Answer 16.8 x = 4.6”, y = 2.96”
16.3 Moments of Inertia
• When we considered the centroids/center of mass we assumed that the force acting on each chunk was the same (uniform gravity).
• If we have a force that varies over a surface linearly we use the second moment of inertia.
• Consider a simple case of a beam (not quite accurate). The material in the beam acts as small springs. As the beam is bent the material in the center is almost neutral, but towards the sides it is compressed or stretched.
• If we consider the beam above as a continuous force, instead of distributed springs it changes the solution (by analogy you might recognize that with centroids we could lump forces, but with second moments we cannot).
• Another example of a linearly varying force is hydrostatic pressure based on water depth. Basically we add the mass of the water directly above to find the pressure on an area.
• The general forms of moments resulting from linearly varying forces are,
• We generally calculate second moments of inertia using,
• as an example we can consider a triangle. The calculations will be to find the second moments of inertia about the x and y axis: not the centroid.
• The polar moment of inertia may be simple to calculate with rotational parts, but it can also be found using a simple relationship (note: the polar ‘J’ and the ‘o’ for the origin)
• After we have found the second moment of inertia about an axis, we can find it about another parallel axis using the parallel axis theorem.
• Consider the application of the parallel axis theorem to the triangle seen before, To find the moment of inertia about the y centroid, when all we have is the y moment of inertia about the x axis.
• NOTE: when using the parallel axis theorem, the centroid should always be used as a reference.
• Like centroids, we can calculate moments of inertia for simple areas using weighted sums (a slightly different technique to finding centroids). The basic steps are,
1. For each simple shape find the moment of inertia about some global axis. This may require the use of centroids and areas to move the axis.
2. Add moments of inertia (or subtract for cut out areas.
• As an example, let’s consider a problem (9.26 pg. 64 Beer and Johnston)
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Problem 16.9 Find the Second Moment of Inertia about the centroid for the beam section shown below.
16.4 Product of Inertia
• When we want to find the maximum moment of inertia of an area we can start by finding the product of inertia.
• This value can be either positive or negative.
• The value will be zero if either x or y is an axis of symmetry. This will allow us to find the product of inertia if we know the area and location of the centroid.
• If we want to find the maximum and minimum moment of inertia we can use the derived relationship below. We can also find the axis of these maxima and minima,
• We can also shift the moment of inertia values to other arbitrary axes using the following relationships,
• If we plot out the values as we rotate the axis, we generate a plot called Mohr’s force circle.
16.5 Internal Forces in Members
• Up to this point beams have been treated as completely rigid, in truth they are not.
• Even a simple beam with only two forces applied can have a variety of forces and moments internally. These include,
Normal force: This is the tension/compression force along the axis of the member.
Shear force: This is the force acting across the axis of the member.
Bending moment: This moment attempts to bend the beam.
• To start, we need to know the forces and moments acting on the beam, as well as the geometry of the member. The forces on the beam are often found using techniques such as the method of members.
• Internal forces and moments will be found by cutting a beam at the point of interest, and using the internal forces as a reaction to external forces. Note that these can be found at any location in the beam as they vary between pins and other supports.
• Consider problem 7.12a) in [Beer and Johnston],
• We can be a little more formal when describing the applied forces. Consider a small part of a beam as shown below. The external shear and bending moments are indicated. Both the shear force V and the bending moment M are drawn in equilibrium. And by convention, both are shown as positive. This is best remembered by looking at the reactions on the left hand side. If we are to draw the internal forces at some point in the beam they would look like those below.
• It can be useful to draw diagrams for both shear and bending moments. These can be done easily by calculating the internal forces for an arbitrary point on the beam.
• As an example lets consider problem 7.38 in [Beer and Johnston]
• NOTE: don’t section at a point load or couple on a beam. At these points the function is undefined. Instead section just before or after.
• You may recognize in the last example that the slope of the bending moment diagram was the corresponding value on the shear force diagram.
• Consider a simpler technique for developing shear force and bending moment diagrams. First draw the shear force diagram from left to right. The direction of the forces should be added directly to the diagram. Then using the values on the shear force diagram, draw the bending moment diagram.
Problem 16.10 Solve the following example using the relationship between shear and bending moments,
• When considering distributed forces, you can lump the force to find reactions, but you must not lump the forces when drawing the shear force diagrams.
• 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 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 refering to tables of values for particular materials..
16.7 Types of Stress
• Consider an example of using a pair of scissors (“shears”) to cut a piece of metal,
Problem 16.11 Given the 1” diameter bar to the left, what is the internal 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?
