The problem of transmitting a torque or rotary motion from one plane to another is frequently encountered in machine design. Typically, circular bars are used for such transmissions chiefly because in these bars a plane section before twisting remains plane after twisting, i.e. there is no warping of the section after loading. We will be interested in determining the maximum load a circular bar is capable of sustaining, the stresses in the bar as well as the deformation or twist of the bar. To do this, we need to establish the relationship between the applied torque, the stresses and the deformation of the bar.

Consider the bar of circular cross-section twisted by couples T at the ends. Because the bar is subjected to torsion only, it is said to be in pure torsion.

Assuming that the end A is fixed, then the torque will cause end B to rotate through a small angle f, known as the angle of twist. Thus the longitudinal line mn on the surface of the bar will rotate through a small angle to position mn¢.

where r is the radius of the circular bar and df is the angle of rotation of one cross-section with respect to another. df/dx is the rate of change of the angle of twist and representing this by q, we obtain

For the case of pure torsion, the angle of twist per unit length is constant along the length of the bar, hence

Note that the above expression is based only on geometric concepts and is therefore valid for any circular bar in pure torsion, irrespective of its material behaviour (elastic or inelastic, linear or non-linear).

For linear elastic materials, shear stresses are proportional to shear strains and the constant of proportionality is the modulus of rigidity, G. Hence
 

ie.

If the shear stress in the elemental area is t, then the shear force on this area is tdA. Moment of this force about the axis of the shaft/bar is (tdA)r = Gqr²dA. That is
dF = tdA
dM = (tdA)r = Gqr²dA

Total resisting torque about the axis of the shaft is the summation taken over the entire x-sectional area, of the moments of all individual elements, i.e.
 

i.e.

For a circular x-section of radius r    
 
 

Summary

We have obtained that
 
 
 
 

Note that
 
 

Units: T is in N.m

I_p is in m⁴

t is in Pa (i.e. N/m²)

r is in m

G is in Pa

q is in radians/m
 
 

The product GI_p is known as torsional rigidity while GI_p/L is called the torsional stiffness, defined as the torque required to produce a unit angle of rotation if one end of the bar with respect to the other end. The torsional flexibility is the reciprocal of the torsional stiffness and is defined as the angle of rotation required to produce a unit torque.

Observe that t is maximum at the outer boundary and zero at the centre of the bar.
 
 

Example 3.1:

Determine the polar moment of inertia for a solid circular cross-section of radius r
 
 
 
 

3.2 Hollow Circular Bars

Observe that the shear stress distribution is a solid bar varies from zero at the centre of the bar to a maximum at the outer boundary of the bar. Hence material near the centre of the bar will be grossly under-stressed and is not being used to full capacity. Where weight reductions and savings of material are important, it is advisable to use hollow shafts to transmit torque.

The analysis of the torsion of a hollow circular bar is essentially the same as that for a solid bar except that the limits of integration for determining I_p changes from 0 to r and becomes r₁ to r₂ where r₁ is the internal radius and r₂ is the outer radius. Hence, for a hollow circular bar of internal radius r₁ and outer radius r₂, the polar moment of inertia is given by:
 
 

Example 3.2:

A solid steel shaft of diameter d = 60 mm and length L = 4 m is to be designed using t_allowable = 40 MPa and allowable angle of twist per unit length q = 1° per m.

3.3 Non-uniform Torsion

Non-uniform torsion arises when a bar made up of two or more segments of different diameters is subjected to torsion at several cross-sections. For this type of case, each region of the bar between applied torques or between changes in x-sections is considered to be in pure torsion and the effect of the applied torque can be determined for each part or segment of the bar, using the equations obtained earlier.

The total angle of twist f of one end of the bar with respect to the other end is obtained by summation, using the general formula
 
 

If the torque and/or the x-sectional area of the bar changes continuously along the axis of the bar, then the total angle of twist f is given by
 
 

Example 3.3a:

Required: (a) Tmax in shaft; (b) f in degrees at free end

Solution
 
 
 
 

Example 3.3b:

A tapered bar AB of solid circular x-section is twisted by torques T applied at the ends. The diameter of the bar varies linearly from d_a at the left-hand end to d_b at the right-hand end. Derive an expression for the angle of twist f of the bar.

