SIMPLE
STRESS
JANNA FEB B. SIMENE, CE
SHEAR STRESS
-force applied is parallel to the resisting area
τ = Shear Force (V)
Area
EXAMPLES
What force is required to
punch
a
20-mmdiameter hole in a plate
that is 25 mm thick? The
shear strength is 350
MN/m2 .
SOLUTION:
τ=
V
A
350MPa =
V
(π)(20)(25)
V = (350)( (π)(20)(25)
= 549,778 N
The bell crank shown is in
equilibrium. (a) Determine
the required diameter of
the connecting rod AB if
its axial stress is limited
to
100MN/m2.
(b)
Determine the shearing
stress in the pin at D if its
diameter is 20mm.
a) Determine the required diameter of the connecting
rod AB if its axial stress is limited to 100MN/m2.
SOLUTION:
∑MD=0
(P)(200)=(30sin60)(240)
P=31.18KN
σ=P
A
100MPa = (31.18KN)(1000)
(π)(D²/4)
D = 19.92mm
30cos60
30sin60
(b) Determine the shearing stress in the pin at D if its
diameter is 20mm.
SOLUTION:
∑Fx=0
P + 30cos60= RD
RDH=46.18KN
∑Fy=0
RDV = 30sin60
RDV=25.98KN
RD² = RDV² + RDH²
= 52.986KN
RDH = 46.18KN
RDV = 25.98KN
(b) Determine the shearing stress in the pin at D if its
diameter is 20mm.
SOLUTION:
τ=
=
V
A
(52.986KN)(1000)
(π)(20)²(2)
4
= 84.33MPa
30cos60
RD
30sin60
BEARING STRESS
-developed when there’s contact pressure between two bodies
EXAMPLES
In the figure shown, assume that a 20mm
diameter rivet joins the plates which are
each 100mm wide.
(a) If the allowable stresses are
140MN/m² for bearing in the plate
material and 80MN/m² for shearing
of the rivet, determine the minimum
thickness of each plate.
(b) Under the conditions specified in
part (a) , what is the largest average
tensile stress in the plate.
100mm
In the figure shown, assume that a 20mm diameter
rivet joins the plates which are each 100mm wide.
(a) If the allowable stresses are 140MN/m² for
bearing in the plate material and 80MN/m² for
shearing of the rivet, determine the minimum
thickness of each plate.
P
A
140MN/m² =
P
(20)(t)
τPLATE
=
=
τRIVET
80MN/m² =
t
P
A
P
(π)(20)²
4
P = 25,133N
P = 2800t
D=20mm
SHEAR ON PLATE
P
P
D=20mm
25,133N = 2800t
t= 8.98mm
SHEAR ON RIVET
(b) Under the conditions specified in part
(a) , what is the largest average tensile
stress in the plate.
D=20mm
τPLATE
=
P
A
=
25,133N
(100x8.98)-(20x8.98)
= 35MPa
t= 8.98mm
W=100mm
SHEAR ON PLATE
The lap joint shown in the figure is
fastened by three 20mm diameter rivets.
Assuming P= 50KN , determine
(a) The shearing stress in each rivet
(b) The bearing stress in each plate
(c) The maximum average tensile stress
in each plate.
Assume that the axial load P is
distributed equally among the three
rivets.
(a) The shearing stress in each rivet
τRIVET
=
P
A
=
(50KN)(1000)
(π)(20)² (3)
4
= 53.05MPa
50KN
50KN
(b) The bearing stress in each plate
τPLATE
=
P
A
=
(50KN)(1000)
(20) (3)(25)
= 33.33MPa
50KN
50KN
(c)The maximum average tensile stress in
each plate.
τPLATE
=
P
A
=
(50KN)(1000)
(130-20) (25)
= 18.18 MPa
D=20mm
t= 25mm
W=130mm
SAMPLE PROBLEMS
THIN-WALLED CYLINDERS OR THIN-WALLED PRESSURE VESSELS
-A tank or pipe carrying a fluid or gas under a pressure is subjected to tensile
forces, which resist bursting, developed across longitudinal and transverse
sections.
TANGENTIAL STRESS/
CIRCUMFERENTIAL STRESS
LONGITUDINAL STRESS
SPHERICAL SHELL
TANGENTIAL/CIRCUMFERENTIAL STRESS
-
Consider the tank shown being subjected to an internal pressure p. The length
of the tank is L and the wall thickness is t.
σt = inside pressure(p) x radius (r) = (p) (D)
thickness (t)
2(t)
= (pi - po) (r)
(t)
where:
T are total tension in the walls
F are the total forces from the
internal pressure, p
Take Note!
T = (σt)(Awall) = (σt)(tL)
F = pA = pDL
EXAMPLES
Calculate the minimum wall
thickness for a cylindrical vessel
that is to carry a gas at a
pressure 1400psi. The diameter
of the vessel is 2ft, and the
stress is limited to 12ksi.
SOLUTION:
σt = (p) (r)
t
(12)(1000) = (1400)(1 x 12)
t
t= 1.4in
1400psi
AXIAL OR LONGITUDINAL STRESS
The total force acting at the rear of the tank F must equal to the total
longitudinal stress on the wall PT. Due to the forces on the ends of the cylinder,
walls will stretch along the axial direction and this will induce longitudinal
stress on the cylindrical walls.
σL = pressure on the wall (p) x radius (r) = (p) (D)
2 x thickness (t)
4t
Take Note!
Take Note!
σt =2(σL)
PL= (σL)(Awall) =(σL)(Dt)
F = pA = pπD²
4
=pπr²
EXAMPLES
A cylindrical steel pressure vessel 400
mm in diameter with a wall thickness of
20 mm, is subjected to an internal
pressure of 4.5 MN/m2 . Calculate the
longitudinal stresses in the steel.
SOLUTION:
σL = (p) (D)
4t
= (4.5MPa)(0.4)
(4)(0.02)
= 22.5MPa
SPHERICAL SHELL
σs = internal pressure(p) x radius (r)
2 x thickness (t)
Take Note!
PT= (σt)(Awall) =(σt)(Dt)
F = pA = pπD²
4
=pπr²
EXAMPLES
The wall thickness of a
4ft diameter spherical
tank
is
5/16inch.
Calculate the allowable
internal pressure if the
stress is limited to
8000psi.
SOLUTION:
σs = (p) (r)
2(t)
8000psi = (p) (2 x 12)
(2)(5/16in)
= 208.33psi
SAMPLE PROBLEMS
END OF CHAPTER 1