IV.) Torsion on Circular Shafts A.) Torque (T)

IV.) Torsion on Circular Shafts

A.) Torque (T)

B.) Torsional Shearing Stress (  v ) -Under Pure Torsional Shear, experiments have shown that for circular shafts of constant cross-section: 1.) Plane section remains plane. 2.) A straight line radius remains a straight line.

B.) Torsional Shearing Stress (  v )

-Under Pure Torsional Shear, experiments have shown that for circular shafts of constant cross-section:

1.) Plane section remains plane.

2.) A straight line radius remains a straight line.

T A B B’ O FIXED END

T Plane Sections Straight Line Radius  L T

From Hooke’s Law, Stress is directly proportional to strain. T  v

From Hooke’s Law, Stress is directly proportional to strain.

Consider a very small square area of a circular cross-section. T  v r c Area=a

Consider a very small square area of a circular cross-section.

The Stress on area “a” is:  =   V (r) c The Force on area “a” is: F =  a  =   V r(a) c The moment produced by this force on area “a” is: M = F(r)  =   V ra(r) =  V ar 2 c c

The Stress on area “a” is:

 =   V (r)

c

The Force on area “a” is:

F =  a  =   V r(a)

c

The moment produced by this force on area “a” is:

M = F(r)  =   V ra(r) =  V ar 2

c c

The total moment produced by the forces on all the small areas gives you the torqe resistance of the cross-section: T =  M =   V ar 2 =  V (  ar 2 ) c c From Statics, recall that the Polar Moment of Inertia (J) is defined as: J =  ar 2 Therefore: T =  V J c

The total moment produced by the forces on all the small areas gives you the torqe resistance of the cross-section:

T =  M =   V ar 2 =  V (  ar 2 ) c c

From Statics, recall that the Polar Moment of Inertia (J) is defined as:

J =  ar 2

Therefore: T =  V J

c

Therefore the torque resistance of a section of a given diameter at a maximum stress level  V is: T =  V J c Or, the stress produced by a torque T, on a section of a given diameter is:  V = Tc J

Therefore the torque resistance of a section of a given diameter at a maximum stress level  V is:

T =  V J

c

Or, the stress produced by a torque T, on a section of a given diameter is:

 V = Tc

J

B.) Angle of twist,  Shearing strain is defined in Ch.9 in terms of the shear deformation,  v as:  v =  v /L  L T

B.) Angle of twist, 

Shearing strain is defined in Ch.9

in terms of the shear deformation,  v as:

 v =  v /L

In the sketch shown, the shear deformation is the distance BB’, therefore:  v =BB’/L T O B B’  c

In the sketch shown, the shear deformation is the distance BB’, therefore:

 v =BB’/L

From geometry, we know that the arc length BB’ = c  therefore:   v = c  L  c T O B B’

From geometry, we know that the arc length BB’ = c  therefore:



 v = c 

L

Since Hooke’s Law states: G=  v /  v we can substitute in the above expression for  v : G=  v __ =  v L c  /L c   c T O B B’

Since Hooke’s Law states:

G=  v /  v

we can substitute in the

above expression for  v :

G=  v __ =  v L

c  /L c 

Solving for  :  =  v L (One formula for  G c  V = Tc J  =  v L = ( T c ) L = TL ( 2 nd formula for  G c (J)G c JG

Solving for  :

 =  v L (One formula for 

G c

 V = Tc

J

 =  v L = ( T c ) L = TL ( 2 nd formula for 

G c (J)G c JG

Summary of Torsion Formulas  = Angle of twist (radians)  V = Shear Stress (psi) L = Length (inches) T = Torque (in-lb) G = Modulus of Rigidity (psi) c = Outside radius of shaft (inches) J = Polar Moment of Inertia (in 4 )   =  v L = TL G c JG  V = Tc J

 = Angle of twist (radians)

 V = Shear Stress (psi)

L = Length (inches)

T = Torque (in-lb)

G = Modulus of Rigidity (psi)

c = Outside radius of shaft (inches)

J = Polar Moment of Inertia (in 4 )



A solid steel circular shaft is 4.5 feet long and has a diameter of 5 inches. If the shaft is fixed at one end a torque of 15,000 ft-lb is applied at the free end, find: 1.) the maximum shear stress in the shaft and show where it occurs (on the sketch provided). 2.) the shear stress at the center of the cross-section (point O). 3.) the angle of twist at the free end.

A solid steel circular shaft is 4.5 feet long and has a diameter of 5 inches. If the shaft is fixed at one end a torque of 15,000 ft-lb is applied at the free end, find:

1.) the maximum shear stress in the shaft and show where it occurs (on the sketch provided).

2.) the shear stress at the center of the cross-section (point O).

3.) the angle of twist at the free end.