Solution
 
 
 
 
 

3.4 Pure Shear

A object that is subjected to only shear stresses is said to be in pure shear. Consider the small volume element that is subjected to pure shear as shown

Force on positive x-face = ?1bc
Force on negative x-face = ?2bc
Force on positive y-face = ?3ac
Force on negative y-face = ?4ac
 
 

Considering equilibrium of forces

SF_x = 0 ® t₃ac = t₄ac ® t₃ = t₄

SF_y = 0 ® t₁bc ® t₂bc ® t₁ = t₂

SM_z = 0 ® t₁abc ® t₃abc ® t₁ = t₃

Therefore t₁ = t₂ = t₃ = t₄ = t

Question:

Determine the expressions for normal and shear stresses on an oblique plane whose normal is inclined at q to the longitudinal-axis of the member which is subjected to pure shear.
 
 

The stresses on the inclined plane can be determined by considering equilibrium of all forces on the three sides of the triangular element. If A₀ is the cross-sectional area and A_q is the oblique/inclined plane area, we have that
 
 

Also the area of the bottom face of the triangle A_b is obtained from
 
 

Considering equilibrium of forces in a direction normal to the inclined plane,
 
 

Considering equilibrium of forces in a direction parallel to the inclined plane
 
 

Hence for an object that is in a state of pure shear, normal and shear stresses s_q and t_q will act on a plane inclined at q to the x-section where
 
 

Note that for q = 0°

s_q = 0 and t_q = t

For q = 90°, i.e. the top face of the element, s_q = 0 and t_q = -t.

The normal stress s_q reaches a maximum value at q = 45° where s_q = -t (tensile) and t_q = 0

Similarly, the normal stress gq is minimum (maximum compressive value) at q = -45° where

s_q = -t (compressive) and t_q = 0.

Hence an element oriented at an angle of 45° is acted upon by equal tensile and compressive stresses in perpendicular directions but no shear stresses are present.
 
 

Note that the foregoing implies that the state of pure shear at the surface or x-section of an object is equivalent to equal tensile and compressive stresses acting on an element of the object which is oriented at an angle of 45° to the cross-section. If a material is weaker in shear on longitudinal plane than on cross-sectional planes, e.g. a circular bar made of wood with grains in the longitudinal direction, the first cracks (shear failure) due to twisting load will occur on the surface in the longitudinal direction. (Qn: Why on the Surface and in the longitudinal Direction). However, if the material is very brittle (ie. much weaker in tension than in shear) failure will occur along the plane of maximum tensile stress, ie.e. along a helix inclined at 45° to the longitudinal axis. (Qn: Why along a helix inclined at 45° to the longitudinal axis; Sketch a typical failure surface). This type of failure is easily demonstrated by twisting a piece of classroom chalk.

Next, consider the strains associated with pure shear. Recall that the effect of shear is to produce a shear distortion (change in shape) but no change in volume. Considering a stress element that is oriented at 45°. Only tensile and compressive stresses t and -t respectively, act on this stress element. The effect of the tensile stress is to elongate the element in the 45° direction and to shorten it in the perpendicular direction (135° or -45°) directions. Similarly, the compressive stress -t, will shorten the element in the 135° or -45° direction and elongate it in the 45° direction.

If the material is linearly elastic, the shear strain for the element at q = 0° is given by g = t/G.

For the element at q = 45°, Normal strain due to s_max = t.

e_x = t/E (at 45° direction) and e_y = 1 n t/E (at 135° or -45° direction).

Normal strains due to s_min = -t.

e_y = -t/E (at 135° or -45° direction) and e_x= n t/E (at 45° direction)

Hence the resultant normal strain at 45° direction.

e_max = t/E + n t/E = t/E(1+n)

Similarly resultant normal strain at 135° or -45° direction.

e_min = -t/E - n t/E = -t/E(1+n)

Hence pure shear produces elongation in the 45° direction and shortening in the 135° or -45° direction. This is consistent with the deformed shape of the element in pure shear. Furthermore, it can be shown that

3.5 Relationship between the Elastic Constants E, G, and n

Consider the square stress element of side h. The effect of pure shear is to distort the square element into a rhombus as shown.
 
