Load and Stress Analysis

1. Load and Stress Analysis Section III

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. 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. 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. 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. 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. , sis 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 Qis 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 Tis the applied external torque; r is the radial direction from the shaft center; Jis the polar second moment of area of shaft cross-section; r is the shaft radius; and tis 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, szall+ve, txy, tyx, tzy, tyz, txz, tzxall+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. Mohr’s Circle • 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. Example: 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