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Shear Stress (Contd)

1. The bolt of the clevis carries the load P across two crosssectional areas,the shear force being V = P/2 on each cross section.

. ere ore, t e o t s sa to e n a state o ou e s ear.

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1. The distribution of direct shear stress is usually complex and not

easily determined.2. It is common practice to assume that the shear force V is uniformly

distributed over the shear area A, so that the shear stress can be

computed from

. ,

average shear stress.

. ,

such as rivets, bolts, and welds.

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Bearing Stress

. ,

developed on the area of contact.

. .

3. Examples of bearing stress are the soil pressure beneath a pier and

the contact pressure between a rivet and the side of its hole.4. If the bearing stress is large enough, it can locally crush the

material, which in turn can lead to more serious problems.

5. To reduce bearing stresses, engineers sometimes employ bearingplates, so that the contact forces are distributed over a larger area.

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Bearing Stress (Contd)

1. Consider the lap joint formed by the two plates that are riveted

together as shown.

2. The bearing stress caused by the rivet is not constant; it actually

varies from zero at the sides of the hole to a maximum behind

the rivet.

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Bearing Stress (Contd)

1. The difficulty is avoided by assuming that the bearing stress

b is uniformly distributed over a reduced area.

2. The reduced area Ab is taken to be the projected area of rivet

. b

4. The bearing stress becomes

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Example 1 (Shear and Bearing Stress)

e ap o n s as ene y our r ve s o

n. ame er. n emaximum load P that can be applied if the working stresses are 14 ksi for

.

applied load is distributed evenly among the four rivets, and neglect friction

.

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Solution:We see that the equilibrium condition is V =P/4.

Given problem

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Design for Shear Stress in Rivets

The value of P that would cause the shear stress in the rivets to reach its

working value is found as follows:

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Design for Bearing Stress in Plate

. e s ear orce = a ac s on e cross sec on o one r ve sequal to the bearing force P b due to the contact between the rivet and

.

2. The value of P that would cause the bearing stress to equal its working

Comparing, the maximum safe load P that can be applied to the lap joint is = , w e s ear s ress n e r ve s e ng e govern ng es gn

criterion.

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Example 2: (Normal and Bearing Stresses)

The shaft is subjected to the axial force of 40 kN. Determine the averagebearing stress acting on the collar C and the normal stress in the shaft

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Solution:

232s m10225.003.04

A

232 b m104.004.04A

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Example 3: (Bearing Stress)

e assem y

cons s s

o

ree

s s

,

, an

a

are

use

o

suppor

e

load of 140 kN. Determine the smallest diameter of the top disk, the diameter

, .

allowable bearing stress for the material is ( allow )b = 350 Mpa, and allowable

allow .

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Solution

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Solution

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Example 5

e a owa e ear ng s ress or e ma er a un er e suppor s a an

is ( allow )b = 15 Mpa determine the size of square bearing plates A' and B'

. .

P = 100 kN.

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FBD

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Example 6 .

magnitude of the allowable suspended load P if the allowable bearing stress is ( allow )b = 220 MPa, the allowable tensile stress is ( allow )t =

, allow .mm, a = 5 mm, and b = 25 mm.

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Allowable Shear Stress: The pin is subjected to double shear. Therefore V = P/2

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