Week 1-2 slides
16
Dr. Li Chang Room s503, Building J07 Tel: 9351-5572 e-mail: [email protected] Mechanics of Solids Review Statics
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Transcript of Week 1-2 slides
Dr. Li Chang Room s503, Building J07 Tel: 9351-5572 e-mail: [email protected]
Mechanics of Solids
Review Statics
Equilibrium: a state of no acceleration, in either translational or rotational sense.
3. Free-body diagram
4. Equations of equilibrium
F
F
Fy
x
y M Fx
notation
Review: Statics
2. Support reactions
1. External Loads
Example: Determine the forces acting on the connecting pin of each joint. (all angles are 45o or 90o and the length of AB is L).
Free-body diagram
Statics: Example – Joint Equilibrium (Method of Joints)
a. Equations of Equilibrium: Sign/notation
b. Equations of Moment: The reference point
Example: Determine the forces acting on the connecting pin of each joint. (all angles are 45o or 90o and the length of AB is L).
Statics: Example – Joint Equilibrium (Method of Joints)
Free-body diagram
2-Force Members
Statics: Example – Joint Equilibrium (Method of Joints)
Example: Determine the forces acting on the connecting pin of each joint. (all angles are 45o or 90o and the length of AB is L).
Pin-joint D: FDA = FAD = 0
FDB = FBD = 10 kN
FDE = FED = 10 kN
Statics: Example – Joint Equilibrium (Method of Joints)
Example: Determine the forces acting on the connecting pin of each joint. (all angles are 45o or 90o and the length of AB is L).
Example: determine the forces carried by members DE, DB, and AB.
Statics: Example – Method of Sections
2-Force Members
Free-body diagram Method of Sections:
Method of Joints:
Equations of equilibrium
Mechanics of Solids
Internal Resultant Loadings
F
a
F
a
2. Support reactions
3. Free-body diagram
4. Equations of equilibrium
where everything in a body (part of it) is in equilibrium Equilibrium: a state of no acceleration, in either translational or rotational sense.
a
a
Fy
M Fx
x
y
Internal resultant loadings
1. External Loads
Mechanics of materials: a branch of mechanics that studies the internal effects of stress and strain in a solid body that is subjected to an external loading.
a
a
Fy
M Fx
Fx : Normal Force (N, FN) a
a
Coplanar Loadings
Fy : Shear Force (V, FV)
M : Bending Moment
x
y
Mechanics of materials: external loads and intensity of internal forces
Internal resultant loadings
a
a
Fz : Normal Force (N, FN)
3-D
Mx : Bending Moment a
a
z
y
x
o
Fx : Shear Force (V)
Fy : Shear Force (V, FV)
Mx
My Fz
My: Bending Moment
Mz : Torsional Moment or torque (T)
Mechanics of materials: external loads and intensity of internal forces
Fy
Mz Fx
Internal resultant loadings
2. Support reactions
3. Free-body diagram
4. Equations of equilibrium
a
a
Mechanics of materials: external loads and intensity of internal forces
where everything in a body (part of it) is in equilibrium Equilibrium: a state of no acceleration, in either translational or rotational sense.
a
a
z
y
x
o Mz
My
Fx
Fy
Mx
Fz
F
Internal resultant loadings
1. External Loads
1-27 The pipe has a mass of 12kg/m. Determine the resultant internal loadings acting on the cross section at B. (p. 21)
z B
60 N
60N
Free-body diagram
Equations of equilibrium A
A
Internal resultant loadings
1-23 The pipe has a mass of 12kg/m. Determine the resultant internal loadings acting on the cross section at B. (p. 19)
z B
60 N
60N
Free-body diagram
Equations of equilibrium
x y
z
MBx MBz
FBy
FBz
MBy
FBx
Fw1
Fw2
Fw1=12 x 0.2 x 9.8= 23.52 [N]
Fw2=12 x 0.4 x 9.8= 47.04 [N]
FBx = 0
FBy = 0
FBz= 70.56 N
Internal resultant loadings
1-23 The pipe has a mass of 12kg/m. Determine the resultant internal loadings acting on the cross section at B. (p. 19)
z B
Free-body diagram
Equations of equilibrium
x y
z
MBx MBz
FBy
FBz
MBy
FBx
Fw1
Fw2
Fw1=12 x 0.2 x 9.8= 23.52 [N]
Fw2=12 x 0.4 x 9.8= 47.04 [N]
TBx – 47.04 x 0.2 = 0
TBx =9.41 [Nm]
MBy + 23.52 x 0.1+ 47.04 x 0.2 – 18 = 0
MBy + 23.52 x 0.1+ 47.04 x 0.2 – 18 = 6.24 [Nm]
MBz = 0
Internal resultant loadings
Example Determine the resultant internal loadings at D, E and F.
4 kN
4 kN
4 kN
D
E
F
B
A
1.2 m 1.2 m
1.8 m
1.8 m C
Internal resultant loadings
