Date post: | 15-Mar-2016 |
Category: |
Documents |
Upload: | marsden-lopez |
View: | 60 times |
Download: | 3 times |
Analysis of Basic Load Cases
• Axial Stress• Tension and Compression
• Shear Stress• Examples
• Bending•Tension/Compression & Shear
• Torsion• Shear stress
Axial Stress
axial FA
FLAE
;
F= Axial Force (Newtons, N)A = Cross-Sectional Area Perpendicular to “F” (mm2)E = Young’s Modulus of Material, MPaL = Original Length of Component, mm
= Average Stress (N/mm2 or MPa) = Total Deformation (mm)
AE = “Axial Stiffness of Component”
Direct Shear Stress
average
PA
average = Avearge Shear Stress (MPa)P = Shear LoadA = Area of Material Resisting “P”
Examples of Direct Shear Stress
Bolted Joint withTwo Shear Planes.
P = 50 KND = 13 mm
avg = ?
Area of bolt (Ab) = D2 / 4 = 13)2 / 4 = 132.7 mm2
A resisting shear = 2 Ab
avg = P / 2Ab = 50000 N/ 2(132.7) mm2 = 188.4 MPa
D
Direct Shear II
175
150
1313
Fillet WeldFind the load P, such that the stress in the weld does not exceed the allowable stress limit of 80 MPa.
9.2
Solution:
avg = P / Aw = 80 MPa
Aw = Throat x Total Length = (9.2)(175)(2) = 3217 mm2
P / 3217 = 80 MPaP = (3217)(80) N = 257386 N = 257 kN
BENDING
C
TNeutral Plane
Before
After
Displacement in Beamy
xv
Curvature of Beam = d vdx
MEI
2
2
M = Moment; EI = Bending stiffness of Beam
Bending Stress
M y
I
max (C)
cc
ct
max (T)
y
max M c
I
Source of Internal Moment
F1
F21 m M
V
x
Fc
Ft
d
Fc = Ft
M= Ft d
Equivalent FBD
F1
F21 m
V
x
dF
F
M = Fd
Bending Stiffness
EI = Bending StiffnessE = Young’s Modulus (Material Dependant)I = Moment of Inertia (2nd Moment of Area)
XN eu tra l A x is
X Ixx = bh3 / 12 (mm4)h
Iyy= hb3 / 12 (mm4)
y
yb
Moment of Inertia
X
A
yx ' x '
X
Parallel Axis TheoremIXX = Ix’x’ + A y2
AIf X-X is the neutral axis:
A y = 0
Locate the Neutral Axis and find the Moment of Inertia for the “T” section shown below. Consider the XX axis, all dimensions are mm.
X X
300
50
60
250
Ct
Ans: Ct = 200 mm Ixx = 2.50x108 mm4
Try it!
Neutral Axis
X X
300
50
60
250
Ct
Ok...
Ct/2
2/60 tt CC
2/25060250 tt CC
0
2502550300
tC
mmCt 200
Mech 422 – Stress and Strain Analysis
X X
300
5060
250
200
Ok...
75
48
7677
23
23
23
1050.2
1044.81013.31044.81081.7
)75)(50)(300(12/)50)(300(
)75)(250)(60(12/)250)(60(
12/
mmx
xxxx
AybhI xx
75
Determine the Bending Stiffness of beamswith this cross section made of:
1) Steel, E=203 000 MPa 2) Aluminum Alloy, E= 72 000 MPa
3) Glass Reinforced Polyester, E = 30 000 MPa
Determine the Bending Stiffness of beamswith this cross section made of:
1) Steel, E=203 000 MPa2) Aluminum Alloy, E= 72 000 MPa3) Glass Reinforced Polyester, E = 30 000 MPa
Ans:1) EI = 5.08 x 1013 Nmm2
2) EI = 1.80 x 1013 Nmm2
3) EI = 0.75 x 1013 Nmm2
In the ratio 1 : 0.36 : 0.15
Bending Stress:
3 m
1 m
30 0 0 K N
100 KN200 KN
M max = 200 KN.m max M c
I
tension = 200x106 N.mm (200) mm / 2.50 x108 mm4
= 160 MPacompression = 200x106 N.mm (100) mm / 2.50 x108 mm4
= 80 MPa
MAX
300 kN
Bending Shear
VA
F1
F21 m
V
x
Bending Shear
VA
F1
F21 m
V
x Fy = 0
V
Bending Shear
VV
V
V
A
F1
F21 m
V
x
Fy = 0 Fx = 0 M = 0
Shear Stress - Bending
b
h
max
max = 1.5 V / A avg = V / A
A = b h
yFor a Rectangle:
Max Shear Stress is at N.A.