 

Recall that the diagonal bd will elongate while ac will shorten. If E_max is the normal strain f is the 45° direction, then the new length of the diagonal bd, L_bd is given by
 
 

where h 2 is the original length (h 2 )e_max is the elongation. Also the shear distortion reduces the angle adc from to where g is the shear strain. Now Ðadb = Ðabd = _Ðadc.
 
 

Using the Cosine Rule
 
 

Hence 1 + 2 e_max + e _ax = 1 + sin g
 
 

Because t_max and g are very small quantities, e _ax is neglected and sin g ? g.
 
 

Hence 1 + 2 e_max = 1 + g

e_max = g/2
 
 

This shows that the normal strain e_max in the 45° direction is pure shear is equal to one-half the shear strain g.
 
 

Recall that g = t/G. Hence e_max = g/2 = t/2G.
 
 

But the normal strain in the 45° direction e_max is given by e_max = t/E + nt/E = t/E (1 + n)
 
 

Hence e_max = t/2G = t/E (1 + n) ® G = E/2(1 + n)
 

Thus, the the elastic constants E, G, and V are not idependant properties as one of them can be determined, using the equation above, if the other two are known.
 

Summary:
 
 

Note that this expression implies that the three elastic constants, the Young’s modulus E, the modulus of rigidity G and the Poisson’s ratio n are not independent properties as one of them can be determined, using the equation above, if the other two are known.
 
 
 
 

Example 3.4: (Problem 3.4-4)

A hollow circular steel bar (G = 80 GPa) is twisted by a torque T that produces a maximum shear strain g_max = 750 x 10^-6 rad. The bar has outside and inside radii of 75 and 60 mm, respectively. What is the maximum tensile stress s_max in the bar. What is the magnitude of the applied torque, T?

Solution
 
 
 
 

Example3.5: (Problem 3.4-2)

A hollow circular shaft of aluminum (G = 28 GPa) has an outside diameter of 100 mm and an inside diameter of 50 mm. When twisted by a torque T, the bar has an angle of twist per unit length q = 0.03 rad/m. What is the maximum tensile stress s_max in the shaft? What is the magnitude of the applied torque T?

Solution
 
 
 
 

Example 3.6: (Problem 3.5-2)

A hollow shaft of outside diameter 80 mm and inside diameter 50 mm is made of aluminum having G = 27 GPa. When the shaft is subjected to a torque T = 52 kN.m, what are the values of the maximum shear strain g_max and the maximum normal strain e_max.

Solution
 
 
 
 

Example 3.7: (Problem 3.5-5)

A circular bar is subjected to a torque that produces a tensile stress s = 8000 psi at 45° to the longitudinal axis. Determine the maximum normal strain e and maximum shear strain g in the bar assuming that E = 11.5 x 10⁶ psi and n = 0.30.

Solution
 
 
 
 

3.6 Transmission of Power by Circular Shafts

By definition, power is the rate of doing work, i.e.
 
 

The work done by a torque T is equal to the product of T and the angle through which it rotates f, i.e. W = Tf where f = angular displacement (measured in radians). Hence,
 
 
 
 
 
 

where w (omega) is the angular speed, i.e. the rate of change of angular displacement with time.
 
 

Note that the angular speed w is often expressed in terms of the frequency, f of revolution, i.e. the number of revolutions per unit time.

w = 2 p f where f is in Hetz (Hz) = s^-1
 
 

Hence P = 2 p fT
 
 

If the angular speed is expressed in rpm (revolutions per minute), denoted by n; then n = 60 f and this implies that f = n/60 and
 
 

Observe that these expressions relate Power transmitted to the Torque T and that Power is a function of frequency of rotation or number of revolutions per minute.

Units: The SI unit for

Power is the watts (W);

Torque is N.m and

w is rad/sec.