• These calculations are commonly used when examining rivets and bolts. Consider the figure below with the bolt in single shear.
• Now consider a bolt that is exposed to double shear. Below we can see a section view of a pin through two members.
• 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.
16.8 Stress Analysis
• Let’s consider an example problem [1.15, pg. 16, Beer and Johnston]. For this problem we will examine all links for maximum stresses.
16.9 Stress on Oblique Planes
• 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.
16.10 Generalized Stress
• 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.
16.11 Factor of Safety
• When we do basic design work we typically use a process of the basic format,
• In particular the design process often involves a factor of safety that allows for,
variations in material properties
variations in expectations load
the non-uniform distribution of stress
• The factor of safety is applied as,
• The values vary for various systems. The values given below are reasonable for static systems.
• For example let us consider a two member mechanism 1.44 from [Beer and Johnston]
• If we apply a force to a material, we create a stress, as shown before. The the effect is that it stretches or compresses somewhat. This deformation is know as strain. In very brief terms, stress causes strain.
16.13 Strain Caused by Axial Loads
• 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 lengh.
16.14 Stress Strain Curves
• For low carbon steel,
• 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 Young’s modulus value increases.
• 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.
• Creep is an effect that can lead to permanently elongated specimens,
16.15 Analysis of Members
Problem 16.12 If the rod is 6” before the load is applied, what is the new length? What load P would result in plastic deformation? What load would result in rupture?
• When a load is applied and removed in cycles the material may fatigue. This means the ultimate strength is effectively reduced. Design curves that can take this into account are found in handbooks, and look like those shown below,
• As another example, let’s consider problem 2.26 in [Beer and Johnston]
16.16 Poisson’s Ratio
• An isotropic material has the same properties in all directions. (materials such as fiber glass do not).
• 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.
Problem 16.13 What would the diameter of the bar become after the load P has been applied?
16.17 Generalized Hooke’s Law
• 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 interact.
16.18 Shear Strain
• 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 16.14 The block to the left is under a shear load. Find the shear strain if it is made of 2014 aluminum. What is the Poisson’s ratio for the material?
• Consider the example shown below from Beer and Johnson, 1992, pg. 89 #2.86,
16.19 Stress Concentrations
• Saint Venant’s Principle states that regardless of how a force is applied, when we move far enough away, the distribution becomes even. For example, if we are applying forces as point loads, they will have very high stress concentrations near the point of application. But, as we move away from the point of application, the force distribution evens out. Consider the pin in the hole where the pin applies a load of P.
• Stress concentrations are hard to predict, and this must often be done using experiments, Finite Element Analysis (FEA), or other techniques.
• During design we must pay attention to the stress concentrations. At some points the stress will be higher that the average stress.
• One technique for estimating maximum stress concentrations is to use tables derived experimentally.
Problem 16.15 Find the maximum stress that will be found in this bar with a hole in it.
• If we consider a cylindrical shaft with torques applied it will tend to rotate an angle proportional to the torque.
• We can find the torque exerted using simple stress equations for a plane cutting through the cylinder,
• Next, consider the entire length of the cylinder and deformation of an element in the shaft.
Problem 16.16 What is the maximum shear stress in the bar? If the bar is made of cold-rolled yellow brass, what is the angle of deflection along the bar?
16.21 Torsion Stress Concentrations
• We can estimate the maximum stress for two shafts connected with a fillet using the following graph,
Problem 16.17 • What is the maximum stress in the shaft shown below? The shafts shown are joined with a fillet of radius 1/16”. What is the maximum shear stress if a torque of 100 lb.in. is applied?
16.22 Pure Bending
• Consider a section of beam that has two moments applied, thus inducing bending,
• Consider a cross section of the beam being bent about the z-axis,
• Now consider the overall geometry of the beam,
Problem 16.18 For the beam below find the maximum stress and radius of curvature. The beam above is under pure bending. If the beam has the triangular cross section shown, find the maximum stress, and the radius of curvature.
16.23 Transverse Loading
• When loads are applied to beams they tend to deflect
• For a beam under a transverse load the stress is a function of the distance from the centroid,
• We can consider the shear on some axis other than the neutral axis of the beam,
Problem 16.19 Try the problem shown below from Beer and Johnson 1992, Pg. 291 #5-1. The end of the beam is loaded with a vertical shear of 250lb. This beam is made of a set of three 2x4 studs nailed together. Find the shear force in each nail.