Tutorial 4
MECH 101
Liang Tengfei
tfliang@ust.hk
Office phone : 2358-8811
Mobile : 6497-0191
Office hour : 14:00-15:00 Fri
1
Outline
 Example about normal stress and normal strain
 Example about shear stress, shear strain, and Hooke’s
law in shear
Example 1
The two bar truss ABC has pin supports at points A and C, which are 2.0m
apart. Members AB and BC are steel bars, pin connected at joint B. The
length of BC is 3m. A sign weighing 5.4kn is suspended from bar BC at D
and E, which are located 0.8m and 0.4m, respectively, from the ends of the
bar. Determine the required cross-sectional area of AB and the required
diameter of pin at C if the allowable stresses in tension and shear are 125
MPa and 45 Mpa.
Draw Free Body Diagram
Supposing counterclockwise moments are positive,
 Mc  0, R
AH
(2.0m)  (2.7 kN )(0.8m)  (2.7 kN )(2.6m)  0
RAH  4.590kN
F
M
 0, RCH  RAH  4.590kN
horiz
B
 0,  Rcv (3.0m)  (2.7 kN )(2.2m)  (2.7 kN )(0.4m)  0
RCV  2.340kN
Return to free body diagram of the entire truss,
F
vert
 0, RAV  RCV  2.7kN  2.7kN  0
RAV  3.060kN .
Reaction force at A and C,
RA  ( RAV )2  ( RAH )2  5.516kN
Rc  ( RCV )2  ( RCH )2  5.152kN
Tensile force in bar AB
Shear force acting on the pin at C
FAB  RA  5.516kN
Vc  Rc  5.252kN
Required area of the bar
AAB 
FAB
 allow
 44.1mm 2
Required diameter of the pin
Apin 
d pin 
Vc
2 allow
 57.2mm 2
4 Apin /   8.54mm
practice
 c  5.21MPa
Shear strain

: change in the shape of the element
Hook’s law in shear
  G
Example 2
A flexible connection consisting of rubber pads( t=9mm) bonded to steel
plates is shown in the figure. The pads are 160mm long and 80mm wide.
(a)Find the average shear strain γaver in the rubber if the force P=16kN and
the shear modulus for the rubber is G=1250kPa.
(B) Find the relative horizontal displacement δ between the interior plate and
the outer plates.
 aver 
Shear strain :
V
ab

 aver
Ge

V
abGe
In most practical situations the shear strain γ is a small angle, and in such cases
tan γ can be replaced by γ.
Shear force V=P/2,
Shear Strain:
V
 aver  ab
 aver
 

Ge
V
abGe
 0.5
Horizontal displacement δ,
hV
d  h 
 4.5mm
abGe
Shear stress and Bearing stress
Shear stress : tangential to the surface
V


Where
Average shear stress:
aver
A
F
Average bearing stress:  b  b Where
Ab
A
Fb  aL  P
Ab  dL
V
m
n
a
L
p
q
V
d 2
aL P
V

2
2
p
m
n

A
d 2
4
4
One shear
surface
Two shear
surfaces
V
F
L
p
q
V
V
F
 
A 1 d2
4
m
n
F
L
p
q
V
F
V
2
 
A 1 d2
4
Example 3
A steel strut S serving as a brace for a boat hoist transmits a compressive forces
P=54KN to the deck of a pier. The strut has a hollow square cross section with wall
thickness t=12mm, and angle θ is 40 °. A pin through the strut transmits the
compressive force from the strut to two gussets G that welded to the base plate B.
Four anchor bolts fasten the base plate to deck. The diameter of the pin is dpin=18mm,
the thickness of the gussets is tG=15mm, the thickness of the base plate is tB=8mm, the
diameter of the anchor bolts is dbolt=12mm.
Determine the following stress: (a)the bearing stress between the strut and the pin;(b) the
shear stress in the pin;(c) the bearing stress between the pin and the gussets,(d) the bearing
stress between the anchor bolts and the base plate,(e) the shear stress in the anchor bolts.
Solution
 (a)the bearing stress between the strut and the pin
= The force in the strut / the total bearing area of the strut
P
54kN
 b1 

 125MPa
2td pin 2(12mm)(18mm)
Bearing
surface
 (b) the shear stress in the pin( the pin tends
to shear the two planes between the gussets and
the strut)
Shearing
surface
 (c) the bearing stress between the pin and the gussets
The pin bear against the gussets at two locations, so the
bearing area is twice the thickness of the gussets times the pin
diameter;
Bearing
surface
(d) the bearing stress between the anchor bolts and the base plate
(The vertical component of force is transmitted to the pier by
direct bearing between the base plate and the pier; the
horizontal component, is transmitted through the anchor bolts.)
 bolt
P cos 40 (54kN )(cos 40)


 108MPa
4tB dbolt
4(8mm)(12mm)
Bearing
surface
Bearing
force
 (e) the shear stress in the anchor bolts
 bolt
P cos 40 (54kN )(cos 40)


 91.4MPa
2
2
4 dbolt / 4 4 (12mm) / 4
Shearing
surface
Shearing force
Practice
 aver  31.8MPa;
 b  41.7MPa
Practice
 max  7330 psi;
 b  12800 psi