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Problem 2: A circular shaft transmits power as shown with pulley loads. The shaft carries a torque, bending, shear and axial
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Answer #1

Given

d=25\ mm

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

\\ R_{Dy}*0.6=3.2*.15+4.2*.45 \\ R_{Dy}=3.95\ kN

Taking force balance in the y-direction

\\ R_{Ay}+R_{Dy}=3.2+4.2=7.4 \\ R_{Ay}=3.45\ kN

The force balance in the x-direction

\\ R_{Ax}=2\ kN

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

\\ M_{max}=0.5925 kNm=592500 \ Nmm

The torque diagram

2.5 kNm A B с D

The maximum torque also at C

\\ T_{max}=2.5 kNm= 2500000 \ Nmm

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

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