A5-cm diameter solid shaft is simultaneously subjected to an axial load of 80 kn and a torque of 400 nm.
a) determine the principal stresses at the surface assuming elastic behavior.
b) find the largest shear stress.

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 ±      ..........1

principal stress σ = 40.74 /2 ±  

principal stress σ = 46.46 MPa and -5.72 MPa

and

maximum shear stress is

maximum shear stress =      ..........2

maximum shear stress =    

maximum shear stress = 26.08 MPa

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

                 = 40.74 MPa

shear stress

                   = 16.3 MPa

principle stress is given as

               =46.46 MPa, -5.72 MPa

b) maximu shear stress

               = 26.08 MPa

this a false

explanation:

measure their densities same density, same sub