The horsepower is unit for Power in the English (U.S. Customary) System of Units

1 Horse Power = 550 ft-lb/s ? 33000 ft-lb/min ? 746 Watts
 
 
 
 

Example 3.8: (Problem 3.6-9)

A hollow propeller shaft of outside diameter 6.0 inches is required to transmit 4000 hp while rotating at 1500 rpm. The material is steel with G = 11.5 x 10⁶ psi. The allowable shear stress t is 7500 psi and the allowable angle of twist per unit length f is 0.01 degrees per inch. (a) Calculate the required inside diameter d of the propeller shaft. (b) specify to the nearest  inch the actual inside diameter to be used.

Solution
 
 
 
 

Example 3.9 (Problem 3.6-2)

How much power P(kW) may be transmitted by a solid circular shaft of diameter 80 mm turning at 0.75 Hz if the shear stress is not to exceed 40 MPa?

Solution
 
 

Example 3.10 (Problem 3.6-12)

A circular shaft ABC is driven by a motor at A, which delivers 300 kW at a rotational speed of 3.2 Hz. The gears at B and C take out 120 and 180 kW, respectively. The lengths of the two parts of the shaft are L_AB = 1.5 m and L_BC = 0.9 m respectively. Calculate the required diameter of the shaft if the allowable shear stress is 50 MPa, the allowable angle of twist in the shaft between points A and C is 0.02 rad, and G = 75 GPa.

Solution
 
 
 
 

3.7 Statically Indeterminate Torsional Members

Consider a shaft consisting of two prismatic sections, loaded by a torque T_o at C with the ends fixed at A and B. The applied Torque T_o will give rise to reactive torques T_a and T_b at ends A and B respectively. Because only one fixed end is needed to ensure equilibrium, the extra fixed end makes the problem to be statically indeterminate to the first degree (since there is one extra/redundant reactive torque that is present but is not needed for equilibrium).

Using the flexibility method, the statically indeterminate structure is "released", i.e. split-up into a statically determinate primary structure with the externally applied load acting on it and a secondary structure with only the redundant force acting on it.

Considering the reactive torque at end B to be the redundant force, the structure is released as shown below:
 
 

The angle of twist f_b produced by T_o at end B is given by
 
 

Considering the secondary structure, the angle of twist produced by T_b at end B is given by
 
 

We know that the angle on twist at end B is equal to zero, since this end is fixed. Hence, for compatibility at end B, (f_b)_o = (f_b)_r; i.e.
 
 
 
 
 
 

From equilibrium
 
 

Substituting T_b from above, we obtain
 
 

Note that the shear stress on each part of the bar is different and is given by
 
 

An alternative approach to solving the same problem is to consider that the angle of twist on each part of the bar must be the same, i.e. f_a = f_b = f₀
 
 

Using equilibrium,
 
 

which is the same as was obtained using the flexibility method. Note that we have utilized both equilibrium and compatibility equations except that we enforced compatibility at C instead of at B.

The second approach should be used in handling composite bars consisting of an inner core of different material from the outer tube.
 
 

From compatibility of rotation, (2)

Using these two equations, T_a and T_b are determined and the maximum shear stresses in each bar is given by
 
 
 
 
 

Example 3.11 (Problem 3.7-16)

A solid shaft is formed of two materials, an outer sleeve of steel (G_s = 80 GPa) and an inner rod of brass (G_b = 36 GPa). The outer diameters of the two parts are 75 mm and 60 mm. Assuming that the allowable shear stresses are t_s = 80 MPa and t_b = 48 MPa in the steel and brass respectively, determine the maximum permissible torque T that may be applied to the shaft.

3.8 Strain Energy In Pure Shear

When a load is applied to a structure, the structure deforms. The applied load does work as the structure deforms and strain energy is absorbed by the structure. The strain energy for a bar in torsion is equal to the area under the torque - rotation diagram. If the bar is linearly elastic, then the torque T will be proportional to the angle of twist f.
 
 

U = Area under the T - f diagram

Alternatively, consider a square element of side h subjected to pure shear. Under the action of the shear stresses, t, the element is distorted by g where g is the shear strain.
 
