chapter 5 torsion · 2014-12-29 · chapter 5: torsion example 5.8 the two solid steel shafts are...
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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