stress ,strain & Hooke's Law

Stress, Strain & Hooke's Law - II

In our first topic, Static Equilibrium, we examined structures in which we assumed the members were rigid - rigid in the sense that we assumed that the member did not deform due to the applied loads and resulting forces. In real members, of course, we have deformation. That is, the length (and other dimensions) change due to applied loads and forces. In fact, if we look at a metal rod in simple tension as shown in diagram 1, we see that there will be an elongation (or deformation) due to the tension. If we then graph the tension (force) verses the deformation we obtain a result as shown in diagram 2.
In diagram 2, we see that, if our metal rod is tested by increasing the tension in the rod, the deformation increases. In the first region the deformation increases in proportion to the force. That is, if the amount of force is doubled, the amount of deformation is doubled. This is a form of Hooke's Law and could be written this way: F = k (deformation), where k is a constant depending on the material (and is sometimes called the spring constant). After enough force has been applied the material enters the plastic region - where the force and the deformation are not proportional, but rather a small amount of increase in force produces a large amount of deformation. In this region, the rod often begins to 'neck down', that is, the diameter becomes smaller as the rod is about to fail. Finally the rod actually breaks.

The point at which the Elastic Region ends is called the elastic limit, or the proportional limit. In actuality, these two points are not quite the same. The Elastic Limit is the point at which permanent deformation occurs, that is, after the elastic limit, if the force is taken off the sample, it will not return to its original size and shape, permanent deformation has occurred. The Proportional Limit is the point at which the deformation is no longer directly proportional to the applied force (Hooke's Law no longer holds). Although these two points are slightly different, we will treat them as the same in this course.

Next, rather than examining the applied force and resulting deformation, we will instead graph the axial stress verses the axial strain (diagram 3). We have defined the axial stress earlier. The axial strain is defined as the fractional change in length or Strain = (deformation of member) divided by the (original length of member) , Strain is often represented by the Greek symbol epsilon(e), and the deformation is often represented by the Greek symbol delta(d), so we may write: Strain (where L_o is the original length of the member) Strain has no units - since its length divided by length, however it is sometimes expressed as 'in./in.' in some texts.

As we see from diagram 3, the Stress verses Strain graph has the same shape and regions as the force verses deformation graph in diagram 2. In the elastic (linear) region, since stress is directly proportional to strain, the ratio of stress/strain will be a constant (and actually equal to the slope of the linear portion of the graph). This constant is known as Young's Modulus, and is usually symbolized by an E or Y. We will use E for Young's modulus. We may now write Young's Modulus = Stress/Strain, or: . (This is another form of Hooke's Law.)
The value of Young's modulus - which is a measure of the amount of force needed to produce a unit deformation - depends on the material. Young's Modulus for Steel is 30 x 10⁶ lb/in², for Aluminum E = 10 x 10⁶ lb/in², and for Brass E = 15 x 10⁶ lb/in². For more values, select: Young's Modulus - Table.
To summarize our stress/strain/Hooke's Law relationships up to this point, we have:
The last relationship is just a combination of the first three, and says simply that the amount of deformation which occurs in a member is equal to the product of the force in the member and the length of the member (usually in inches) divided by Young's Modulus for the material, and divided by the cross sectional area of the member. To see applications of these relationships, we now will look at several examples.


