Chapter 3: Torsion  Tutorial...
Embed Size (px)
Transcript of Chapter 3: Torsion  Tutorial...

Chapter 3:
TorsionChapter 5:
Stresses
In beams
Chapter 4:
Shear and
Moment
Diagram

Torsion
Torsion or Torque, T, put simply, is referred to as a
twisting moment.

The derived formulas are:
Where:
Torsional Shearing Stress
T  Torque
r Radial Distance
J  Polar Moment of Inertia
Shafts

TORSION ON CIRCULAR SHAFTS
1. For Solid Shafts:
Ix
Iy
r

TORSION ON CIRCULAR SHAFTS
1. For Hollow Shafts:
Ix
Iy
R
r

The derived formulas are:
Where:
Angle of Twist
T Torque
L Length of Shaft over which the
torque is acting
J  Polar Moment of Inertia
G Modulus of Rigidity

In solving torsion problems particularly with circular shafts with the
derived formulas, the following assumptions must be observed:
1. Circular sections, remain circular.
2. Stresses do not exceed the proportional limit.
3. Plane sections remain plane and do not warp.
4. The projection upon a transverse section of straight radial lines in
the sections remain straight.
5. Shaft is loaded by twisting couples in planes that are perpendicular
to the axis of the shaft.

100 Nm1000 Nm 900 Nm
500 Nm700 Nm 800 Nm
Torque Diagram

Example : What is the minimum diameter of a solid steel
shaft that will not twist through more than 3o , in a 6meter length when subjected to a torque of 14 kNm? What maximum shearing stress is developed? Ifthe modulus of rigidity of the shaft is 83 GPa.

Example : A stepped steel shaft consists of a hollow shaft
2m long, with an outside diameter of 100 mm andan inside diameter of 70 mm, rigidly attached to asolid shaft 1.5 m. long, and 80 mm in diameter.Determine the maximum torque which can beapplied without exceeding a shearing stress of 70MPa or a twist of 2.5o in the 3.5 m length. Use G = 83GPa.
T T
2 m 1.5 m

Example : A steel shaft with a constant diameter of 60 mm
is loaded by torques applied to gears attached to itas shown in the figure. Using G = 83 GPa, determinethe relative angle of twist of gear D relative to gearA.
2m
3m
3mD
C
B
A
1000 N.m
1200 N.m
1000 N.m
800 N.m

Example : Determine the maximum shearing stress
developed in each material and the angle of rotationof the free end of the shaft. Use Gal = 28 GPa, GSt= 83GPa and GBr = 35 GPa.
Aluminum Steel Bronze
3m2m 1.5 m
d = 100 mm
d = 75 mm
1. 5 KNm4 KNm

Example : The compound shaft shown is attached to rigid
supports. For the bronze segment AB, the diameteris 75 mm, < or = 60MPa, and Gbr = 35 GPa. For thesteel segment BC, the diameter is 50 mm, < or =80 MPa, and Gst = 83 GPa. Compute the maximumtorque T that can be applied.
2 m 1.5 m
Bronze Steel
A B C
T

Example : A shaft composed of segments AC, CD and DB
is fastened to rigid supports and loaded as shown.For steel, Gst = 83 GPa, for aluminum, Gal = 28GPa,and for bronze, Gbr = 35 GPa. Determine themaximum shearing stress developed in eachsegment.
2 m 1.5 m
BronzeSteel
A C D
Aluminum
B
1 m
25 mm 25 mm
50 mm
300 Nm 700 Nm

Example : A round steel shaft 3 meters long tapers
uniformly from a 60 mm diameter at one end to a 30mm diameter at the other end. Assuming that nosignificant discontinuity results from applying theangular deformation equation over eachinfinitesimal length, compute the angular twist forthe entire length when the shaft is transmitting atorque of 170 N.m. Use G = 83 GPa.
60 mm 30 mm
3 meters

Example : The steel shaft and an aluminum tube are
connected to a fixed support and to a rigid disk asshown in the cross section. Knowing that the initialstresses are zero, determine the maximum torque Tthat can be applied to the disk if the allowablestresses are 120 MPa in the steel shaft and 70 MPa inaluminum tube. Use Gst = 77 GPa and Gal = 27 GPa.
76 mm
8 mm
50 mm
500 mm

Example : The two steel shafts shown in the figure, each with
one end built into a rigid support, have flanges rigidlyattached to their free ends. The shafts are to be boltedtogether at their flanges. However, initially there is a 6omismatch in the location of the bolt holes, as shown inthe figure. Determine the maximum shearing stress ineach shaft after the shafts are bolted together. Use G =83 GPa and neglect deformations of the bolts andflanges.
2 m 1 m
50 mm 40 mm
6o