A bar is shown in (a) in pure shear due to torsion. Pure shear means that the only forces acting on the element are shear stresses. The directions of these stresses depend on the direction of the applied torques. Assuming the right hand rule end cb will rotate in the sense shown in (b). The shear stresses shown in (b) are by sign convention positive. A shear stress on a positive face is positive if it acts in the direction of one of the positive axes.
Transcript
A bar is shown in (a) in pure shear due to torsion. Pure shear means that the only forces acting on the element are shear stresses. The directions of these stresses depend on the direction of the applied torques. Assuming the right hand rule end cb will rotate in the sense shown in (b). The shear stresses shown in (b) are by sign convention positive.
A shear stress on a positive face is positive if it acts in the direction of one of the positive axes.
To determine the state of stress as a function of angle we will cut an oblique section at an angle this results in a triangle with AB unchanged but all other lengths altered. The state of pure shear is shown in (a). In (b) we take an oblique section and resolve the resultant force into two components and . The resolved components can be rewritten as shown in (c) where we have equated the forces and used the relationship that F= A= .L.t since t is constant for the element A=A0.L
(NB Secant=h/a)
costansinsec 000 AAA 2sincossin2
sintancossec 000 AAA
2cossincos 22
2cos
2sin
The stress as a function of angle is shown below in the figure. Maximum shear stress occurs at 0°±90° and the maximum normal stresses at ±45°.
At 45° there are equal and opposite normal stresses acting on the element. This often leads to torsion failures at 45° to the torsion axis.
To determine the deformation of the bar we need to consider shear strains and normal strains. The shear distortion tends to deform a rectangle to a parallelepiped.
The normal stresses at 45° will elongate the section but lateral contraction will also occur due to poisson contraction. Similarly the compressive stress at 135° will lead to a compression of the element but will also extend the element at 45°.
The total normal strain in the 45° direction is given by
)1(max EEE
)1(2 max hLbd
After deformation
Using cosine rule c2=a2+b2-2abcos
2
cos2 2222 hhhL bd
2
cos11 2max
sin121 2maxmax
G22max
For small strains
)1(max EEE
Earlier we derived
Therefore
12
EG
TWWork Done = Force x distance = torque x angle moved
Power =Rate of Work Done
Tdt
dT
dt
dWP
Is in rads.s-1
Very often rate is given in r.p.m and Power is given in hp
1 hp = 550 ft-lb/s.
A solid steel shaft AB is shown in the Figure. It is to be used to transmit 5 hp from the motor M to which it is attached. If the shaft rotates at =175 rpm and the steel has an allowable shear stress of allow=14.5 ksi, determine the required diameter of the shaft to the nearest 1/8 in.
A solid steel shaft AB is shown in the Figure. It is to be used to transmit 5 hp from the motor M to which it is attached. If the shaft rotates at =175 rpm and the steel has an allowable shear stress of allow=14.5 ksi, determine the required diameter of the shaft to the nearest 1/8 in.
slbftTP /.2750550*5 srad /33.1860
2175
lbftT .1.15033.18
2750
allow
T
c
c
c
J
4
2
inT
callow
429.0)14500(
)12)(1.150(223/13/1
Therefore d of the shaft should be greater than 0.858in i.e. 7/8=0.875in
A solid shaft ABC of 50 mm diameter is driven at A by a motor that transmits 50 kW to the shaft at 10 Hz. The gears at B and C drive machinery requiring power equal to 35 kW and 15 kW respectively. Compute the maximum shear stress in the shaft and the angle of twist ac between the motor at A and the gear at C. (Use G=80 GPa).
