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

are their multiple choices ?

explanation:

weight of the box is given as

here this is placed on the inclined of angle 42 degree

now this weight will have two components one is along the incline other is perpendicular to it

so here we need to apply a force parallel to incline opposite to its component of weight

so here the component of weight is given as

here we have

so it requires 75 n force to move it upwards