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Shear & Torsion

Shear stressIf two equal and opposite parallel forces Q, not in the

same line, act on parallel faces of a member then it is

said to be loaded in shear.

A shear stress () is defined as a stress which is

applied parallel or tangential to a face of a material.

Shear stressThe formula to calculate average shear stress is:

where = shear stress;F = force applied;A = cross sectional area.

Shear stress

AF

load resisting area

loadshear )( stress shear

The shear stress will always be tangential to the area on which it acts.

Example I:In a guillotine, Q is the total force exerted bythe blade. This force is balanced by an equaland opposite force provided at the edge of thetable.

The area resisting shear is measured by theplate thickness multiplied by the length of theblade.

Shear stress

Example II:

In a punching operation the area of the

resisting shear would be the plate thickness

multiplied by the perimeter of the hole

punched.

Shear stress

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The rivet is commonly loaded in shear:

Riveted Joints

In single-shear the area resisting

shear is the cross-sectional area of

the rivet, d2/4, where d is the

diameter of the rivet.

In double-shear the resisting area

is twice the area of section of the

rivet, and the load which it can be

carried is theoretically twice that in

the single shear.

Riveted Joints

The single shear takes place on the single plane and the

shear area is the cross - sectional of the rivet, whereas the

double shear takes place in the case of Butt joints of rivets

and the shear area is the twice of the X - sectional area of

Clevis

Riveted Joints

Torque

Torque, also called moment or moment of force, is the

tendency of a force to rotate an object about an axis,

fulcrum, or pivot. A torque can be thought of as a twist.

The magnitude of torque depends on three quantities:

First, the force applied; second, the length of the lever

arm connecting the axis to the point of force

application; and third, the angle between the two.

L) & Fbet (angle sinx L Length,x F , Force T ,Torque

Power

(kW) 1000

T*N*2 ower P where

N – number of revolutions per second;

T – torque (Nm).

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The ultimate shear stress or strength is defined

as:

Ultimate shear stress

shear resisting areaload punch maximum

Allowable working stress-factor of safety

The most suitable strength or stiffness criterion for

any structural element or component is normally

some maximum stress or deformation which must

not be exceeded. In the case of stresses the value

is generally known as the maximum allowable

working stress.Because of uncertainties of loading conditions, designprocedures, production methods, etc., designers generallyintroduce a factor of safety into their designs, defined asfollows:

stress working allowablestress maximum safety of factor

Factor of Safety

A factor of safety of 3 implies that the design is

capable of carrying three times the maximum stress

to which it is expected the structure will be

subjected in any normal loading condition.

Factor of Safety

Values are normally selected on the basis of a

consideration of the social, human safety and

economic consequences of failure. Typical values

range from 2.5 (for relatively low consequence,

static load cases) to 10 (for shock load and high

safety risk applications).

Load Factor

In some loading cases, e.g. buckling of struts, neither the

yield stress nor the ultimate strength is a realistic criterion for

failure of components. In such cases it is convenient to

replace the safety factor, based on stresses, with a different

factor based on loads. The load factor is therefore defined

as:

load working allowablefailureat load factor load

This is particularly useful in applications of the so-called plastic limit

design procedures.

Coupling Flanges

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Shear Strain

The object is rigidly fixed at one face and subject to a stress

q on the other parallel face. The object deforms sideways.

The shear strain is defined as the angle of deformation, θ in

radians. Given that the angle Φ is small,

lx

The shear strain is defined as theratio of the distance deformed tothe height.

Module of Rigidity

The modulus of rigidity, G is given by:

strain,shear stress,shear

G

G

A shaft is to be fitted with a flanged coupling having 8

bolts on a circle of diameter 150 mm. The shaft may be

subject either to a direct tensile load of 400 kN or to a

twisting-moment of 18 kNm. If the maximum direct and

shearing stresses permissible in the bolt material are 125

MN/ m2 and 55MN/m2 respectively, find the minimum

diameter of bolt required. Assume each bolt takes an

equal share of the load or torque.

Using this bolt diameter and assuming only one bolt to

carry the full torque what would then be the shearing

stress in the bolt?

Torsion of a Thin Tube

Torsion of a thin tube

Consider the thin tube of mean radius r, having a wall

thickness (t is very small compared with r).

If a torque T is applied to both ends of the tube, one end

will twist relative to the other. A strip AB parallel to the

tube axis will distort to AB'.

Torsion of a thin tubeIf it is assumed that the displacement BB' is small

compared with the length of tube AB, then AB' will be

approximately straight.

Then angle < BOB' is the angle of twist θ of the length

AB. The shear strain is:rad r

ABBB' strain,Shear

l

since BB' = rθ , and AB = length of tube l.

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Torsion of a thin tube

rT

F radiustorque

force,Shear

rt2

areaforceshear

stress,Shear

F

The shear force on the cross-section of the tube is:

This force acts on area 2rt, since the tube is thin.

Therefore

tr2T

rt2 2

rT /

then,/r and rigidity, of modulus theG, / since Also l

G /r

l

r l

GHence,

Notes:

For a given torque the angle of twist varies directly with the length;

In these formulae, the twist must be in radians.

