1. Solid Mechanics Strength of Materials Mechanics of Materials
The relations between externally applied load and their internal effects on bodies.
Deformations
Engineering Mechanics
deals with the external effects of forces on rigid bodies
Rigid body
The body for which the change in shape (deformation) can be neglected

2. Learning Objectives Identify compressive and tensile forces
Identify brittle and ductile characteristics
Calculate the moment of inertia
Calculate the modulus of elasticity
Define Normal Stress and Strain
Calculate stresses and deformations

3. Elasticity
When a material returns to its original shape after removing the load
Example: rubber bands

4. Elastic Material Properties
Unstressed Wire
Apply Small Load
Remove the load
Material Returns to
Original Dimensions

5. Inelastic (Plastic) Material Properties
Bottle Undergoing
Compressive
Stress
Inelastic
Response
Unstressed
Bottle

6. Compression Applied load that squeezes the material
Example: compressive stresses can crush an aluminum can
Students become more involved if you bring in samples that demonstrate the stress. Bring plastic bottles, aluminum cans, sponges, rags, etc for students to participate with the lecture.
This is a fairly intuitive stress that can be easily identified in everyday structures such as bridges or buildings. You might ask students to name building materials in the room that are under compressive stress.

7. Compression Example
Unstressed Sponge
Sponge in Compression

8. Compressive Failure
This paper tube was crushed, leaving an accordion-like failure
This manila file folder tube was compression stress characterized. A load was continually added until the tube crushed.
Characterizations like this are constantly done inside factories to understand how their material will perform in the field. These measurements are then used by the Civil Engineers so they can design structures that will be able to easily withstand expected forces.

9. Tension Applied load that stretches a material
Example: tensile stresses will cause a rubber band to stretch
Students become more involved if you bring in samples that demonstrate the stress. Bring rubber bands, balloons, etc.
This is a less intuitive stress for students to grasp.

10. Tension Example Steel cables supporting I-Beams are in tension.
This freeway overpass is under construction. The vertical steel supports and horizontal I-Beams support the concrete based roadway until they can be supported by the concrete pillar seen above. The end I-Beam is held in place by support cables pulling in opposite directions. These steel cables are in tension.

11. Tensile Failure Frayed rope Most strands already failed
Prior to catastrophic fail
This frayed rope is a great example of a many strands having failed a tensile stress and about to have a catastrophic failure.

12. Tensile Failure
This magnesium test bar is tensile strained until fracture
Machine characterizes the elastic response
Data verifies manufacturing process control
Visited an industrial site where they do injection molding of magnesium. The tester clamps onto the fatter ends of the dogbone shaped rod, then slowly pulls the material apart. The tester records the amount of force (stress) and measures the amount of stretching (strain) on the test rod. The stress/strain diagram shows the rod’s elastic region, the yield stress point, and finally to failure. The yield stress point is recorded and used for statistical process control and to record experimental results.

13. Axial Stress on the Vertical Post
Force Directions
AXIAL: an applied force along the length or axis of a material
Axial and transverse describe force orientations.
An axial force can be in either compression (a vertical post supporting the center span of a bridge) or in tension (a tuned piano wire pulled taut).
A transverse force is easily visualized as a downward force in the center of a beam supported at each end. During this stress, the beam is experiencing both a compressive and tensile stress. See the next slide for an example.
Axial Stress on the Vertical Post

14. Transverse Stress on the Horizontal Aluminum Rod
Force Directions
TRANSVERSE: an applied force that causes bending or deflection
The left hand picture depicts an axial stress on the post is due to the trellis load. The stress is along the length, or axis, of the post. This example is of a compressive axial stress.
The right hand picture shows the effects of a transverse stress (weights in the can) on a length of a horizontal aluminum rod. Note a transverse stress will subject the material to both a tensile and compressive stress. The bottom of the rod is in tension (forces are stretching the outer skin material) while the top of the rod is in compression (forces are pushing skin material together).
Transverse Stress on the Horizontal Aluminum Rod

