# A Torque Or Twisting Mechanics Essay

When a shaft is subjected to a torque or twisting, a shearing stress is produced in the shaft. The shear stress varies from zero in the axis to a maximum at the outside surface of the shaft.

The shear stress in a solid circular shaft in a given position can be expressed as:

Ïƒ = T r / Ip (1)

where Shear Stress in the Shaft

Ïƒ = shear stress (MPa, psi)

T = twisting moment (Nmm, in lb)

r = distance from center to stressed surface in the given position (mm, in)

Ip = "polar moment of inertia" of cross section (mm4, in4)

The "polar moment of inertia" is a measure of an object's ability to resist torsion.

### Circular Shaft and Maximum Moment

Maximum moment in a circular shaft can be expressed as:

Tmax = Ïƒmax Ip / R (2)

where

Tmax = maximum twisting moment (Nmm, in lb)

Ïƒmax = maximum shear stress (MPa, psi)

R = radius of shaft (mm, in)

Combining (2) and (3) for a solid shaft

Tmax = (Ï€/16) Ïƒmax D3 (2b)

Combining (2) and (3b) for a hollow shaft

Tmax = (Ï€/16) Ïƒmax (D4 - d4) / D (2c)

### Circular Shaft and Polar Moment of Inertia

Polar moment of inertia of a circular solid shaft can be expressed as

Ip = Ï€ R4/2 = Ï€ D4/32 (3)

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where

D = shaft outside diameter (mm, in)

Polar moment of inertia of a circular hollow shaft can be expressed as

Ip = Ï€ (D4 - d4) /32 (3b)

where

d = shaft inside diameter (mm, in)

### Diameter of a Solid Shaft

Diameter of a solid shaft can calculated by the formula

D = 1.72 (Tmax/Ïƒmax)1/3 (4)

### Tensional Deflection of Shaft

The angular deflection of a torsion solid shaft can be expressed as

Î¸ = 584 L T / (G D4) (5)

Where

Î¸ = angular shaft deflection (degrees)

L = length of shaft (mm, in)

G = modulus of rigidity (Mpa, psi)

The angular deflection of a torsion hollow shaft can be expressed as

Î¸ = 584 L T / (G (D4- d4)) (5b)

### TORSION OF A CIRCULAR SHAFT:

Twisting moment or torque are forces acting through a distance so as to promote rotation. Torsion is the twisting moment of bar when it is loaded by torque that tends to rotate the circular shaft.

Moment is the general tendency of one or more applied forces to rotate an object an axis. Torque is the special case if the applied net force is equal to zero.

### DERIVATION OF TORSION:

To prove :TJ=GÎ¸l=Ø­r

In solid mechanics, torsion is the twisting of an object due to an applied torque. In circular sections, the resultant shearing stress is perpendicular to the radius.

For solid or hollow shafts of uniform circular cross-section and constant wall thickness, the torsion relations are:

where:

• R is the outer radius of the shaft.
• Ï„ is the maximum shear stress at the outer surface.
• Ï† is the angle of twist in radians.
• T is the torque (NÂ·m or ftÂ·lbf).
• â„“ is the length of the object the torque is being applied to or over.
• G is the shear modulus or more commonly the modulus of rigidity and is usually given in gigapascals (GPa), lbf/in2 (psi), or lbf/ft2.
• J is the torsion constant for the section . It is identical to the polar moment of inertia for a round shaft or concentric tube only. For other shapes J must be determined by other means. For solid shafts the membrane analogy is useful, and for thin walled tubes of arbitrary shape the shear flow approximation is fairly good, if the section is not re-entrant. For thick walled tubes of arbitrary shape there is no simple solution, and FEA may be the best method.
• the product GJ is called the torsional rigidity.

### Twisting moments (torques) and torsional stiffness

Torsion is the twisting of a beam under the action of a torque (twisting moment). It is systematically applied to screws, nuts, axles, drive shafts etc, and is also generated more randomly under service conditions in car bodies, boat hulls, aircraft fuselages, bridges, springs and many other structures and components. A torque, T , has the same units (Nm) as a bending moment, M . Both are the product of a force and a distance. In the case of a torque, the force is tangential and the distance is the radial distance between this tangent and the axis of rotation.

### Torsion of a Cylindrical Bar

Torsion of a cylindrical bar is illustrated in the figure. It can be seen that the shear strain in an element of the bar is given by

This equation applies both at the surface of the bar, as shown, and also for any other radial location, using the appropriate value of r . Clearly, the shear strain varies linearly with r , from zero at the centre of the bar to a peak value at the free surface.

The shear stress, Ï„, at any radial location, is related to the shear strain by

where G is the shear modulus. It follows that

The torque, T , can therefore be written as

As for the beam bending case, the geometrical integral is represented as a (polar) second moment of area

For a solid cylinder of diameter w , this can be written as

The torque is thus given by

Therefore,

### SHAFT IN SERIES:

In order to form a complete shaft sometimes two shafts are connected in series, the angle of twist is the sum of angle of twist is the sum of angle of twist of two shafts connected in series...i.e .Î¸ = Î¸1+Î¸2

### SHAFT IN PARALLEL:

The shaft is said to be in parallel with a driving torque is applied at the junction of shaft and the resisting torque at the other end. Here the angle of twist is same for each shaft. i.e. Î¸1 = Î¸2

### QUESTION:

Q. A shaft must transmit 20 kW of power at 300 rev/min. The shear stress must not exceed 150 MPa. Calculate a suitable diameter.

Sol: P = 2Ï€NT/60 = 2P(300)(T)/60 = 20 000 W

T = 636.6 Nm

T/J=Ï„/R

(636.6) (32)/(Ï€D4) = (150 x 106)/(D/2)

D3 = (636.6)(32)/(Ï€)(2)(15 x 106) = 216 x 10-6 D = 0.0278 m