.5925 kNm at C

The torque diagram

2.5 kNm A B с D

The maximum torque also at C

At this section

Y -C BU Z D

Shear stress

$\\ \tau=\frac{16T_{max}}{\pi d^3}=814.87\ MPa$

Bending stress(-ve means compressive stress)

$\\ \sigma_{bending}=-\frac{64M_{max}y}{\pi d^4}$

Normal stress due to axial force

$\\ \sigma_{axial}=-\frac{2*1000}{\pi d^2/4}=-4.0744\ MPa$

Normal stress

$\\ \sigma=\sigma_{bending}+\sigma_{axial} \\ \sigma=- \frac{64M_{max}y}{\pi d^4}-4.0744$

At point A

$\\ \sigma_A=-4.0744-\frac{64M_{max}d/2}{\pi d^4}= -390.32\ MPa$

$\\ \tau_A=814.87\ MPa$

The stress element

814.87 MPa х ſz -390.32 MPa

Tresca shear stress

$\\ \tau_{max,A}=\sqrt{(\sigma_A/2)^2+\tau_A^2}=837.92\ MPa$

Von mises stress

$\\ \sigma_{m,A}=\frac{1}{2}\sqrt{(\sigma_A-0)^2+(0-\sigma_A)^2+6\tau_A^2}= 1035.5\ MPa$

At point B

$\\ \sigma_B=-4.0744-\frac{64M_{max}*0}{\pi d^4}=-4.0744\ MPa$

$\\ \tau_B=814.87\ MPa$

The stress element

814.87 MPa у 4.0744 MPa Z

Tresca shear stress

$\\ \tau_{max,B}=\sqrt{(\sigma_B/2)^2+\tau_B^2}= 814.88\ MPa$

Von mises stress

$\\ \sigma_{m,B}=\frac{1}{2}\sqrt{(\sigma_B-0)^2+(0-\sigma_B)^2+6\tau_B^2}= 998.02\ MPa$