MODULE 4
Bearing Stress
This module is consists of concepts, principles, definition of terms,
illustrations, solved problems and activity to enhance the skills of the learners in
analyzing and solving problems.
Concepts/Principles:
Bearing Stress is a contact pressure between separate bodies.
Examples of Bearing Stress
1. Soil pressure beneath piers
2. Forces on bearing plates
3. Contact pressure between an axle and its bearing
4. Contact pressure between rivet or bolt
5. Contact surface of the plate against with it pushes
Exaggerated Bearing Deformation of Upper Plate
Pb
Pb
Pb
Pb
Pb
t
d
projected area of
the rivet hole
In the figure, the result of an excessive bearing stress is to cause yielding of the
plate or of the rivet, or both. The intensity with which the rivet bear against the
rivet hole is not constant but actually varies from zero (o) at the edges of the
hole to a maximum directly in back of the rivet. The difficulty inherent in a
variable stress distribution is availed by the common practice of assuming the
bearing stress (δb) to be uniformly distributed over a reduced area which is the
projected area of the rivet hole.
FORMULA:
Pb
δb = ----------Ab
or
Pb = Ab δb or
Where:
Pb = bearing load
Ab = projected area of the rivet hole
t
= thickness of the plate
d = width of the projected area
δb = bearing stress
Pb = (t x d) δb
EXAMPLE 1: For the lap joint shown in the figure is fastened by three 20-mm Φ
rivets. Assuming that P=50 KN, determine the maximum a.) average tensile
stress in each plate, b.) shearing stress in each rivet, and c.) bearing stress in
each plate. Assume that the applied load P is distributed equally among the three
rivets.
P
P
130mm
20 mm
P
25 mm
P
Solution:
Given:
P = 50 KN
Φr = 20 mm
a.)
P
Max. tensile stress in each plate, δs = -------Anet
50 KN (1000 N/1 KN)
δs = ---------------------------------------------------(130 mm) (25 mm) – (20 mm) (25 mm)
50 000 N
= ---------------------------------= 18.18 MPa
2750 mm2
b.) Determine the shearing stress in each rivet:
P
δs = -------------A
50 KN (1000 N/ 1 KN)
δs = ---------------------------------π (20)2 (3 rivets) mm2
---------4
= 53.05 MPa ans
c.) Determine the bearing stress in each plate:
Pb
t
d
Pb
δb = ---------A
=
50 KN (1000 N/KN)
-------------------------------------- = 33.33 MPaans
(20 mm) (25 mm) (3 rivets)
EXAMPLE 2: For the Lap Joint shown in problem No. 1, determine the Maximum
load P which may be applied if the shearing stress in the rivets is limited to 60
MPa, the bearing stress in the plates to 110 MPA, and the average tensile stress
in the plate to 140 MPa.
P
P
130mm
20 mm
P
25 mm
P
Pb
t
d
a. From Shear:
P
δs = ------ ; P = 60 N/mm2 (π )(20)2 mm2 / 4) (3 rivets)=56548.67 N=56.55 KN
A
b. From Bearing Stress:
Pb
δb = ------- ; Pb = 110 N/mm2 (20 mm) (25mm) (3 rivets) = 165000N = 165 KN
Ab
c. From tensile stress:
Pt
δt = ------ ; Pt = (140 N/mm2) x (130 mm)(25 mm) – (20 mm)(25 mm)
A
= (140 N/mm2) (3250 mm2 – 500 mm2)
= 385000 N = 385 KN
Therefore: the maximum load P which may be applied to the plate is 56.55KNans
EXERCISES:
DIRECTION: On short bond paper, answer the following activity correctly and
honestly. Submit the Photo or e-copy of your answer sheet via Google
Classroom on or before the due date stated in the instruction on the Classwork of
the Google Classroom.
Activity 1: Read with understanding and solve the following problem carefully
and honestly: Write your solution on a separate sheet of paper.
1. In the figure, assume that a 20-mm diameter rivet joins the plates which are
100 mm wide.
a. If the allowable stresses are 140 MPa for bearing in the plate material and 80
MPa for shearing of the rivet, determine the minimum thickness of each plate.
(Ans. t = 9 mm)
b. Under the conditions specified in part (a), what is the largest average tensile
stress in the plates?
(Ans. δt = 34.91 MPa)
Figure:
Pb
Pb
t=?
Pb
Pb