The 2-in.-diameter drive shaft ab on the helicopter is subjected to an axial tension of 1 lb and a torque of 300 lb ft. determine the principal stresses and the maximum in-plane shear stress that act at a point on the surface of the shaft:

σ1 = 4.38 ksi

σ2 =-1.2 ksi  

τ =  2.7903 ksi

Explanation:

diameter (d) = 2 inch

radius (r) = 1 inch

axial tension = 10,000 lb

torque (T) = 300 lb.ft

Determine the principal stresses and the maximum in plane shear stress

normal stress (σₙ) = force / area

σx = \frac{10000}{π x r^{2}}

σx = 3.18 ksi

σy = 0

shear stress (τ) = \frac{Tc}{J}

where c =  radius

           J = polar moment of inertia = \frac{πc^{4}}{2}

τ = \frac{T}{ \frac{πc^{3}}{2}}

τ = \frac{300 x 12}{ \frac{π x 1^{3}}{2}} = 2.2918 ksi

now that we have σx, σy and τ we have to apply the formula below to find the principal shear stresses

σ1,2 = \frac{σx + σy}{2} ±

σ1,2 = \frac{3.18 + 0}{2} ±

σ1,2 = 1.5915 ± 2.7903

σ1 = 1.5915 + 2.7903 = 4.38 ksi

σ2 = 1.5915 - 2.7903 = -1.2 ksi  

maximum in plane shear stress (τ) =  

maximum in plane shear stress (τ) =  2.7903 ksi

float.

archimedes' principle

denser than water sinks in water.

denser than mercury would sink in it