Question

A 5-cm diameter solid shaft is simultaneously subjected to an axial load of 80 kN and a torque of 400 Nm.

Answer 1

Answer:

principal stress σ = 46.46 MPa and -5.72 MPa

maximum shear stress = 26.08 MPa

Explanation:

given data

diameter = 5 cm = 50 mm

axial load = 80 kN = 80000 N

torque = 400 Nm - 400000 N-mm

to find out

principal stresses and shear stress

solution

we find first axial stress that is force / area

axial stress = 80000 / ( π/4×50² )

axial stress σ = 40.74 MPa

and

shear stress = torque × radius  / area

shear stress = 400000 × 25  / ( π/4×50² )

shear stress τ = 16.3 MPa

so

principal stress will be

principal stress = σ/2 ±  $\sqrt{(\sigma/2^{2}+\tau^{2} )}$    ..........1

principal stress σ = 40.74 /2 ± $\sqrt{(40.74/2^{2}+ 16.3^{2} )}$  

principal stress σ = 46.46 MPa and -5.72 MPa

and

maximum shear stress is

maximum shear stress =  $\sqrt{(\sigma/2^{2}+\tau^{2} )}$    ..........2

maximum shear stress =  $\sqrt{(40.74/2^{2}+ 16.3^{2} )}$  

maximum shear stress = 26.08 MPa

Answer 2

Answer:

principle stress    =46.46 MPa, -5.72 MPa

maximu shear stress 26.08 MPa

Explanation:

d = 5 cm = 50 mm

axial load = 80 kN

torque = 400Nm

we know that

axial stress =  force/area

                  $= \frac{80*10^3}{\frac{\pi}{4} 50^2}$

                 = 40.74 MPa

shear stress$= \frac{ Tr}{J}$

                   $= \frac{400*10^3 *25mm}{\frac{\pi}{32}50^4}$

                   = 16.3 MPa

principle stress is given as

$\sigma_p = \frac{\sigma_a}{2} \pm \sqrt{ (\frac{\sigma_a}{2})^2 +z^2}$

               $=\frac{40.74}{2} \pm \sqrt{ (\frac{40.74}{2})^2 +16.3^2}$

               =46.46 MPa, -5.72 MPa

b) maximu shear stress

$\sigma_p = \sqrt{ (\frac{\sigma_a}{2})^2 +z^2}$

$= \sqrt{ (\frac{40.74}{2})^2 +16.3^2}$

               = 26.08 MPa