In the structure shown in Diagram 1, members ABC and BDE are assumed to be solid rigid members. Member BDE is supported by a roller at point E, and is pinned to member ABC at point B. Member ABC is pinned to the wall at point A. Member ABC is an aluminum rod with a diameter of 1 inch. (Young's Modulus for Aluminum is 10 x 10⁶ lb/in²)
For this structure we would like to determine the axial stress in member ABC both in section AB and section BC, and to determine the movement of point C due to the applied loads.
Part I. To solve the problem we first need to determine the external support forces acting on the structure. We proceed using our static equilibrium procedure (from Topic 1 - Statics)
1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.(See Diagram 2)
2: Resolve forces into x and y components.
3: Apply the equilibrium conditions.
Sum F_x = A_x = 0
Sum F_y = A_y + E_y -16,000 lbs - 12,000 lbs = 0
Sum T_E = (16,000 lbs)(12 ft) + (12,000 lbs)(8 ft) - A_y(12 ft) = 0
Solving for the unknowns: A_y = 24,000 lbs; E_y = 4000 lbs
Part II. An interesting aspect to this problem is that member ABC is not an axial member, and so it is not in simple uniform tension or compression. However, we are fortunate in that it is not a complex non-axial member. It is not in shear, but rather simply is in different amounts of tension above and below point B. Therefore to determine the amount of stress in each part of ABC, we first make a free body diagram of member ABC and apply static equilibrium principles.
1: FBD of member ABC. (Diagram 3)
2. Resolve all forces into x and y components.
3. Apply equilibrium conditions.
Sum F_x = B_x = 0
Sum F_y = 24,000 lb. - B_y - 16,000 lbs = 0
Solving for the unknownsB_y = 8,000 lb.
Now to find the force in section AB of member ABC. Cut the member between points A and B,and analyze the top section. We can do this since if a member is in static equilibrium, then any portion of the member is also in static equilibrium.
Looking at Diagram 4 (which is the free body diagram of the upper section of member ABC), we see that for the section of AB shown to be in equilibrium, the internal force (which becomes external when we cut the member) must be equal and opposite to the 24,000 lb force of the wall on the member at point A.
Once we know the tension in section AB of member ABD, we find the stress from our relationship Stress = F/A = 24,000 lb/(3.14 x .5²) = 30,600 psi.
We then use the same approach with section BC of member ABC. We cut member ABC between point B and point C, and apply static equilibrium principles to the top section. Diagram 5 is the free body diagram of that section, and by simply summing forces in the y-direction, we see that the internal force BC (which becomes an external force when we cut the member) must be 16,000 lb. for equilibrium.
The stress in section BC is then given by Stress = F/A = 16,000 lb/(3.14 x .5²)
Stress (BC) = 20,400 psi.
Part III. To determine the movement of point C is a relatively simple problem when we realize that the movement of point C will be equal to the deformation (elongation) of section AB plus the deformation (elongation) of section BC.
That is, the Movement of C = Def_AB + Def_BC
Movement of. C = [ (FL / EA)_AB + (FL / EA)_BC ]
Movement of C = [ (24,000 lbs)(72 in) / (10*10⁶ psi)(3.14*(.5 in)²)]_AB + [ (16,000 lbs)(48 in) /
(10*10⁶ psi)(3.14*(.5 in)²)]_BC
Movement of C = (.220 in) + (.0978 in) = .318 in


Example 3

In the structure shown in Diagram 1, member BCDFG is assumed to be a solid rigid member. It is supported by a two cables, AB & DE. Cable AB is steel and cable DE is aluminum. Both cables have a cross sectional area of .5 in². For this structure we would like to determine the axial stress in cable AB and DE. We would also like to determine the movement of point F due to the applied loads. [Young's Modulus for steel : E_st = 30 x 10⁶ psi, Young's Modulus for Aluminum: E_al = 10 x 10⁶ psi.]

Part I. To solve the problem we first need to determine the external support forces acting on the structure. We proceed using our static equilibrium procedure (from Topic 1 - Statics)
1: Draw a free body diagram showing and labeling all load forces and support (reaction) forces, as well as any needed angles and dimensions.
2: Resolve forces into x and y components. All the forces are acting in the y-direction, as is shown in the FBD.
3: Apply the equilibrium conditions.
Sum F_y = A_y + E_y - 30,000 lb. - 10,000 lb. = 0
Sum T_B = (-30,000 lb.)(4 ft) + E_y(10 ft) - (10,000 lb.)(12ft) = 0
Solving for the unknowns: E_y = 24,000 lb.; A_y = 16,000 lb.

Part II. Since both the steel member AB and the aluminum member DE are single axial members connected to the supporting ceiling, the external forces exerted by the ceiling on the members is also equal to the internal forces in the members. Thus F_AB = 16,000 lb. (tension), F_DE = 24,000 lb. (tension). To find the stress in each cable is now straight forward. We apply the stress equation

(from the Stress / Strain / Hooke's Law relationships shown to the right).
So, Stress AB = F/A = 16,000 lbs/.5 in² = 32,000 psi., Stress DE = F/A = 24,000 lbs/.5 in² = 48,000 psi.


Part III To find the movement of point F requires us to use a bit of geometry. Point F moves since both member AB and ED deform and member BCDFG moves downward according to these deformations.
1: Calculate deformation of members AB and ED.
Def_AB = (FL / AE)_AB = (16,000 lbs)(120 in) / (30*10⁶ lbs/in²)(.5 in²) = .128 in
Def_ED = (FL / AE)_ED = (24,000 lbs)(120 in) / (10*10⁶ lbs/in²)(.5 in²) = .576 in

2: Movement of point F. In Diagram 3 we have shown the initial and final position (exaggerated) of member BCDFG.

Point B moves down .128 inches (the deformation of member AB), point D moves down .576 inches (the deformation of member DE). Point F moves down an intermediate amount. To determine this we have drawn a horizontal line from the final position of point B across to the right side of the beam as shown. From this we see that the distance point F moves down is .128 inches + "x" (where x is the distance below the horizontal line as shown in the diagram). We can determine the value of x from proportionality (since similar triangles are involved), and write: (x / 12 ft) = (.448 in / 10 ft). Solving for x we find: x = .5376 inches.
So the Movement of F = .128 inches + .5376 inches = .666 inches