Shear & Bending Moment Diagrams
T13.1
Internal Resultant & Stress Pairings
Force [F]
Internal resultant
1 normal component
2 tangential components
P V
1 normal component
t
normal stress [F/L2]
Moment [F-L]
Internal resultant
1 normal component
2 tangenti
Rotation due to Torsion
fA fAB = absolute rotation @ A = relative angle of twist between A & B
T12.2
Angle of Twist, f
L
c
Special Case
Homogeneity in G, J, and T
G(x)=G J(x)=J T(x)=T
A 0.75T
G1, J1
T ( x) T TL dx dx J(x)G(x) JG 0 JG 0
L
L
B 0.25T
G2, J
Stress due to Torsion
T12.1
Internal Resultant & Stress Pairings
Force [F]
Internal resultant
1 normal component
2 tangential components
P V
1 normal component
normal stress [F/L2]
Moment [F-L]
Internal resultant
1 normal component
2 tangential componen
Pressure Vessels
T11
Pressure Vessels
t r
p 0 0 0 p 0 0 0 p
p
Shear Stresses in a Pressure Vessel?
Cylindrical Pressure Vessels
h
F
x
p[2r dx] 2 h [t dx] 0
h
pr h t
l
F
x
l [2 rt ] p[ r 2 ] 0
l
pr l 2t
Cylindrical Pressure Vessel Failure Surfaces
S
Superposition
Statically Indeterminate
T10
Examples of Superposition
A,E,r
L
F F vend,F 1 vend,1
F
F Look at the back inside cover of Hibbeler
g
Superposition is Valid if
1) Material is linear elastic [material linearity] 2) Small displacements [geometric
(Axial) Stress Concentrations
T09.1
Saint Venants Principle
Maximum Stress
Photoelasticity [CE 281]
Finite Element Method (FEM) [CE 254]
xx xy xz fx 0 x y z yx yy yz fy 0 x y z zx zy zz fz 0 x y z yz zy
Analytical [CE 201]
zx xz xy yx
Stress Concentrati
Connections
T08
P
Define Failure
s (t)
e (g)
Example
How can this connection fail? For each failure mode determine the relationship between the applied load, P, and the failure stress on the failure surface.
P
dr
h
P
dh
w t
P
Tension
Shear
Bearing
Shear
T
Constitutive Relations
T07
Lab Experiments
Tension Test (E) Torsion Test (G)
Red vectors depict loading applied by test apparatus on specimen.
Tension Test
P
s=P/A D A
Not a well defined yield plateau
L
s
Well defined yield plateau
e
e=D/L
Typical Tension
Strain
T06
Basics
Strain is not a vector
xx xy xz
(gij is not a component of strain, but gij is) Units of strain = Length per unit length [L/L]
in/in, m/m units of are often provided, i.e., unitless
% is often used: 0.2% strain = 0.002
xy xz xx 1 y
Mohrs Circle
T05.2
Does this bar have shear stresses in it?
Does this bar have shear stresses in it?
Does this bar have shear stresses in it?
Does this bar have shear stresses in it?
Cant always do what we just did nor do you generally want to!
yy
xy
'
Stress
T05.1
Basics
Stress is not a vector
xx xy xz xx xy xz xy yy yz xy yy yz xz yz zz xz yz zz
x
z
y
Units = Force per unit Area (F/L2)
psi = lbs/in2 ksi = kips/in2 = 1000 psi Pa = N/m2 MPa = 106 Pa
Symmetry of Stress
y
x
Can Solve Directly for Str
Internal Forces
Free Body Diagrams
T04.5
Example
0.5 kN/m
Determine the internal forces acting (on a plane perpendicular to the axis of the member) at C.
1.5 kN/m
2 kN
C
3m 6m
Internal Resultant & Stress Pairings
Force [F]
Internal resultant
1 normal comp
Frames
Free Body Diagrams
T04.4
Dont Panic!
(Hibbeler, SMoM1)
(Sheppard & Tounge, Statics, 2005)
(Hibbeler, SMoM1)
Example
P=1.8 kN. Determine the moment M required to maintain equilibrium.
Example
The motion of the backhoe bucket is controlled by the hyd
Stress due to Bending Moment
T13.2
Internal Resultant & Stress Pairings
Force [F]
Internal resultant
1 normal component
2 tangential components
P V
1 normal component
normal stress [F/L2]
Moment [F-L]
Internal resultant
1 normal component
2 tangential c
Stress due to Shear Force
Part II: The Second Path
Do you recall t=V/A? That was Part I
T13.3
Internal Resultant & Stress Pairings
Force [F]
Internal resultant
1 normal component
2 tangential components
P V
1 normal component
t
normal stress [F/L2]
Momen
Beam Deflections
T13.4
Recall from T13.1:
1050
V
(lbs)
-400 -950 3912
-150
M
(lb-in)
-1600
deflected shape
Moment-Curvature Relationship
Recall from T13.2:
y
M
1 y 1 yE 1 My / I yE M EI M( x ) E ( x ) I( x )
1
Governing Differential Equation
From MATH 1
Example
~
~
A 75 kg person is climbing a 5 m long ladder. The coefficient of friction for both surfaces is 0.25. Neglect the weight of the ladder. How far across (d) can the person's center of mass move before the ladder slips?
1)'t1'l : -T :rs-C'.f~ I) J
Example
The t wo steel shafts each have a diameter o f i -in. and are supported by bearings at A, B, and ( , which allow free rotation. I f t he support at D is fixed, determine the rotation o f end B when the torque o f 60 Ib-ft is applied t o gear G. Gs
Review
T16
Review Problem
A cylindrical, thin-walled, aluminum, pressure vessel and its contents are lifted by cables as shown below. The mean diameter of the cylinder is 600 mm and its wall thickness is 6 mm. The vessel is pressurized to 0.50 MPa and the