Chapter 5 Torsion - libvolume4.xyzlibvolume4.xyz/.../torsionofcircularshaftspresentation1.pdf · Chapter 5: Torsion Example 5.8 The two solid steel shafts are coupled together using

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

TorsionalTorsional Deformation of a Circular ShaftDeformation 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 torque

J(x) = shaft’s polar moment of inertiaG = 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 5.13

The 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

ThinThin--Walled Tubes Having Closed Cross SectionsWalled 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 section

t = 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