A stepped brass shaft (G = 40 GPa) ABCD shown in Fig. 4 is fixed in both ends an

Question

A stepped brass shaft (G = 40 GPa) ABCD shown in Fig. 4 is fixed in both ends and loaded by torques T1 - 400 Nm and T2 = 220 Nm at point B and C respectively. The segment AB and BC have a solid circular cross sections with 60mm and 40mm diameter respectively, and segement CD is a circular tube of outside diameter of d(out) = 50mm and d(in) = 30mm.(A) Determinethe maximum shear stresses in each segment of the shaft. (B) Determine the angle of twist of segment BC.

-d AB-60 mm Problem No. 4 (25%): d-40 mm -din= 30 mm A stepped brass shaft (G-40 GPa) ABCD shown in Fig. 4 is fixed in both ends and loaded by torques T 400 Nm and Tà- 220 Nm at point B and C respectively The segments AB and BC have a solid circular cross sections with 60 mm and 40 mm diameters respectively, and segment CD is a circular tube of outside diameter dout 50 mm and inside diameter d= 30 mm. (a) Determine the maximum shear stresses in each segment of the shaft. (b) Determine angle of twist of the segment BC dout 50 mm T1 -0.8 m 1.2m Fig. 4 Fi

Explanation / Answer

First use this equation to know the max shear stress in AB and CD :

S=Td/J

T is the momentum

d is the max distance

J is (1/4)piR^4

So for AB:

T1=400 Nm

J=(1/4)pi(30x10^-3)^4 = 6.36 x10^-7 m^4

d1=60/2=30 mm= 30x10^-3 m

S=(400)( 30x10^-3)/(6.36 x10^-7) = 1.88 x10^7 Pa= 18.86 MPa

For CD:

T2=220 Nm

d2=50/2=25 mm= 25x10^-3 m

J2=(1/4)pi(R^4) - (1/4)pi(r^4) = 2.67x10^-7 m^4

S=(220)(25x10^-3)/(2.67x10^-7) = 2.06 x10^7 Pa= 20 MPa

angle=TL/JG =

T=T1-T2= 400-220=180 Nm

L=0.8 m

G=40 x10^9 Pa

J=(1/4)piR^4 = (1/4)pi(20x10^-3)^4 = 1.25 x10^-7 m^4

angle=(180)(0.8)/(40 x10^9 )(1.25 x10^-7) = 0.0288 rad

good luck!!

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