1. Load and Stress Analysis
Section III
Load and Stress Analysis

2. Talking Points Introduction about stresses
Shearing force and bending moment diagrams
Bending, Transverse, & Torsional stresses
Compound stresses and Mohr’s circle
Stress concentration
Stresses in pressurized cylinders, rotating rings, curved beams, & contact

3. i. Static Equilibrium and Free-Body Diagram
Introduction about stresses
i. Static Equilibrium and Free-Body Diagram
Assume downward force as negative and upward force as positive; and counterclockwise moment as positive and clockwise as negative.
Loads may act on multiple planes.

4. ii. Direct Normal Stress & Strain
Introduction about stresses – Cont.
ii. Direct Normal Stress & Strain
Assuming elasticity
The load is applied along the axis of the bar (perpendicular to the cross-sectional area) and it is uniformly distributed across the cross-sectional area of the bar.
The normal stress can be tensile (+) or compressive (-) depending on the direction of the applied load P.
The stress unit in N/m2 or Pa or multiple of this unit, i.e. MPa, GPa.
Hooke’s Law

5. Introduction about stresses – Cont.
Sometimes, a body is subjected to a number of forces acting on its outer edges as well as at some other sections, along the length of the body. In such case, the forces are split up, and their effects are considered on individual sections. The resulting deformation of the body is equal to the algebraic sum of the deformation of the individual sections. Such a principle of finding out the resultant deformation is called the principle of superposition.
Principle of Superposition: L1 L2 L3 d3 d1 d2 L1 L2 L3 P3 P1 P2 P4 d3 d1 d2

6. Introduction about stresses – Cont.
Example on Principle of Superposition:
A brass bar, having cross sectional area of 10 cm2 is subjected to axial forces as shown in the figure. Find the total elongation of the bar (dL). Take E = 80 GPa.
dL = -150 mm

7. Introduction about stresses – Cont.
iii. Poisson’s Ratio
For 1D stress system ( )
1D stress system or
From Hooke’s Law:
For 2D stress system ( )
and
For engineering materials, n = 0.25 to 0.33.
For a rounded bar, the lateral strain is equal to the reduction in the bar diameter divided by the original diameter.

8. Introduction about stresses – Cont.
Example on Poisson’s Ratio:
A 500 mm long, 16 mm diameter rod made of a homogenous, isotropic material is observed to increase in length by 300 mm, and to decrease in diameter by 2.4 mm when subjected to an axial 12 kN load. Determine the modulus of elasticity and Poisson’s ratio of the material.
E = 99.5 GPa
n = 0.25

9. iii. Direct Shear Stress & Strain
Introduction about stresses – Cont.
iii. Direct Shear Stress & Strain
Assuming elasticity Q
The load, here, is applied in a direction parallel to the cross-sectional area of the bar. Q
G is known as modulus of rigidity
Relation between E, G, and n
Single & Double Shear
The rivet is subjected to single shear
The rivet is subjected to double shear

10. Relationship between shear force and bending moment
Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams
Sign Convention
Simply supported beam
Cantilever beam
Relationship between shear force and bending moment Or

11. Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples
i. Concentrated Load Only:

12. Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples
ii. Distributed Load Only:

13. Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples
iii. Combination of Concentrated and Distributed Load:

14. Shearing Force (S.F.) and Bending Moment (B.M.) Diagrams - Examples
iv. If Couple or Moment is Applied:

15. Bending, Transverse, & Torsional stresses
i. Bending Stress
where, M is the applied bending moment (B.M.) at a transverse
section, I is the second moment of area of the beam cross-section
about the neutral axis (N.A.), i.e , s is the stress
at distance y from the N.A. of the beam cross-section.

16. Bending, Transverse, & Torsional stresses – Cont.
ii. Transverse Stress or
where Q is the applied vertical shear force at that section; A
is the area of cross-section “above” y, i.e. the area between y
and the outside of the section, which may be above or below
the neutral axis (N.A.); y is the distance of the centroid of
area A from the N.A.; I is the second moment of area of the
complete cross-section; and b is the breadth of the section at
position y. R b d

17. Bending, Transverse, & Torsional stresses – Cont.
iii. Torsional Stress
where T is the applied external torque; r is the radial direction
from the shaft center; J is the polar second moment of area of
shaft cross-section; r is the shaft radius; and t is the shear
stress at radius r.
Note: when torsion is present
Ductile materials tends to break in a plane perpendicular to its longitudinal axis; while brittle material breaks along surfaces perpendicular to direction where tension is maximum; i.e. along surfaces forming 45o angle with longitudinal axis.
Solid section
Hollow shaft
Ductile material
Brittle material

18. Compound stresses and Mohr’s circle
Machine Design involves among other considerations, the proper sizing of a machine member to safely withstand the maximum stress which is induced within the member when it is subjected separately or to any combination of bending, torsion, axial, or transverse load.

