STRUCTURAL MECHANICS: CE203

Chapter 5

Torsion

Notes are based on Mechanics of Materials: by R. C. Hibbeler, 7th Edition, Pearson

Dr B. Achour & Dr Eng. K. El-kashif

Civil Engineering Department, University of Hail, KSA

(Spring 2011)

Chapter 5: Torsion

Chapter 5: Torsion

Torsional Deformation of a Circular ShaftTorsional Deformation of a Circular Shaft

Torque is a moment that twists a member about its longitudinal axis.

If the angle of rotation is small, the length of the shaft and its radius will remain unchanged.

Chapter 5: Torsion

The Torsion FormulaThe Torsion Formula

When material is linear-elastic, Hooke’s law applies. A linear variation in shear strain leads to a

corresponding linear variation in shear stress along any radial line on the cross section.

Chapter 5: Torsion

The Torsion FormulaThe Torsion Formula

If the shaft has a solid circular cross section,

If a shaft has a tubular cross section,

Chapter 5: Torsion

Example 5.2Example 5.2

The solid shaft of radius c is subjected to a torque T. Find the fraction of T that is resisted by the material contained within the outer region of the shaft, which has an inner radius of c/2 and outer radius c.Solution:Stress in the shaft varies linearly, thus

The torque on the ring (area) located within the lighter-shaded region is

For the entire lighter-shaded area the torque is

Chapter 5: Torsion

Solution:Solution: Using the torsion formula to determine the maximum stress in

the shaft, we have

Substituting this into Eq. 1 yields

Chapter 5: Torsion

Example 5.3Example 5.3 The shaft is supported by two bearings and is subjected to three

torques. Determine the shear stress developed at points A and B, located at section a–a of the shaft.

Solution:From the free-body diagram of the left segment,

The polar moment of inertia for the shaft is

Since point A is at ρ = c = 75 mm,

Likewise for point B, at ρ =15 mm, we have

Chapter 5: Torsion

Power TransmissionPower Transmission

Power is defined as the work performed per unit of time.For a rotating shaft with a torque, the power is

Since , the power equation is

For shaft design, the design or geometric parameter is

Chapter 5: Torsion

Example 5.5Example 5.5 A solid steel shaft AB is to be used to transmit 3750 W from the motor M

to which it is attached. If the shaft rotates at w =175 rpm and the steel has an allowable shear stress of allow τallow =100 MPa, determine the required diameter of the shaft to the nearest mm.Solution:The torque on the shaft is

Since

As 2c = 21.84 mm, select a shaft having a diameter of 22 mm.

Chapter 5: Torsion

Angle of TwistAngle of Twist

Integrating over the entire length L of the shaft, we have

Assume material is homogeneous, G is constant, thus

Sign convention is determined by right hand rule,

Φ = angle of twistT(x) = internal torqueJ(x) = shaft’s polar moment of inertia G = shear modulus of elasticity for the material

Chapter 5: Torsion

Example 5.8Example 5.8 The two solid steel shafts are coupled together using the meshed gears.

Determine the angle of twist of end A of shaft AB when the torque 45 Nm is applied. Take G to be 80 GPa. Shaft AB is free to rotate within bearings E and F, whereas shaft DC is fixed at D. Each shaft has a diameter of 20 mm.Solution:From free body diagram,

Angle of twist at C is

Since the gears at the end of the shaft are in mesh,

Chapter 5: Torsion

Solution:Solution: Since the angle of twist of end A with respect to end B of shaft AB

caused by the torque 45 Nm,

The rotation of end A is therefore

Chapter 5: Torsion

Example 5.10Example 5.10 The tapered shaft is made of a material having a shear modulus

G. Determine the angle of twist of its end B when subjected to the torque.Solution:From free body diagram, the internal torque is T.

Thus, at x,

For angle of twist,

Chapter 5: Torsion

Example 5.11Example 5.11 The solid steel shaft has a diameter of 20 mm. If it is subjected to

the two torques, determine the reactions at the fixed supports A and B.Solution:By inspection of the free-body diagram,

Since the ends of the shaft are fixed,

Using the sign convention,

Solving Eqs. 1 and 2 yields TA = -345 Nm and TB = 645 Nm.

Chapter 5: Torsion

Solid Noncircular ShaftsSolid Noncircular Shafts

The maximum shear stress and the angle of twist for solid noncircular shafts are tabulated as below:

Chapter 5: Torsion

Example 5.13Example 5.13The 6061-T6 aluminum shaft has a cross-sectional area in the shape of

an equilateral triangle. Determine the largest torque T that can be applied to the end of the shaft if the allowable shear stress is τallow = 56 MPa and the angle of twist at its end is restricted to Φallow = 0.02 rad. How much torque can be applied to a shaft of circular cross section made from the same amount of material? Gal = 26 GPa.Solution:By inspection, the resultant internal torque at any cross section along the shaft’s axis is also T.

By comparison, the torque is limited due to the angle of twist.

Chapter 5: Torsion

Solution:Solution: For circular cross section,

we have

The limitations of stress and angle of twist then require

Again, the angle of twist limits the applied torque.

Chapter 5: Torsion

Thin-Walled Tubes Having Closed Cross SectionsThin-Walled Tubes Having Closed Cross Sections

Shear flow q is the product of the tube’s thickness and the average shear stress.

Average shear stress for thin-walled tubes is

For angle of twist,

= average shear stressT = resultant internal torque at the cross sectiont = thickness of the tubeAm = mean area enclosed boundary

Chapter 5: Torsion

Example 5.14Example 5.14 Calculate the average shear stress in a thin-walled tube having a

circular cross section of mean radius rm and thickness t, which is subjected to a torque T. Also, what is the relative angle of twist if the tube has a length L?Solution:

The mean area for the tube is

For angle of twist,

Chapter 5: Torsion

Example 5.16Example 5.16 A square aluminum tube has the dimensions. Determine the

average shear stress in the tube at point A if it is subjected to a torque of 85 Nm. Also compute the angle of twist due to this loading. Take Gal = 26 GPa.Solution:

By inspection, the internal resultant torque is T = 85 Nm.

The shaded area is

For average shear stress,

Chapter 5: Torsion

Solution:Solution: For angle of

twist,

Integral represents the length around the centreline boundary of the tube, thus

Chapter 5: Torsion

Stress ConcentrationStress Concentration

Torsional stress concentration factor, K, is used to simplify complex stress analysis.

The maximum shear stress is then determined from the equation

Chapter 5: Torsion

Example 5.18Example 5.18

The stepped shaft is supported by bearings at A and B. Determine the maximum stress in the shaft due to the applied torques. The fillet at the junction of each shaft has a radius of r = 6 mm.

Solution:By inspection, moment equilibrium about the axis of the shaft is satisfied

The stress-concentration factor can be determined by the graph using the geometry,

Thus, K = 1.3 and maximum shear stress is

Chapter 5: Torsion

Inelastic TorsionInelastic Torsion

Considering the shear stress acting on an element of area dA located a distance p from the center of the shaft,

Shear–strain distribution over a radial line on a shaft is always linear.

Perfectly plastic assumes the shaft will continue to twist with no increase in torque.

It is called plastic torque.

Chapter 5: Torsion

Example 5.20Example 5.20 A solid circular shaft has a radius of 20 mm and length of 1.5 m. The

material has an elastic–plastic diagram as shown. Determine the torque needed to twist the shaft Φ = 0.6 rad.

Solution:The maximum shear strain occurs at the surface of the shaft,

The radius of the elastic core can be obtained by

Based on the shear–strain distribution, we have