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CHAPTER OUTLINE1. Torsional Deformation of a Circular Shaft2. The Torsion Formula3. Power Transmission4. Angle of Twist5. Statically Indeterminate Torque-Loaded
Members6. *Solid Noncircular Shafts7. *Thin-Walled Tubes Having Closed Cross
EXAMPLE 5.3Shaft shown supported by two bearings and subjected to three torques.Determine shear stress developed at points A and B, located at section a-a of the shaft.
EXAMPLE 5.3 (SOLN)Internal torqueBearing reactions on shaft = 0, if shaft weight assumed to be negligible. Applied torques satisfy moment equilibrium about shaft’s axis.Internal torque at section a-a determined from free-body diagram of left segment.
• Power is defined as work performed per unit of time
• Instantaneous power is• Since shaft’s angular velocity ω = dθ/dt, we can
also express power as
5.3 POWER TRANSMISSION
P = T (dθ/dt)
P = Tω
• Frequency f of a shaft’s rotation is often reported. It measures the number of cycles per second and since 1 cycle = 2 radians, and ω = 2πf T, then power
EXAMPLE 5.5Solid steel shaft shown used to transmit 3750 W from attached motor M. Shaft rotates at ω = 175 rpm and the steel τallow = 100 MPa.Determine required diameter of shaft to nearest mm.
If shaft is subjected to several different torques, or x-sectional area or shear modulus changes suddenly from one region of the shaft to the next, then apply Eqn 5-15 to each segment before vectorially adding each segment’s angle of twist:
5.4 ANGLE OF TWISTProcedure for analysisAngle of twist• When circular x-sectional area varies along
shaft’s axis, polar moment of inertia expressed as a function of its position x along its axis, J(x)
• If J or internal torque suddenly changes between ends of shaft, φ = ∫ (T(x)/J(x)G) dx or φ = TL/JGmust be applied to each segment for which J, Tand G are continuous or constant
• Use consistent sign convention for internal torque and also the set of units
EXAMPLE 5.950-mm-diameter solid cast-iron post shown is buried 600 mm in soil. Determine maximum shear stress in the post and angle of twist at its top. Assume torque about to turn the post, and soil exerts uniform torsional resistance of t N·mm/mm along its 600 mm buried length. G = 40(103) GPa
EXAMPLE 5.9 (SOLN)Internal torqueMagnitude of the uniform distribution of torque along buried segment BC can be determined from equilibrium of the entire post.
EXAMPLE 5.9 (SOLN)Maximum shear stressLargest shear stress occurs in region AB, since torque largest there and J is constant for the post. Applying torsion formula
EXAMPLE 5.9 (SOLN)Angle of twistAngle of twist at the top can be determined relative to the bottom of the post, since it is fixed and yet is about to turn. Both segments AB and BC twist, so
5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS
• A torsionally loaded shaft is statically indeterminate if moment equation of equilibrium, applied about axis of shaft, is not enough to determine unknown torques acting on the shaft
5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS
• From free-body diagram, reactive torques at supports A and B are unknown, Thus,
Σ Mx = 0; T − TA − TB = 0
• Since problem is statically indeterminate, formulate the condition of compatibility; end supports are fixed, thus angle of twist of both ends should sum to zero
5.5 STATICALLY INDETERMINATE TORQUE-LOADED MEMBERS
Procedure for analysisEquilibrium• Draw a free-body diagram• Write equations of equilibrium about axis of shaftCompatibility• Express compatibility conditions in terms of
rotational displacement caused by reactive torques
• Use torque-displacement relationship, such as φ = TL/JG
• Solve equilibrium and compatibility equations for unknown torques
EXAMPLE 5.136061-T6 aluminum shaft shown has x-sectional area in the shape of equilateral triangle. Determine largest torque T that can be applied to end of shaft if τallow = 56 MPa, φallow = 0.02 rad, Gal = 26 GPa.How much torque can be applied to a shaft of circular x-section made from same amount of material?
