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Tutorial 7_Review MECH 101. Liang Tengfei [email protected] Office phone : 2358-8811 Mobile : 6497-0191 Office hour : 14:00-15:00 Fri. A chance to show what you have learned:. Statics structure in Equilibrium → the forces atcing on it Mechanics of material - PowerPoint PPT Presentation
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Tutorial 7_Review MECH 101
Liang [email protected]
Office phone : 2358-8811Mobile : 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
Question
0F
0M
choose
solve build
draw
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 → assume a direction.
Build up equilibrium equationsBuild 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.
The 100N weight of the rectangular plate acts at its midpoint. Determine the reactions exerted on the plate at B and C.
B
A
C
100N
45O
Solution:
Notice AB is a two-force member, so the reaction at B must be directed along the line between A and B.
4m
Example
B C
100N
45O
BFCYF
CXF
Apply the equilibrium equation:
045cos CXo
BX FFF
010045sin CYo
BY FFF
021004 CYB FM
NFCY 50
NFCX 50
NFB 7.70
4m
Solution
Other things in staticsReplace the
distributed load
Two force member
0
L
F f x dx 0
( )L
f x xdx
dF
Mechanics of material
Internal force stress
deformation strain
statics
observation
Equilibrium equation
Equation of compatibility
Hook’s law
Normal Stress and Normal StrainNormal Stress and Normal Strain
A
P
L
Normal stress: force per unit area
Normal strain: elongation per unit length
This equation is valid only if the stress is uniformly distributed over the cross section of the bar.
A
PP
P
L
A
P
Remind strain is a dimensionless quantity
Hook’s Law and Poisson’s RatioHook’s Law and Poisson’s Ratio
E
strain axial
strain lateral
Hook’s law
Poisson’s ratio
Poisson’s ratio is also a constant, a property of the material, and dimensionless
PP
Note: A permanent strain exists in the specimen after unloading from the plastic region.
Dashed means the original shape with out 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.
δ =PL
AE
Elongation → +δ, Contraction → -δ Tension → + P, Compression → - P
ExampleExample
δA = δA/B + δB/C + δC/D = PABLAB
AABE PBCLBC
ABCE + 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
1200mm2(210)(103)kN/m2 = 0.104 mm
ExampleExample
Shear Stress - single shear
22
4, is the diameter of the bolt
14
V P Pd
A dd
Shear stress:
Bearing stress: , is the thickness of the bar or flangebb
b
F Ph
A d h
Shear Stress and Bearing StressShear Stress and Bearing Stress
A
Vaver
b
bb A
F PaLFb dLAb
Shear stress acts tangential to the surface of the material.
Average shear stress:
Average bearing stress: Where
22
PaLV
4
2dA
Where
n
V
m
q
V
p
aLnm
d
4
2dA
Shear Strain and Hooke’s law in ShearShear Strain and Hooke’s law in Shear
G
Shear strain : change in the shape of the element
is smallWhen
Hook’s law in shear
Stress on inclined Plane
)2cos1(2
1cos2 )2(sin
2
1cossin
Thermal effect
T T
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?
B
A
C
100N45
O
4m
Example
NFCY 50
NFCX 50
NFB 7.70
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?