One end of the hollow aluminum shaft 120 cm (not 120 cm) long (Figure 2) is fixe

Question

One end of the hollow aluminum shaft 120 cm (not 120 cm) long (Figure 2) is fixed, and the other end is connected to a gear with an outside diameter of d = 46 cm (not 40 cm) as shown. The gear is subjected to a tangential gear force of P = 42,446 N (not 45000 N). The shear modulus of the aluminum is G = 2 x 1010 Pa. What is the maximum shear stress in the shaft? (Use inner and outer shaft diameters provided on Figure 2) {Answer in MPa to 3 significant figures}.

Explanation / Answer

Maximum shear stress can be calculated as

max = torsion + direct

torsion = T r / J

direct = 2*V/A

= shear stress

T = twisting moment

r = distance from center to stressed surface in the given position

J = Polar Moment of Inertia of an Area

V = Direct Shear force

A= Area

Here

T = F*d'/2 = 42446 * 0.23

T = 9762.58 N-m

r = Do/2 = 5 cm = 0.05 m

J = 3.14*(D4o - D4i) / 32  

J = 6.711*10-6 m4

torsion = 9762.58*0.05 / 6.711*10-6

torsion = 72.733*106 Pa = 72.733 MPa

A = 3.14* (Do2 - Di2) / 4 =3.436*10-3 m2

V = 42446 N

direct = 2*42446 / 3.436*10-3 = 24.706 MPa

max = torsion + direct =72.733+ 24.706

max = 97.439 MPa

Related Questions

Navigate

• Browse (All)
• Browse O
• Subjects

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.