Transcript Slide 1
Shear stress versus shear strain
For homogeneous & isotropic materials pure shear causes angular distortion of a
material as shown above.
Pure shear is most often studied by torsional tests of thin bars. Behavior is completely
analogous to normal stresses.
Shear stress-strain diagram
G is shear modulus or
modulus of rigidity
G
E
G
2(1 )
The shear stress-strain
diagram for a
titanium alloy bar
tested in torsion is
shown in the Figure.
Determine
i)
Shear modulus
ii) Proportional Limit
iii) Ultimate stress
iv) Yield stress
v) Fracture stress
A bearing pad used to support machines is shown. The following formulae can be
used to determine shear stress, shear strain and displacement.
aver
aver
G
V
ab
V
abG
V
d h tan h tan
abG
For small angles
d h
hV
abG
Problem (4): The plastic block shown of 50 mm depth is bonded to a rigid support and to a
vertical plate to which a 240-kN load P is applied. Knowing that for the plastic used G=1000
MPa, determine the deflection of the plate.
80 mm
120 mm
Rigid
support
P
Spring 2006 Mid Term Question
Problem (4): The plastic block shown of 50 mm depth is bonded to a rigid support and to a
vertical plate to which a 240-kN load P is applied. Knowing that for the plastic used G=1000
MPa, determine the deflection of the plate.
80 mm
aver
120 mm
V
240 ,000
40 MPa
ab 0.05 * 0.12
aver
40
0.04 rads
G
1000
Rigid
support
P
Spring 2006 Mid Term Question
d h 80* 0.04 3.2mm
Allowable Stress
Factor of safety
fail
F.S.
allow
fail
F .S.
allow
allow
allow
F.S.=1
fail
F.S.
fail
F.S.
F.S.=3
Designing for allowable stress
Bearing
(compression)
A
P
allow
Tension
A
V
allow
Shear
Problem (6): Determine the intensity w of the maximum distributed load that can be
supported by the hanger assembly so that an allowable shear stress of allow = 15 ksi is
not exceeded in the 0.40-in.-diameter bolts at A and B, and an allowable tensile stress of
σallow = 25 ksi is not exceeded in the 0.5-in.-diameter rod AB.
Spring 2006 Mid Term Question Solve at home – will cover in help session on Sept 7th