1. Analysis of Basic Load Cases Axial Stress
Tension and Compression
Shear Stress
Examples
Bending
Tension/Compression & Shear
Torsion
Shear stress

2. s = Average Stress (N/mm2 or MPa)
Axial Stress
F= Axial Force (Newtons, N)
A = Cross-Sectional Area Perpendicular to “F” (mm2)
E = Young’s Modulus of Material, MPa
L = Original Length of Component, mm
s = Average Stress (N/mm2 or MPa)
D = Total Deformation (mm)
AE = “Axial Stiffness of Component”

3. Direct Shear Stress taverage = Avearge Shear Stress (MPa)
P = Shear Load
A = Area of Material Resisting “P”

4. Examples of Direct Shear Stress
Bolted Joint with
Two Shear Planes.
P = 50 KN
D = 13 mm
tavg = ?
Area of bolt (Ab) = p D2 / 4 = p (13)2 / 4 = mm2
A resisting shear = 2 Ab
tavg = P / 2Ab = N/ 2(132.7) mm2 = MPa

5. Direct Shear II Fillet Weld13 13
150
175
Fillet Weld
9.2
Find the load P, such that the stress in the weld does
not exceed the allowable stress limit of 80 MPa.

6. tavg = P / Aw = 80 MPa Solution: Aw = Throat x Total Length
= (9.2)(175)(2) = 3217 mm2
P / 3217 = 80 MPa
P = (3217)(80) N = N
= 257 kN

7. BENDING
Before C
After T
Neutral Plane

8. Displacement in Beam Curvature of Beam = y x v
M = Moment; EI = Bending stiffness of Beam

9. Bending Stress
smax (C) y cc s ct
smax (T)

10. Source of Internal MomentFc d Ft
Fc = Ft
M= Ft d

11. Equivalent FBD
M = Fd F d F

12. Bending Stiffness Ixx = bh3 / 12 (mm4) Iyy= hb3 / 12 (mm4)
EI = Bending Stiffness
E = Young’s Modulus (Material Dependant)
I = Moment of Inertia (2nd Moment of Area) y
Ixx = bh3 / 12 (mm4) h
Iyy= hb3 / 12 (mm4) y b

13. Moment of Inertia IXX = Ix’x’ + A y2 Parallel Axis Theorem A
If X-X is the neutral axis:
S A y = 0

14. Try it! Locate the Neutral Axis and find the Moment of
Inertia for the “T” section shown below. Consider the
XX axis, all dimensions are mm.
300
Ans: Ct = 200 mm
Ixx = 2.50x108 mm4 50 X X
250 Ct 60

15. Ok...
Neutral Axis
300 50 X X
Ct/2
250 Ct 60

16. Mech 422 – Stress and Strain Analysis
300
Ok... 50 75 X X 75
250
200 60
Mech 422 – Stress and Strain Analysis

17. Determine the Bending Stiffness of beams
with this cross section made of:
1) Steel, E= MPa
2) Aluminum Alloy, E= MPa
3) Glass Reinforced Polyester, E = MPa

18. Determine the Bending Stiffness of beams
with this cross section made of:
1) Steel, E= MPa
2) Aluminum Alloy, E= MPa
3) Glass Reinforced Polyester, E = MPa
Ans:
1) EI = 5.08 x Nmm2
2) EI = 1.80 x Nmm2
3) EI = 0.75 x Nmm2
In the ratio 1 : 0.36 : 0.15

19. Bending Stress: 300 kN M max = 200 KN.m
s tension = 200x106 N.mm (200) mm / 2.50 x108 mm4
= 160 MPa
MAX
s compression = 200x106 N.mm (100) mm / 2.50 x108 mm4
= 80 MPa

20. Bending Shear A V

21. Bending Shear A V V
S Fy = 0

22. Bending Shear V A V V V
S Fy = 0
S Fx = 0
S M = 0

23. Shear Stress - Bending t avg = V / A t max = 1.5 V / A t avg t max
For a Rectangle:
t max y
t max = 1.5 V / A
t avg = V / A
A = b h b h
t avg
Max Shear Stress is at N.A.

24. t y = V Q / I b Shear in Bending Q = 1st Moment of Area
In General:
top of web
t y = V Q / I b X X
Q = 1st Moment of Area
N.A.
b = width of the X-section
at the plane of interest.

25. 1st Moment of Area Qy = A d A=zw w Consider all of the
X -section above (or below)
the plane of interest. z d y NA
A=zw
Qy = A d

26. Find The 1st Moment of Area at the Neutral Axis. (dimensions are mm.)
300 50 X X
250 Ct 60

27. Q = S Ay The 1st Moment of Area at the Neutral Axis:
(dimensions are mm.)
Q = S Ay
300
= (300)(50)(75)
+ (60)(50)(25)
= 1.2 x 106 mm3
Note:
At the N.A. b=60 mm 50 75 X X 25
250
200 60

28. Shear Stress: t y = V Q / I b t max = (200x103 N) (1.2x106 mm3)
300 kN
100 kN
200kN
t y = V Q / I b
V max = 2000 KN
t max = (200x103 N) (1.2x106 mm3)
(2.50x108 mm4 (60) mm
t avg = V/Aweb = (200x103 N) / (60)(300) mm2
= 11.1 MPa
= 16 MPa

29. Bending Shear The “Maximum” Stress Distributions in the beam are: 300
Compression
80 MPa
tmax=16 MPa 50
N.A. X X
250 60
160 MPa
Tension

30. Torsion r T = Torque, G = Shear Modulus of Elasticity
L = Length of Shaft, J = Polar Moment of Inertia
q is in Radians!

31. Shear Stress Distribution
tmax r D

32. Polar Moment of InertiaDo Di

33. Shear Modulus, G E = Young’s Modulus v = Poisson’s Ratio
Example: Steel
E = MPa, v = 0.3
G = MPa

34. Shear Stress-Strain Curve
Shear Stress, t
Shear Modulus, G
Shear Strain, g

35. 500 mm
Torque
150 mm
50 mm
25 KN

36. T = 25x103 N (150 mm) = 3.75x106 N.mm
L = 500 mm
= p (50)4 / 32 = 6.14x105 mm4
3.75x106 N.mm (500) mm =
= 3.9x10-2 = 2.2o
6.14x105 mm4 (78000 N/mm2)
3.75x106 N.mm (25 mm) =
= MPa
6.14x105 mm4
FS = 200/152.7 = 1.31

37. Superposition
Assume 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 ?
300 kN
F = 5000 KN
200 kN
100 kN

38. Result: Axial Net + Bending = Tension Tension Compression 80 MPa
F/A = 167 MPa
80 MPa
83 MPa (Tension)
Tension
N.A. + =
Tension
< 350 MPa
No Yield!
160 MPa
327 MPa
Tension

39. 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.