19. Compound stresses and Mohr’s circle – Cont.
Maximum & Minimum Normal Stresses
For 2D Case:
Stress State
3D General Stress State
Where:
sx is a stress at a critical point in tension or compression normal to the cross section under consideration, and may be either bending or axial load, or a combination of the two.
sy is a stress at the same critical point and in direction normal to the sx stress.
txy is the shear stress at the same critical point acting in the plane normal to the Y axis (which is the XZ plane) and in a plane normal to the X axis (which is the YZ plane). This shear stress may be due to a torsional moment, transverse load, or a combination of the two.
sn(max) and sn(min) are called principal stresses and occurs on planes that are at 90° to each other, called principle planes also planes of zero shear.
Note: sx, sy, sz all +ve, txy, tyx, tzy, tyz, txz, tzx all +ve. Due to static balance, txy = tyx, tzy = tyz, and txz = tzx.
Clockwise (CW)
2D Stress State
Counterclockwise (CCW)

20. Compound stresses and Mohr’s circle – Cont.
Maximum Shear Stresses (tmax)
tmax at the critical point being investigated is equal to half of the greatest difference of any of the three principal stresses. For the case of two-dimensional loading on a particle causing a two-dimensional stresses; The planes of maximum shear are inclined at 45° with the principal planes.
The planes of maximum shear are inclined at 45° with the principal planes.
The angle between the principal plane and the X-Y plane is defined by:

21. Compound stresses and Mohr’s circle – Cont.
It is a graphical method to find the maximum and minimum normal stresses and maximum shear stress of any member.
From the diagram:
sx = OA, txy = AB, sy =OC, and tyx = CD. The line BED
is the diameter of Mohr's circle with center at E on the s
axis. Point B represents the stress coordinates sx, txy on
the X faces and point D the stress coordinates sy, tyx on
the Y faces. Thus EB corresponds to the X-axis and ED to
the Y-axis. The maximum principal normal stress smax
occurs at F, and the minimum principal normal stress
smin at G. The two extreme-value shear stresses one
clockwise and one counterclockwise, occurs at H and I,
respectively. We can construct this diagram with
compass and scale and find the required information
with the aid of scales. A semi-graphical approach is
easier and quicker and offer fewer opportunities for
error.
2-D

22. Compound stresses and Mohr’s circle – Cont.
tmax is equal to half of the greatest difference of any of the three principal stresses. In the case of the below figure:
where,
Principal Element
Min
Max
True views on the various faces of the principal element
3-D

23. Compound stresses and Mohr’s circle – Examples.
A machine member 50 mm diameter by 250 mm long is supported at one end as a cantilever. In this example note that sy = 0 at the critical point.
Case 2: Bending only:
Case 1: Axial load only:
In this case all points in the member are subjected to
the same stress.

24. Compound stresses and Mohr’s circle – Examples.
Case 3: Torsion only:
Case 4: Bending & Axial Load :
In this case the critical point occur along the outer
surface of the member.

25. Compound stresses and Mohr’s circle – Examples.
Case 5: Bending & Torsion:
Case 6: Torsion & Axial Load :

26. Compound stresses and Mohr’s circle – Examples.
Case 7: Bending, Axial Load, and Torsion:

27. Compound stresses and Mohr’s circle – Examples.
Example on Mohr’s circle:
The stress element shown in figure has sx = 80 MPa and txy, = 50 MPa (CW). Find the principal stresses and directions.
Locate sx = 80 MPa along the s axis. Then from sx,
locate txy = 50 MPa in the (CW) direction of the t axis to
establish point A. Corresponding to sy = 0, locate tyx = 50
MPa in the (CCW) direction along the t axis to obtain point
D. The line AD forms the diameter of the required circle
which can now be drawn. The intersection of the circle
with the s axis defines smax and smin as shown.

28. Stress Concentration
Occurs when there is sudden changes in cross-sections of members under consideration. Such as holes, grooves, notches of various kinds.
The regions of these sudden changes are called areas of stress concentration.
Stress-concentration factor (Kt or Kts)
The analysis of geometric shapes to determine stress-concentration factors is a difficult problem, and not many solutions can be found.
Theoretically

29. Stresses in pressurized cylinders, rotating rings, curved beams, & contact