EXAMPLE 5.13 (SOLN)Circular x-sectionWe need to calculate radius of the x-section.Acircle = Atriangle; ... c = 14.850 mm
Limitations of stress and angle of twist require
τallow = Tc/J; ... T = 288.06 N·m
φallow = TL/JGal; ... T = 33.10 N·m
Again, torque is limited by angle of twist.Comparing both results, we can see that a shaft of circular x-section can support 37% more torque than a triangular one
*5.7 THIN-WALLED TUBES HAVING CLOSED CROSS SECTIONS
IMPORTANT• Shear flow q is a product of tube’s thickness and
average shear stress. This value is constant at all points along tube’s x-section. Thus, largestaverage shear stress occurs where tube’s thickness is smallest
• Both shear flow and average shear stress act tangent to wall of tube at all points in a direction to contribute to resultant torque
EXAMPLE 5.16Square aluminum tube as shown.Determine average shear stress in the tube at point A if it is subjected to a torque of 85 N·m. Also, compute angle of twist due to this loading.Take Gal = 26 GPa.
Since t is constant except at corners, average shear stress is same at all points on x-section.Note that τavg acts upward on darker-shaded face, since it contributes to internal resultant torque T at the section
5.8 STRESS CONCENTRATIONIMPORTANT• Stress concentrations in shafts occur at points of
sudden x-sectional change. The more severe the change, the larger the stress concentration
• For design/analysis, not necessary to know exact shear-stress distribution on x-section. Instead, obtain maximum shear stress using stress concentration factor K
• If material is brittle, or subjected to fatigueloadings, then stress concentrations need to be considered in design/analysis.
EXAMPLE 5.18Stepped shaft shown is supported at bearings at Aand B. Determine maximum stress in the shaft due to applied torques. Fillet at junction of each shaft has radius r = 6 mm.
EXAMPLE 5.18 (SOLN)Internal torqueBy inspection, moment equilibrium about axis of shaft is satisfied. Since maximum shear stress occurs at rooted ends of smaller diameter shafts, internal torque (30 N·m) can be found by applying method of sections
EXAMPLE 5.18 (SOLN)Maximum shear stressFrom experimental evidence, actual stress distribution along radial line of x-section at critical section looks similar to:
EXAMPLE 5.19Tubular shaft made of aluminum alloy with elastic τ-γdiagram as shown. Determine (a) maximum torque that can be applied without causing material to yield, (b) maximum torque or plastic torque that can be applied to the shaft. What should the minimum shear strain at outer radius be in order to develop a plastic torque?
EXAMPLE 5.19 (SOLN)Outer radius shear strainTube becomes fully plastic when shear strain at inner wall becomes 0.286(10-3) rad. Since shear strain remains linear over x-section, plastic strain at outer fibers determined by proportion:
EXAMPLE 5.21Tube made from brass alloy with length of 1.5 m and x-sectional area shown. Material has elastic-plastic τ-γ diagram shown. G = 42 GPa.
Determine plastic torque TP. What are the residual-shear-stress distribution and permanent twist of the tube that remain if TP is removed just after tube becomes fully plastic?
Plastic torqueThen TP is removed, then “fictitious” linear shear-stress distribution in figure (c) must be superimposed on figure (b). Thus, maximum shear stress or modulus of rupture computed from torsion formula,
• If shaft is statically indeterminate, reactive torques determined from equilibrium, compatibility of twist, and torque-twist relationships, such as φ = TL/JG
• Solid noncircular shafts tend to warp out of plane when subjected to torque. Formulas are available to determine elastic shear stress and twist for these cases
• Shear stress in tubes determined by considering shear flow. Assumes that shear stress across each thickness of tube is constant
CHAPTER REVIEW• Shear stress in tubes determined from
τ = T/2tAm
• Stress concentrations occur in shafts when x-section suddenly changes. Maximum shear stress determined using stress concentration factor, K (found by experiment and represented in graphical form). τmax = K(Tc/J)
• If applied torque causes material to exceed elastic limit, then stress distribution is not proportional to radial distance from centerline of shaft
CHAPTER REVIEW• Instead, such applied torque is related to stress
distribution using the shear-stress-shear-strain diagram and equilibrium
• If a shaft is subjected to plastic torque, and then released, it will cause material to respond elastically, causing residual shear stress to be developed in the shaft