 

Observe that h²t is the volume of the element. Hence the strain energy density
 
 

To obtain the total amount of strain energy stored in a circular bar subjected to pure torsion, consider an elemental circular section of radius r and thickness dr, extending along the entire length of the bar. The strain energy in the elemental section dU is given by
 
 

Corresponding expression for non-uniform torsion and continuously varying x-sectional area is
 
 
 
 

Example 3.12 (Problem 3.8-3):

How much strain energy is stored in the stepped steel shaft shown in the figure when the angle of twist of equals 0.015 radians. Use G = 11.8 x 10⁶ psi.

Thin-walled Tubes of Arbitray x-sectional shape (Closed sections)

    Circular x-sectional shape is the most efficient shape for resisting torsion and transmitting power and is frequently utilized in machine parts.  For complex structures however, the need may arise to use structural members with non-circular x-sections, eg. the wings of an airplane.
    Consider a cylindrical tube of arbitrary x-sectional area (the tube has a longitudinal axis of which is a straight line and the x-section at every point along this axis is identical).  The tube is subjected to pure torsion and its thickness t is small compared to its width and t may vary around the x-section.
 
 

    Consider the rectangular element abcd where ab and cd are located || to the longitudinal axis and bc and ad are perpendicular to the axis.  Let us assume that the sheer stress varies from t_b at b to t_c at c since t is free to vary around the x-section.  From equilibrium, identical shear stresses must act on the opposite face ad in the opposite direction.  At corners b and c, the shear stresses or orthogonal faces must be t_b and t_c, respectively.  Because ab and cd are || to the longitudinal axis and since the x-section is identical at every point along this axis, the thickness t is constant along ab and also constant (different value) along cd.  The shear stresses on the face ab and cd are thus constant and have values t_b and t_c, respectively.
    The forces associated with the shear stresses t_b and t_c  are:

                                                            f_b = t_bt_bdx  and  f_c = t_ct_cdx

where t_b  and t_c are the thickness points b and c, respectively.  For equilibrium in the longitudinal direction,  f_b = f_c ie:
 
 

hence at every point on the x-section, the product of the shear stress and the tube thickness is a constant called the shear flow, ¦,
ie:

shear flow, ¦ = tt

Note that if t is constant around the x-section or at certain region, then t is also a constant.  Also, t_max will occur where t is smallest.

Relationship between Torque and Stress
        Consider an element of length ds from one arbitrary x-sectional shape.

This element is located at s from an arbitrarily chosen co-ordinate established through point O.

On the elemenatl area,
                    f = tt ==> t = f/t

if dV is the shear force on this elemental area, then
                    dV = tA = f/t ^. tds = fds

Moment dM of the shear force dV about point O is given by
                    dM = rdV = rfds

The total momen of the shear forces in all elemental areas must be equal to the applied Torque T, ie:
 

where L_m is the length of the midline.
    Observe that the area enclosed by ds at O is 1/2 rds.  Hence the integeral above can be interpreted geometrically as twice the area enclosed by the midline, ie.
 
 
 

Note that A_m is not the x-section if t varies.

Strain Energy of Thin-walled Tubes
            Consider again an elemental volume of dimensions t, ds and dx

Strain energy density for element
Hence, strain energy for element
 
 
 

Example 3.13 (3.4-11 & 13  Problems have been combined)

    A uniformly tapered tube AB of hollow circular cross-section of uniform wall thickness t and length L is loaded by a distributed torque having intensity t(x) per unit length that varies linearly from a maximum value t_A at end A to 0 at end B.  The bar is fixed at end A and free at end B.  (a)  What is the shear stress at the fixed end A in the bar?  )b)  What is the angle of twist at end B?  The average diameters are d_A and d_B = 1/2 d_A.  The polar moment of inertia may be represented by approximate formula

Solution
 
 
 
 

Example 3.14

A uniformly tapered tube AB of hollow circular cross section is shown below. The tube has constant wall thickness t and length L. The average diameters at the ends are dA and dB = 2dA. The polar moment of inertia may be represented by the approximate formula  (Eq. 3-18). Derive a formula for the angle of twist of the tube when it is subjected to torques T acting at the ends.