15. Which Material is Stronger?
Two equal length but different materials
Cross section area of cable 1 is 10 mm2
Cross section area of cable 2 is 1000 mm2
W=500N
Cable 1
W=5000N
Cable 2

16. Axial Loading: Normal Stress
The force intensity on that section is defined as the normal stress.

18. Find the stresses in different column segments.
The steel column in a three-storey building supports the factored loads shown in Fig. 1. If P1= 120 kN and P2= 240 kN Assume E=206 GPa. Column cross section area=85.4 cm2
Find the stresses in different column segments. 4m C D

19. B
2P1=240 kN
R1=240 kN
R1=240 kN
2P2=480 kN C B
R2=720 kN
R2=720 kN
2P2=480 kN D C
R3=1200 kN

21. Examples of axially loaded bar
Usually long and slender structural members
Truss members, hangers, bolts
Prismatic means all the cross sections are the same

22. EXAMPLE Bar width = 35 mm, thickness = 10 mm
Determine max. average normal stress in bar when subjected to loading shown.

23. Internal loading Normal force diagram
By inspection, largest loading area is BC, where PBC = 30 kN

24. Average normal stress

25. Average shear stress over each section is:P A
τavg =
τavg = average shear stress at section, assumed to be same at each pt on the section
V = internal resultant shear force at section determined from equations of equilibrium
A = area of section

26. Case discussed above is example of simple or direct shear
Caused by the direct action of applied load F
Occurs in various types of simple connections, e.g., bolts, pins, welded material

27. Shear Force And Shear Stress
Forces F are applied transversely to the Block AC.
Corresponding internal forces act in the plane of the sections V are called shearing forces.
The corresponding average shear stress is,

28. Shearing Stress Examples
Single Shear F F V F

29. A

30. Double Shear F
V=F/2
V=F/2

31. Single shear
Steel and wood joints shown below are examples of single-shear connections, also known as lap joints.
Since we assume members are thin, there are no moments caused by F

32. Single shear
For equilibrium, x-sectional area of bolt and bonding surface between the two members are subjected to single shear force, V = F
The average shear stress equation can be applied to determine average shear stress acting on colored section in (d).

33. Double shear
The joints shown below are examples of double-shear connections, often called double lap joints.
For equilibrium, x-sectional area of bolt and bonding surface between two members subjected to double shear force, V = F/2
Apply average shear stress equation to determine average shear stress acting on colored section in (d).

34. Shearing Stress Examples
Single Shear
Double Shear

35. Procedure for analysis
Internal shear
Section member at the pt where the τavg is to be determined
Draw free-body diagram
Calculate the internal shear force V
Average shear stress
Determine sectioned area A
Compute average shear stress τavg = V/A

38. EXAMPLE
The wooden strut shown below is suspended from a 10-mm-diameter steel rod, which is fastened to the wall. If the strut supports a vertical load of 5 kN, compute the average shear stress in the rod at the wall and along the two shaded planes of the strut

40. Bearing Stress in Connections
Bolts, rivets, and pins create stresses on the points of contact or bearing surfaces of the members they connect.
The resultant of the force distribution on the surface is equal and opposite to the force exerted on the pin.
Corresponding average force intensity is called the bearing stress,

41. ALLOWABLE STRESS Ffail Fallow F.S. =
When designing a structural member or mechanical element, the stress in it must be restricted to safe level
Choose an allowable load that is less than the load the member can fully support
One method used is the factor of safety (F.S.)
F.S. =
Ffail
Fallow

42. ALLOWABLE STRESS
If load applied is linearly related to stress developed within member, then F.S. can also be expressed as:
F.S. =
σfail
σallow
F.S. =
τfail
τallow
In all the equations, F.S. is chosen to be greater than 1, to avoid potential for failure.
Specific values will depend on types of material used and its intended purpose.

