8/12/2019 SOM Lecture 03
1/25
8/12/2019 SOM Lecture 03
2/25
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.
8/12/2019 SOM Lecture 03
3/25
8/12/2019 SOM Lecture 03
4/25
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.
8/12/2019 SOM Lecture 03
5/25
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.
8/12/2019 SOM Lecture 03
6/25
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.
8/12/2019 SOM Lecture 03
7/25
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
8/12/2019 SOM Lecture 03
8/25
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
.
8/12/2019 SOM Lecture 03
9/25
Solution:We see that the equilibrium condition is V =P/4.
Given problem
8/12/2019 SOM Lecture 03
10/25
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:
8/12/2019 SOM Lecture 03
11/25
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.
8/12/2019 SOM Lecture 03
12/25
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
8/12/2019 SOM Lecture 03
13/25
Solution:
232s m10225.003.04
A
232 b m104.004.04A
8/12/2019 SOM Lecture 03
14/25
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 .
8/12/2019 SOM Lecture 03
15/25
Solution
8/12/2019 SOM Lecture 03
16/25
8/12/2019 SOM Lecture 03
17/25
8/12/2019 SOM Lecture 03
18/25
Solution
8/12/2019 SOM Lecture 03
19/25
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.
8/12/2019 SOM Lecture 03
20/25
FBD
8/12/2019 SOM Lecture 03
21/25
8/12/2019 SOM Lecture 03
22/25
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.
8/12/2019 SOM Lecture 03
23/25
Allowable Shear Stress: The pin is subjected to double shear. Therefore V = P/2
8/12/2019 SOM Lecture 03
24/25
8/12/2019 SOM Lecture 03
25/25
Top Related