Shear Stress
Shear Stress develops when the applied loads cause one
section of the body to slide past its adjacent section. The
force acts parallel to the area.
V
__
τ= A
τ
V
A
P
where V || A
Shear Stress
Resultant Shear
Cross – Sectional Area
Shearing area
parallel to the
load
P
V
∑Fx = 0;
V=P
TYPES OF SHEARING STRESS
SINGLE SHEAR
LAP JOINT
Rivet under Single shear
P
B
P
t
P
Shearing
Area
P
130mm
DOUBLE SHEAR
Rivet under Double shear
BUTT JOINT
t
P/2
splice plate
t
main plate
P
P/2
Shearing
Areas
SINGLE SHEAR
with n number of bolt or rivet;
ex. n = 2
w/ only 1 bolt or rivet
P
P
Lower Plate
Right Side View
P
P
P
Lower Plate
P
P
Right Side View
P
DOUBLE SHEAR
with n number of bolt or rivet;
ex. n = 4
w/ only 1 bolt or rivet
P/2
P
P/2
P/2
P
P/2
Middle Plate (Top View)
Middle Plate (Top View)
P
P
Middle Plate (Bottom View)
Middle Plate (Bottom View)
P
Right Side View
P
Right Side View
P
P
TYPES OF SHEARING STRESS
PUNCHING SHEAR
P
INDUCED SHEAR
P
L = width
x
t
D
a
a
Ɵ
at Section a-a :
x
t
V
L
Ashearing = x.L
Ashearing = πD.t
t
πD
Pcos(Ɵ)
Bearing Stress
Contact pressure exerted by one body upon
another body. The force acts perpendicular to the
area where load is applied.
P
b
__
δb =
A
δb
Pb
A
where P
A
┴
Bearing Stress
Perpendicular Force / Bearing Force
Projected Contact Area
Bearing Stress
END STRESS
LAP JOINT
t
P
B
P
P
130mm
P
130mm
P
P
Enlargement of Holes
t
A = Dt
D
EXAMPLE no. 01
A circular hole is to be punched out of a plate that has a
shear strength of 40ksi The working compressive stress
in the punch is 50ksi. a.) Compute the maximum
thickness of a plate in which a 2.5” diameter hole can be
punched. b.) If the plate is 0.25in thick, determine the
smallest diameter that can be punched. P
answer :
a.) tplate = 0.781”
b.) Dhole = 0.800 ”
EXAMPLE no. 02
The lap joint shown in the figure is fastened by four ¾-in.
diameter rivets. Calculate the maximum safe load P that can
be applied if the shearing stress in the rivets is limited to 14 ksi
and the bearing stress in the plates is limited to 18 ksi. Assume
the applied load is uniformly distributed among the four rivets.
answer :
P = 24,___ N
EXAMPLE no. 03
Compute for the shearing stress in the pin at B for the member
supported as shown in the figure. The diameter of the pin is
20mm.
answer :
94.01 MPa
EXAMPLE no. 04
Member B is subjected to a compressive force of 600 lb. If A
and B are both made of wood and are 1.5 in. thick, determine
to the nearest 1/8” the smallest dimension a of the support so
that the average shear stress along the blue line does not
exceed 50 psi.
answer :
a = 6.50 inches
Homework no. 02 - b
The figure shows a roof truss and the detail of the connection at joint
B. Members BC and BE are angle sections with thickness shown in
the figure. The working stresses are 70MPa for shear in rivet and
140MPa for bearing stress due to the rivets. How many 19-mm
diameter rivets are required to fasten the said members to the gusset
plate?
DETAIL OF JOINT B
14 mm thick
GUSSET PLATE
D
B
G
75X75X13 mm
4m
3m
C
A
3m
E
3m
50KN
3m
F
75X75X6 mm
P BE
3m
3m
75KN
H
50KN
answer :
P BC
0
