Torsion Sample Questions

Topics: Shear stress, Force, Torque Pages: 5 (811 words) Published: June 24, 2013
THE UNIVERSITY OF ZAMBIA
SCHOOL OF ENGINEERING
DEPARTMENT OF MECHANICAL ENGINEERING
ME 332 STRENGTH OF MATERIALS I

SAMPLE PROBLEMS-TORSION OF CIRCULAR SHAFTS AND THIN WALLED SECTIONS

QUESTION 1

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QUESTION 2

………………………………………………………………………………………………………………………………………………………………….

QUESTION 4

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Q6.The drive shaft for a truck (outer diameter 60 mm and inner diameter 40 mm) is running at 2500 rpm (see figure Q6). (a) If the shaft transmits 150 kW, what is the maximum shear stress in the shaft? (b) If the allowable shear stress is 30 MPa, what is the maximum power that can be transmitted?

Figure Q6.
SOLUTION:
Drive shaft for a truck

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Q7.A tubular shaft being designed for use on a construction site must transmit 120 kW at 1.75 Hz. The inside diameter of the shaft is to be one-half of the outside diameter. If the allowable shear stress in the shaft is 45 MPa, what is the minimum required outside diameter d

SOLUTION
Tubular shaft

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Q8.The shaft ABC shown in the figure is driven by a motor that delivers 300 kW at a rotational speed of 32 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are L1 =1.5 m and L2 = 0.9 m. Determine the required diameter d of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist between points A and C is 4.0°, and G =75 GPa.

SOLUTION:

Motor-driven shaft

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Q9.Calculate the shear stress τ and the angle of twist ɸ (in degrees) for a steel tube (G =76 GPa) having the cross section shown in figure Q9. The tube has length L = 1.5 m and is subjected to a torque T =10 KN m.

Figure Q9

SOLUTION
Steel tube

Q10.A thin tubular shaft of circular cross section(see figure Q10) with inside diameter 100 mm is subjected to a torqueof 5000 N m.If the allowable shear stress is 42 MPa, determine the requiredwall thickness t by using (a) The approximate theory for a thin-walledtube, and

(b) The exact torsion theory for a circular bar.

Figure Q10

SOLUTION:
Thin tube

←Q10
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Q11.An aluminum tube, 1.2 m long, has the semicircular cross section shown in the figure. If stress concentrations at the corners are neglected, determine (a) The torque that causes a maximum shear stress of 40 MPa; (b) The corresponding angle of twist of the tube. Use G = 28 GPa for aluminum.

Q12.

SOLUTION

Since the shaft consists of three portions AB, BC, and CD, each of uniform cross section and each with a constant internal torque, the following equation may be used. ɸ=TiLiJiGi……………. (Eq Q1.12)

Angle of Twist.

Using Eq. (Q1.12) and recalling that G 5 =77 GPa for the entire shaft, we have

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Q13.A motor drives a circular shaft through a set of gears at 630 rpm; 20kW are delivered to a machine on the right and 60kW on the left (Figure Q13). If the allowable shear stress of the shaft is 37 MPa, determine the minimum diameter of the shaft.

Figure 13

SOLUTION
The torsion problem is statically determinate. The torques acting on the shaft have to be calculated first by Equation (Q 13). The shaft is then analyzed by Equation (Q13) to determinethe maximum shear stress that depends on the diameter. A comparison of the maximum shearstress with the allowable stress of the material leads to the solution of the minimum diameter.

Torsion of rotating shafts
Members as rotation shafts for transmitting power are usually subjected to torque. The followingformula is used for the conversion of kilowatts (kW), a common unit used in the industry, intotorque applied on a shaft: T=159Pf ……….. (Eq. Q13)

Or
T=9540PN

Where

Where...
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