1. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Stress and Strain: Prismatic metal bar in tension. An axial force is a force that coincides with the longitudinal axis of a member. Tensile forces tend to stretch or elongate a member Compressive forces tend to shorten or compress a member. Under the action of these forces, “stresses” will develop in resistance to applied force. The internal resisting force is distributed over the cross section of the element. This distribution is referenced as a force per unit area, or called “unit stress”. Since the stress is perpendicular to the cross section, it is often called NORMAL or DIRECT stress. It is usually express in psi, ksi, psf, ksf (US Customary) or Pa or Mpa (SI). Stress Equation: s = P A where s = Average stress P= external applied force (lb, k, N) A= cross sectional area over which stress develops (in 2, m 2, mm 2 )

2. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Example of calculating normal stress

3. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Allowable Stress When we are selecting or sizing members that will feel tensile stress, it is necessary to limit the amount of stress on the material. All materials will have a failure (yielding) limit. The equation for sizing a member so that the material will not be overstressed is: A = P S all Where S all is a material property that will be minimized by some factor of safety.

4. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Example: Allowable Stress The steel rod hangers are to support the pipes at a power plant. Each rod is 3/8 th in. diameter and the material has an allowable stress of 24,000 psi. Calculate the allowable tensile load on each rod.

5. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Group Work: Allowable Stress A timber column is subjected to a compressive load of 25,000 as shown. The column is supported by a 2 foot square footing which bears on soil. Compute: a.Bearing stress between column and footing b.Bearing stress under footing. Note: Neglect wt. of column/footing

6. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Stress: Members with holes When a member has holes bored thru it, it’s necessary to subtract the cross-sectional area of the holes from the A value used in calculations. The new area can be considered the “effective area”

7. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Example: Bolt through timber beam. If P=15k, what is tensile stress in bolt and bearing stress between washer and timber beam? Washer ½” 3-1/4”

8. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Shear Stress Tensile and compressive forces along the primary axis will produce stress that are normal to the cross section, as shown in the diagrams. There is yet another kind of shear that may be produced, called “shear” and the stresses will be parallel to the plane of the cross- section and also in the plane of the line of action of the force producing them. The equation for shear stress is: s s = P A Where s is the average shear stress, P is the applied force and A is the area where the shear stress develops Here are some examples of how shear stresses may develop:

9. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Figure 9-10Shear stress examples.

10. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Punching Shear Stress Examples

11. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Shear in Bolted Connections Example

12. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Example Calculation of Shear Stress in a Bolt A tractor drawbar is connected to an implement as shown. The tractor pulls with a force of 50kN that must be transmitted by shear pin. Calculate: a) The shear stress in the bolt if the diameter is 19 mm b) the percent increase in shear stress if the bolt diameter is reduced to 16 mm.

13. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Allowable Shear Stress Similar to stress from tension/compression, all materials will have a shear strength that cannot be exceeded in design. The equation used to size an element being subject to shear is: A = P S s(all)

14. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Strain and Deformation Deformations: Axial strain (tension/compression) Shear Strain (shear) Unitless Axial Strain e =δ L o Unitless Shear Strain e =δ L o

15. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Axial Strain Example For the bar shown, the total strain was measured to be.00056. Calculate the total deformation of the bar.

16. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Shear Strain Example For the block shown, P is applied at the top of the block and displaces the top a horizontal distance of.0041 inches. The height of the block is 2.6 inches. Compute the shear strain.

17. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Material Behavior Ductility is a mechanical property that describes the extent in which solid materials can be plastically deformed without fracture. Examples of ductile material includeplasticallyfracture Malleability, a similar concept, refers to a material's ability to deform under compressive stress; this is often characterized by the material's ability to form a thin sheet by hammering or rolling. Ductility and malleability do not always correlate with each other; for instance, gold is both ductile and malleable, but lead is only malleable Schematic appearance of round metal bars after tensile testing. (a) Brittle fracture (b) Ductile fracture (c) Completely ductile fracture

18. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. Material Behavior For elastic materials, each incremental increase in stress will produce a proportional increase in strain. This results in a straight line on a stress/strain diagram. Material will fully rebound. Proportional Limit is the stress at which this relationship will not longer hold true and the line will begin to curve. Exceeding this limit will also result in permanent deformation (not full rebound). Elastic Limit is stress at which the material will continue to deform plastically with no increase in stress (load). Strain Hardening: Strengthening that occurs because of movements within the crystal structure of the material. Ultimate Stress: Limit of the strain hardening region when necking begins to occur. Fracture: Sample finally ruptures/fractures THE SLOPE OF THE LINE IN THE ELASTIC REGION IS CALLED “E” AND KNOWN AS THE MODULUS OF ELASTICITY OR YOUNG’S MODULUS. E= stress/strain = s/e

19. George F. Limbrunner and Leonard Spiegel Applied Statics and Strength of Materials, 5e Copyright ©2009 by Pearson Higher Education, Inc. Upper Saddle River, New Jersey 07458 All rights reserved. ` Material Behavior A tensile member is subjected to a 5000 lb load. The member is 30 inches long and made from steel. Compute the tensile stress and the total axial deformation. Assume E=30,000,000 psi.