Transcript MECH 101 - Hong Kong University of Science and Technology

Tutorial 7_Review
MECH 101
Liang Tengfei
[email protected]
Office phone : 2358-8811
Mobile : 6497-0191
Office hour : 14:00-15:00 Fri
1
A chance to show what you have learned:
 Statics
structure in Equilibrium → the forces atcing on it
 Mechanics of material
the force → stress and strain in each point →
deformation & break or not
Statics
Free body
choose
Question
draw
The force you want
Free body diagram
solve
force equation +
moment equation
F  0
M  0
build
Which free body should I choose?
 remember which
force you want
 let the target force
appear in you F.D.B
 external force will
appear in the F.B.D
 Specify your free
body
Solve the force from the pin C
acting on the member DC and
AB
How to draw F.B.D?
 Only external force will
appear in the F.B.D
 Search around the F.B.
every thing contacting the F.B.
will give it force. How about
gravity??
 Draw all the force in F.B.D
if you known the direction
draw the real direction.
otherwise
direction.
→
assume a
→
Build up equilibrium equations
 Build up the equation base on the F.B.D
the sign of the force and moment is base on the direction of the force
in F.B.D
usually
force
:
same with the coordinate : +
moment :
counterclockwise : +
solve the force
 Clarify the real direction of the force.
 Use your intuition to check the answer.
Example
 The 100N weight of the rectangular plate acts at its
midpoint. Determine the reactions exerted on the plate at
B and C.
4m
C
B
A
45
O
100N
Solution:
Notice AB is a two-force member, so the reaction at B must be directed
along the line between A and B.
Solution
4m
C
B
45
FCX
O
FB
100N
FCY
Apply the equilibrium equation:
F  F cos45  F  0
F F sin 45  F 100 0
o
X
B
CX
o
Y
M
B
B
CY
 FCY 4  100 2  0

FCX  50N
FCY  50N
FB  70.7 N
Other things in statics
 Replace the distributed
load
L
L
F   f  x dx
0
d
 f ( x) xdx
0
F
 Two force member
Mechanics of material
Equilibrium
equation
statics
Internal
force
stress
Hook’s
law
observation
deformation
Equation of
compatibility
strain
Normal Stress and Normal Strain
Normal stress:
force per unit area
P
P
P

A
A
This equation is valid only if the stress is uniformly distributed
over the cross section of the bar.
Normal strain:
A
elongation per unit length


L
P
P
L

Remind strain is a dimensionless quantity
Hook’s Law and Poisson’s Ratio
Hook’s law
  E
Note: A permanent strain exists in the specimen after unloading from
the plastic region.
lateral strain

 

axial strain

Poisson’s ratio
Poisson’s ratio is also a
constant, a property of the
material, and dimensionless
P
P
Dashed means the
original shape with out
P
Example
PL
δ =
AE
Elongation → +δ, Contraction → -δ
Tension → + P, Compression → - P
Composite A-36 steel bar shown made from two
segments AB and BD. Area AAB = 600 mm2 and
ABD = 1200 mm2.
Est = 210 GPa
Determine the vertical displacement of end A
and displacement of B relative to C.
Example
P L
PABLAB
+ BC BC
ABCE
AABE
δA = δA/B + δB/C + δC/D =
=
+
PCDLCD
ACDE
+75kN x 1m x 106
600mm2(210)(103)kN/m2
+
+
+35kN x 0.75m x 106
1200mm2(210)(103)kN/m2
-45kN x 0.5m x 106
1200mm2(210)(103)kN/m2
= 0.61 mm
Displacement of B relative to C (δBC) =
+35kN x 0.75m x 106
= 0.104 mm
1200mm2(210)(103)kN/m2
Shear Stress - single shear
V
P
4P


, d is the diameter of the bolt
2
1
A
d2 d
4
F
P
Bearing stress:  b  b 
, h is the thickness of the bar or flange
Ab d  h
Shear stress:  
Shear Stress and Bearing Stress
Shear stress acts tangential to the surface of the material.
V
aL P
d 2
A

Where V 
Average shear stress:  aver 
4
A
2
2
F
Average bearing stress:  b  b Where F  aL  P A  dL
b
b
Ab
V
m
d
n
a
L
p
m
n

q
V
d 2
A
4
Shear Strain and Hooke’s law in Shear
Shear strain
When

: change in the shape of the element
is small
Hook’s law in shear
  G
Stress on inclined Plane
1
cos   (1  cos 2 )
2
2
sin  cos  
1
(sin 2 )
2
Thermal effect
T  T
Example
 The 100N weight of the rectangular plate acts at its midpoint. Determine
the reactions exerted on the plate at B and C.
 if the pin at C is connected by a double shear pin. Pin’s lenghth is 2.5cm
(0.5, 1.5 , 0.5), The shear and bearing stress limit of the pin is 100MPa &
150MPa, if the safe factor is 1.5, what the minimum diameter of the pin?
FCX  50N
4m
C
B
A
FCY  50N
45
O
100N
FB  70.7 N
 2. the bar AB has a rectangular cross-section. Its
area is 10 mm2 . AB is glued together at pq, theta =
30 degree. the shear stress limit on this surface is
50MPa, will this bar break?