Question

The drive shaft of a dump truck is constructed of 3.5” OD x 1/4” wall round tubing made of ASTM A242 steel. The driveshaft is 65” long and is subjected to 500 lbf*ft of torque.

• What is the maximum shear stress in the drive shaft?

• What is the safety factor with respect to yielding?

• How much torque would result in a 1° twist of the drive shaft?

Answer 1

Solution:-

Given- Length L = 65inch,

Torque T = 500 lbf x ft = 500 x 12 lbf - inch

Outer Diameter D = 3.5 inch

Maximum Radius R = 3.5/2 = 1.75 inch

internal Diameter d = 3.5 - (2x1/4) = 3 inch

material is ASTM A242 steel.

i) By the equarion of torsion :-

T/J = / R

= TR/J

where J is polar MOI, so rhat J = (D⁴-d⁴)/32

now putting all the values, we have

J = ( 3.5⁴ - 3⁴)/32

J = 6.78 inch⁴

^{so that Maximum shear stress}

^{= 500 x12x 1.75/6.78}

= 1548.67 lbf/inch²

^{ii) Factor of safety FS = yield stress/working stress}

Yield stress for shear of ASTM A242 steel is = 50 ksi =50 x 10³ lbf/inch²

FS = 50x 10³/1548.67 =32.28

iii) for 1^o twist, Torque will be,

for ASTM A242steel G = 11 x10⁶ lbf/inch²

and = 1^o = 1 x /180 = 0.0174

T/J = G/L

T = GJ/L

T = 11 x 10⁶ x 0.0174 x6. 78/65

T = 20930.51 lbf - inch

so that,

Torque T at 1^o = 20930.51 lbf - in