43. ALLOWABLE STRESS Factor of safety considerations:
uncertainty in material properties
uncertainty of loadings
uncertainty of analyses
number of loading cycles
types of failure
maintenance requirements and deterioration effects
importance of member to structures integrity
risk to life and property
influence on machine function

44. DESIGN OF SIMPLE CONNECTIONS
To determine area of section subjected to a normal force, use P
σallow
A =
To determine area of section subjected to a shear force, use
A = V
τallow

45. Cross-sectional area of a tension member
Condition:
The force has a line of action that passes through the centroid of the cross section.

46. Cross-sectional area of a connecter subjected to shear
Assumption:
If bolt is loose or clamping force of bolt is unknown, assume frictional force between plates to be negligible.

47. Required area to resist bearing
Bearing stress is normal stress produced by the compression of one surface against another.
Assumptions:
(σb)allow of concrete < (σb)allow of base plate
Bearing stress is uniformly distributed between plate and concrete

48. Required area to resist shear caused by axial load
Although actual shear-stress distribution along rod difficult to determine, we assume it is uniform.
Thus use A = V / τallow to calculate l, provided d and τallow is known.

49. Procedure for analysis
When using average normal stress and shear stress equations, consider first the section over which the critical stress is acting
Internal Loading
Section member through x-sectional area
Draw a free-body diagram of segment of member
Use equations of equilibrium to determine internal resultant force

50. Procedure for Analysis
Required Area
Based on known allowable stress, calculate required area needed to sustain load from A = P/τallow or A = V/τallow

51. EXAMPLE
The two members pinned together at B. If the pins have an allowable shear stress of τallow = 90 MPa, and allowable tensile stress of rod CB is (σt)allow = 115 MPa
Determine to nearest mm the smallest diameter of pins A and B and the diameter of rod CB necessary to support the load.

52. Draw free-body diagram:

53. Diameter of pins:

54. Diameter of pins: Choose a size larger to nearest millimeter.
dA = 7 mm
dB = 10 mm

55. Diameter of rod: Choose a size larger to nearest millimeter.
dBC = 9 mm

56. Example
The suspender rod is supported at its end by a fixed – connected circular disk as shown below. If the rod passes through a 40-mm-diameter hole, determine the minimum required diameter of the rod and the minimum thickness of the disk needed to support the 20-kN load. The allowable normal stress for the rod is σallow= 60 Mpa, and the allowable shear stress for the disk is Ʈallow= 35 Mpa.

57. The end chord of a timber truss is framed into the bottom chord as shown in Fig.1. Neglecting friction, compute the dimension “b” and “c” if the allowable shearing stress is 3 .2 MPa and the allowable bearing stress is 6.9 MPa.
End Chord
bottom Chord
Detail A

59. Find the stress in the column.
A concrete slab is supported by square column as shown in Fig. 1. The column transfers a load of 200 kN to a 1.2 m X 1.2 m footing.
What is the value of minimum thickness of the slab so that the allowable punching shear stress equal to 0.8 MPa.
Find the stress in the column.
Calculate the bearing stress in the soil below the footing.
Slab
Slab thickness “t”
Column size
400x400 mm
P= 200 kN
Footing
Soil
1.2 m
Section
Footing
Slab
Plan
Figure 1: plan and section for the column-slab-footing

60. Graphical Representation
Force vs. Deformation in the elastic region 5 10 15 20 25
Deformation, y (in x 0.01)
Steel Beam Data
Linear Regression
Materials in the elastic region are characterized by a linear deflection response to applied load force.
The slope of the line will be used in the lab to determine a material’s modulus of elasticity. This line is characterized by the line equation y=mx+b; where y is the Load, P (lbs); m is the line slope; x is the Deflection, y (inx0.01); and b, the y-intercept, is 0 because there is no deflection with no applied load.
A sharp student will ask why the independent variable is on the Y-Axis. The best answer is because the shape of the line becomes too ‘weird’ when characterized beyond the yield point. When the Load is graphed on the Y-Axis like above, the characterization curve will flatten out before finally breaking. If the Load was graphed on the X-Axis, the curve would plot almost straight up which is harder to grasp.