Torsion of Solid Shafts

Assumptions:1. The shaft is composed of a succession of thin

concentric tubes;

2. Each thin tube carries shear forceindependent of, and without interfering with,its neighbours.

3. Lines which are radial before twisting areassumed to remain radial after twisting;

4. The shaft is not stressed beyond the elasticlimit.

Area, dA = 2r x dr

If is the shear stress at radius r, then

Shear force on tube = 2r x dr x

Torque carried by tube = 2r x dr x x r

= 2r2 dr

lr

G

We know that

rl

G

drx rG

x r2 tube by carried Torque 2

l

dr rG

2 3

l

The whole torque T carried by the solid shaft is the sum of

all the elementary torques, ie

dr rG 2 TR

03

l

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Since radial lines before twisting remain radial after

twisting, is the same for all the thin tubes making up the

shaft. Also G and l are constant, therefore

dr r 2 G

TR

03

l pI G l

Ip is the polar second moment of area of a shaft

of circular section.

dr 2 IR

0p 3r

Rearranging G IT

p l

l

G Since

rτ

r G

lpIT

A useful re-arrangement

pGIT l

Another useful re-arrangement Important points:The angle of twist varies directly with length l ;

SincepI

Tr

for a given torque T, the shear stress q is

proportional to the radius r. Thus the

maximum shear stress occurs at the outside

surface where r=d/2, and the shear stress at

the centre of the shaft is zero.

A thin steel tube 90 mm inside diameter is subjected toa torque of 500Nm.

a) If the shear stress is not to exceed 28 MN/m2,calculate the tube thickness.

b) If the twist is not to exceed 2.5mm of arc on a600mm length what would be the thicknessrequired?

G = 84 GN/m2.

A thin tube 1.5 mm thick, 80 mm mean diameter is

subjected to a torque of 350 Nm. Calculate (a) the

shear stress in the tube, (b) the twist on a 1 m length.

G = 84 GN/m2.

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Twisting of Hollow shafts

dr r 2 G

TR

03

l

If d2 and d1 are the outside and inside diameters of

a hollow shaft subject to a twisting moment T then

the equation

may be rewritten as:

dr r 2 G T2/d

2/d

32

1

l

Twisting of Hollow shafts

dr 2 I2

1

d

dp

2

2

3/

/r

32)d - (d

I4

14

2p

The maximum shear stress for a given torque is again

at the outside fibres of the shaft, where r = d2/2.

r

pIT

Since

then

r . pI

T

Comparison of shear stress distributions in a solid and a hollow shaft

Stiffness and Strength

The stiffness or torsional rigidity of a shaft is the

torque to produce unit angle of twist. Thus if a

torque T produces a twist then

lpIG

T stiffness

Stiffness and Strength

The strength of a shaft is measured by the torque it

can transmit for a given permissible value of the

maximum shear stress.

For a given shear stress the strengths of two shafts

are the ratio of the corresponding torques.

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Power and Torque

(kW) 1000

T*N*2 ower P where

N – number of revolutions per second;

T – torque (Nm).

Compare the torsional stiffness of a solid

shaft 50 mm diameter, 300 mm long, with

that of a hollow shaft of the same

material having diameters 75 mm, 50 mm

and length 200 mm.

The propeller shaft of an aircraft engine

is steel tubing of 75 mm external and 60

mm internal diameter. The shaft is to

transmit 150 kW at 1650 rev/min. The

failing stress in shear for this shaft is 140

MN/m2. What is the safety factor?

7.87

Angle of Twist in Elastic Range

GITL

p

Angle of Twist in Elastic Range

i iip

ii

GILT

Angle of Twist in Elastic Range

L

0 p G (x)Idx T(x) φ

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Angle of Twist in Elastic Range

L

0 p G (x)Idx T(x) φ

(x)I J(x) p

Sign Convention

Sign Convention Sign Convention

A shaft is made of a steel alloy having an allowable shear

stress of allow = 120 N/mm2. If the diameter of the shaft is 15

mm, determine the maximum torque T that can be

transmitted.

What would be the maximum torque T′ if a 10-mm-diameter

hole is bored through the shaft? Sketch the shear-stress

distribution along a radial line in each case.

The solid shaft of radius r is subjected to a torque T.

Determine the radius r′ of the inner core of the shaft that

resists one-quarter of the applied torque (T/4). Solve the

problem by using the torsion formula.

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The shaft has an outer diameter of 25 mm and an inner diameter

of 20 mm. If it is subjected to the applied torques as shown,

determine the absolute maximum shear stress developed in the

shaft. The smooth bearings at A and B do not resist torque.

The steel shaft is subjected to the torsional loading shown.

Determine the absolute maximum shear stress in the shaft and

sketch the shear-stress distribution along a radial line where it is

maximum.

Torsion of Noncircular Members

3 - 58

Torsion of Noncircular Members

• At large values of a/b, the maximum shear stress and angle of twist for other open sections are the same as a rectangular bar.

GabcTL

abcT

32

21

max

• For uniform rectangular cross-sections,

• Previous torsion formulas are valid for axisymmetric or circular shafts

• Planar cross-sections of noncircular shafts do not remain planar and stress and strain distribution do not vary linearly