• Stress is the force per unit area. This is because the geometry of the parts are not fixed, and we must account for the size of the cross section that is resisting a force or moment.

• So far we have dealt with tensions and compressions in members somewhat simply. If we consider a simple member as shown below, we use the force divided by the area to give us the force per unit area.

• The normal units for stress are given below.

• The material at the section of interest acts like numerous supports, each exerting internal forces. As a result the stress is statically indeterminate, but a reasonable estimate is made by assuming uniform distribution. In the internal forces are not constant over the cross section of the part.

• For two dimensional objects the internal forces of interest are shear force, normal force and bending moment.

• Using stress we can calculate deformations (strain) and failure loads. This is done by referring to tables of values for particular materials.

• Consider an example of using a pair of scissors (“shears”) to cut a piece of metal,

Problem 4.1 Given the 1” diameter bar to the left, what is the internal (normal) stress if the load P is 100KN? If the bar is made of gray cast iron, what is the difference between the ultimate tensile strength and the induced stress?

• By convention we will always define tension as a positive stress, and compression as a negative stress.

• Single Shear: These calculations are commonly used when examining rivets and bolts. Consider the example below with the bolt in single shear.

• Double Shear: Now consider a bolt that is exposed to double shear. Below we can see a section view of a pin through two members.

• Bearing Stress: We may also consider how the force on a pin effects the surrounding material. In the figure below we are finding the bearing stress. In this case the bearing stress will be a normal stress.

• The bearing stress as shown only needs to be considered when the member is in tension.

• Let’s consider an example problem [1.15, pg. 16, Beer and Johnston]. For this problem we will examine all links for maximum stresses.

• Under simple axial loads metals and other materials act like simple springs. This behavior is known as elastic

• If the material becomes permanently deformed this is known as plastic deformation.

• Consider a generic load-deflection curve,

• This curve is produced by forcing the deformation, then measuring the holding force using a special tensile tester. The specimen pictured above is fastened into a hydraulic or screw based machine. The specimen is then stretched and the applied load measured. The curves can be directly drawn from using load and deflection.

• Secondly, the graph above is often changed as shown to make it independent of geometry.

• Note that Young’s modulus is a spring constant for a unit volume of material. To get it into a traditional spring constant we must multiply it by area, and divide by length.

• While necking, the cross section at one point decreases, thus increasing the stress. In turn this continues rapidly until fracture.

• The strain hardening of some materials occurs as they are stretched, the Ultimate Tensile Strength increases, but Young’s modulus remains the same.

• Each material will have it’s own stress-strain curve and these are determined experimentally, and found in abundance in handbooks.

• If a material is brittle, it does not deform much and simply breaks. Or simply the ultimate and rupture strengths are the same.

• Ductile materials deform quite a bit before the ultimate stress, necking typically occurs before rupture.

• The area under the stress strain curve indicates toughness. A larger area under a stress strain curve will make a material that must be extensively deformed before it will fail.

• Creep is an effect that can lead to permanently elongated specimens,

• Consider the simple example below,

• As another example, let’s consider problem 2.26 in [Beer and Johnston]

• We can consider the more general case of the 3D element,

• If we assume the element is square, then we can replace the stresses with forces,

• Finally, for simplicity, we can look at the element from a single side,

• For statics, the sum of the forces and moments must total zero. As a result we would see a shear stress (τxy) induce another shear that is perpendicular (τyx).

• We also consider the case where we rotate the virtual element,

• The value of these relationships show how stress is spread throughout an object, and will become valuable as stress is examined in greater detail.

4.1 Soustas-Little, R.W. and Inman, D.J., Engineering Mechanics Statics, Prentice-Hall, 1997.

• Consider the arbitrary definition of a force, and how any force can be replaced with components. If we take a force causing an axial stress on a beam, we could replace it with two components that give the same resultant. The only complication being that as we shift the plane of application, the effective section area changes.

Problem 4.2 The two pieces of wood below are to be glued together and will carry a tensile load of 2 kip. What are the shear and axial stresses between the glued faces?

• While Young’s modulus is intended for the stress to strain relationship for Axial or normal strain, we will use the Shear modulus G for shear strain.

• First consider the general square cubic element below,

• Now, Assume a shear stress in the positive x direction that induces the deformation as shown below.

• NOTE: these calculations are based upon the assumption of small angles of deflection.

• How a material strains in shear is related to normal strain by Poisson’s Ratio,

Problem 4.3 The block to the left is under a shear load. Find the shear strain if it is made of 2014 aluminum. How much does the top move to the right if the base is fixed? What is the Poisson’s ratio for the material?

• Consider the example shown below from Beer and Johnson, 1992, pg. 89 #2.86,

• An isotropic material has the same properties in all directions. (materials such as fiber glass do not and are called anisotropic materials).

• When we apply a stress to an isotropic material in one direction, we induce stress in the perpendicular direction. The resulting ratio between perpendicular stresses, and strains, is called Poisson’s Ratio

• In physical terms: as we stretch a bar, it becomes a bit thinner. Stretching of silly putty is a good visual example.

Problem 4.4 What would the diameter of the bar become after the load P has been applied?

• When we load a material in a single direction the effect of Poisson’s ratio is naturally included in Young’s Modulus. But, when there are multiple loads in multiple directions, we must uses Poisson’s ratio to determine how they intersect.

Problem 4.5 The cubical gage block to the right is loaded on two faces with 1 kip of compression. The modulus of elasticity for the material is 1 Mpsi, and the poisson’s ratio is 0.3. What are the new outside dimensions?

4.2 Soustas-Little, R.W. and Inman, D.J., Engineering Mechanics Statics, Prentice-Hall, 1997.