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

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

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

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

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

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.

Compression Example Unstressed Sponge Sponge in Compression

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.

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.

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.

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.

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.

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

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

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

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

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

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

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,

Shearing Stress Examples Single Shear F F V F

A

Double Shear F V=F/2 V=F/2

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,

Bearing Stress Example

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

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

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.

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.

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.

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.

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.

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

Brittle and Ductile Response Graphs 5 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.

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.

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

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

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

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, text, p. 61 Stress & Strain: Axial Loading

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

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