Simple Stress:
Shearing Stress
ES205 Mechanics of Deformable Bodies
Engr. GG Ancog
Shearing Stress
Forces parallel to the area resisting the force cause shearing
stress. Shearing stress is also known as tangential stress. Shear
stress arises whenever the applied loads cause one section of a
body to slide past its adjacent section.
where V is the resultant shearing force which passes through
the centroid of the area Av being sheared (parallel area).
Force
Area
Force
Force
Force
Examples of Direct Shear
a) Single Shear
in a Rivet
b) Double
Shear in a bolt
a) Shear in a metal sheet
produced by a punch
Resisting Area
Sample Problem 1
What force is required to punch a 20-mm-diameter hole in a
plate that is 25 mm thick? The shear strength of the plate is 350
MPa.
Solution
P
350
25 mm
=
(
P = 549,778.71 N
P = 549.78 KN
20 mm
)(
)
Sample Problem 2
The bell crank, which is in equilibrium under the forces shown in
the figure, is supported by a 20mm-diameter pin at D that is in
double shear.
Determine:
(a) the required diameter of
the connecting rod AB,
given that its tensile working
stress is 100 MPa;
(b) the shear stress in the pin.
Solution
RDX
15KN
25.98KN
RDY
(a) Solve for the diameter
of the connecting rod AB
∑MD = 0 +
PAB(200) – 25.98(240) = 0
PAB = 31.177 KN (T)
100
=
.
d = 19.924 mm ~ 20 mm
Solution
31.177KN
RDX
15KN
25.98KN
RDY
(b) Shear stress in the pin
∑FY = 0 +
∑FX = 0 +
RDX – 15 – 31.177 = 0 RDY – 25.98 = 0
RDX = 46.177 KN
RD =
RDY = 25.98 KN
52.98 KN
τ = 84.32 MPa
Sample Problem 3
Two wooden planks, each 22mm thick and 160mm wide, are
joined by the glued mortise joint shown. Knowing that the joint
will fail when the average shearing stress in the glue reached
820 KPa, determine the smallest allowable length d of the cuts
if the joint is to withstand an axial load of magnitude P = 7.6 kN.
Solution
Given: τ = 820 KPA
P = 7.6 KN
Req’d: d = ?
Identify the resisting area:
A = 7[d(22 mm)]
d = 60.20 mm
Surfaces that will
have sliding
P.S. Please check Google Classroom
for your Assignment No. 2.
Have a good day!