avg
Shear in Bending
b = width of the X-section at the plane of interest.
y = V Q / I bQ = 1st Moment of Area
In General:
X X
b @ N.A.
b @ top of web
1st Moment of Area
y
A=zw
w
z
Qy = A d
• Consider all of the X -section above (or below)
the plane of interest.d
NA
Try It!
Find The 1st Moment of Area at the Neutral Axis.(dimensions are mm.)
X X
300
50
60
250
Ct
Solution:The 1st Moment of Area at the Neutral Axis:(dimensions are mm.)
X X
300
50
60
250 200
75
25
Q = Ay= (300)(50)(75) + (60)(50)(25)= 1.2 x 106 mm3
Note: At the N.A. b=60 mm
Shear Stress:
3 m
1 m
3 0 0 0 K N
100 kN200kN
V max = 2000 KN y = V Q / I b
max = (200x103 N) (1.2x106 mm3) (2.50x108 mm4 (60) mm
avg = V/Aweb = (200x103 N) / (60)(300) mm2
= 11.1 MPa
= 16 MPa
300 kN
SummaryThe “Maximum” Stress Distributions in the beam are:
X X
300
50
60
250
Bending Shear
80 MPa
160 MPaTension
Compression
max=16 MPa
N.A.
Torsion
TrJ
TLJG
r
T = Torque, G = Shear Modulus of ElasticityL = Length of Shaft, J = Polar Moment of Inertia is in Radians!
Shear Stress Distribution
max
r
D
Polar Moment of InertiaCircle: J =
32
Tube: J = 32
D
D Do i
4
4 4( )
DoDi
Shear Modulus, G
GE
2 1( )
E = Young’s Modulusv = Poisson’s Ratio
Example: SteelE = 203 000 MPa, v = 0.3G = 78 000 MPa
Shear Stress-Strain CurveSh
ear S
tress
,
Shear Strain,
Shear Modulus, G
Try it!Determine the angle of twist for the steel shaft shownbelow. Calculate the safety factor against yield. The shearstrength of the steel is 200 MPa, and the Young’s Modulusis 203000 MPa.
500 mm
25 KN
150 mm
Torque
50 mm
Solution:T = 25x103 N (150 mm) = 3.75x106 N.mm
L = 500 mm
J = 32
D4
= (50)4 / 32 = 6.14x105 mm4
TLJG
3.75x106 N.mm (500) mm 6.14x105 mm4 (78000 N/mm2)
= = 3.9x10-2 = 2.2o
TrJ
= 3.75x106 N.mm (25 mm)6.14x105 mm4
= 152.7 MPa
FS = 200/152.7 = 1.31
SuperpositionAssume the beam in our example is made of steel with a yield stress of 350 MPa. If it is subjected to an additional Axial Tension of 5000 kN along it N.A., will it yield ?
3 m
1 m
100 kN200 kN
F = 5000 KN
300 kN
Result:
Bending
80 MPa
160 MPaTension
Compression
N.A.
AxialF/A = 167 MPaTension
+ =
83 MPa (Tension)
327 MPa
Tension
Net
< 350 MPaNo Yield!
Rule for Adding Stresses:
Like stresses at a point acting in the same direction and on the same plane can
be added algebraically.
• You can’t add a shear stress to a tensile/compressive stress.• You can’t add a stress in one location to one at another.• The effects of combined shear and tension/compression are covered later in this course.