61. Yield Stress
The stress point where a member cannot take any more loading without failure or large amounts of deformation.
This is the point where the material characteristics leave the linear elastic region. Beyond the yield stress point, a material will either fail catastrophically or deform a large amount with incrementally increased loading.

62. Ductile Response
Beyond the yield stress point, the material responds in a non-linear fashion with lots of deformation with little applied force
Example: metal beams
Metal bends but doesn’t break making it attractive for building material. If stresses are in the metal’s elastic region such as small earthquakes, the building structure will survive with no major damage. If the stress goes beyond the elastic region, as in a major earthquake, the building structure is damaged but it doesn’t fall. Also by becoming permanently changed, it is easy to identify where the structural damage is located.

63. Ductile Example Unstressed Coat Hangar After Applied Transverse
Stress Beyond the Yield
Stress Point
The applied transverse stress exceeded the yield point and caused permanent deformation. Any additional small stress will cause a lot more deformation.

64. Brittle Response
Just beyond the yield stress point, the material immediately fails
Example: plastics and wood
Though these materials have many redeeming values, they fail in a brittle mode. Engineers have found ways to keep utilizing these materials in a safe manner by improving other aspects, such as design improvements like diagonal bracing.

65. Brittle Failure After Applied Stress Beyond the Yield Stress Point
Brittle Example
Unstressed Stick
This wooden beam has a small elastic region but broke when the applied stress was beyond the yield stress point.
Brittle Failure After Applied Stress Beyond the Yield Stress Point

66. Brittle and Ductile Response Graphs5 10 15 20 25 30 45 60
Deflection, y
Ductile Response
Brittle Response
Failure
This graph has a lot of information.
The red squares are the linear load vs. deflection of a brittle material, say a plastic sample. The stress point and the failing point are nearly identical.
The blue triangles represent the load vs. deflection of a ductile material, say a piece of steel. The yield stress point is near 20 pounds. Note by adding a small amount of additional load of four pounds results in a great amount of deflection. Eventually the material reaches its failing point.

67. Modulus of Elasticity
Quantifies a material’s resistance to deformation
Constant for a material, independent of the material’s shape.
Units are in force / area. (PSI or N/m2)
Symbol: E
Each material has its own characteristic modulus of elasticity. Designers select material with the appropriate E and cost tradeoff.
Modulus of elasticity can be found in literature and will be measured in this week’s lab. The slope of the load/deflection plot is a function of length, moment of inertia, and modulus of elasticity. Use the linear regression function in the Excel program to calculate slope, the length can be measured, the moment of inertia can be calculated from the material’s cross sectional measurements, and then solve for the modulus of elasticity.

68. Stress & Strain: Axial Loading
Normal Strain
text, p. 48
Stress & Strain: Axial Loading

69. Stress & Strain: Axial Loading
Stress-Strain Test
text, p. 50
Stress & Strain: Axial Loading

70. Stress-Strain Diagram: Ductile Materials
text, p. 52
Stress & Strain: Axial Loading

71. Deformations Under Axial Loading
From Hooke’s Law:
From the definition of strain:
Equating and solving for the deformation,
With variations in loading, cross-section or material properties,
Stress & Strain: Axial Loading

72. Find the stresses in different column segments.
The steel column in a three-storey building supports the factored loads shown in Fig. 1. If P1= 120 kN and P2= 240 kN Assume E=206 GPa. Column cross section area=85.4 cm2
Find the stresses in different column segments.
what is the total displacement of point A? 4m C D

73. 2P1=240 kN B
R1=240 kN
R1=240 kN
2P2=480 kN B
R2=720 kN C
R2=720 kN
2P2=480 kN C D
R3=1200 kN