BOLT DEFLECTION .1 B ASIC C ONCEPTS.1 BASICCONCEPTS

Let’s apply equal and opposite forces to the ends of a rod of nonuniform diameter, as shown in Figure 5.1. If the tension stress created in the rod is below the proportional limit, we can use Hooke’s law and the relationship between springs in series to compute the change in length of the rod.

The combined change in length of the rod will be equal to the sum of the changes in each section:

DL_C¼ DL1þ DL2þ DL3 (5:1)

Hooke’s law tells that the change in one section will be DL¼FL

EA (5:2)

where

DL¼ change in length (in., mm) A ¼ cross-sectional area (in.², mm²) L ¼ length of the section (in., mm) E ¼ modulus of elasticity (psi, GPa) F ¼ applied tensile force (lb, N)

85

Since the various sections are connected in series, they each see the same force, so we can combine Equations 5.1 and 5.2 above and write

DL_C¼ F L1 Now, the spring constant of a body is defined as

K¼ F

DL (5:4)

where

K ¼ spring constant or stiffness (lb=in., N=mm)

DL¼ change in length of the body under load (in., mm) F F ¼ applied load (lb, N)

The spring constant of a group of bodies, connected in series, is 1

K_T ¼ combined spring constant of the group (lb=in., N=mm)

K1, K2, . . .¼ spring constants of individual members of the group (lb=in., N=mm) Now, the equation for the spring constant of a body can be rewritten as

DL¼ F

Comparing our equation for the spring constant for a group of bodies to the equation for the stretch or change in length of a group of bodies, we see that

1

FIGURE 5.1 Rod of nonuniform diameter, loaded in tension, and equivalent spring model.

Note that the stiffness of either a plain or complex body is very much a function of the ratio between length and cross-sectional area—it’s a function, in other words, of the ‘‘shape’’ of body just as much as it is a function of the material from which the body is made. If we take one piece of alloy steel and make two bolts from it, one a short, stubby bolt and the other long and thin, and we then place the bolts in tension and plot elastic curves for them, we will end up with two curves such as shown in Figure 5.2.

The equations used for a rod having several different diameters are basically the equations we would use for computing change in length and stiffness of bolts. If we can compute or predict the lengths, cross-sectional areas, and modulus of the material, we should be able to compute the deflection under load. There is some ambiguity, however, about each of these factors when we’re dealing with bolts. We’ll take a closer look.

First, though, note that I included the modulus of elasticity in the list of things we’ll need to know. We have not needed that material property until now. Most engineers dealing with bolts and joints don’t need to know the modulus because they’re primarily interested in the tensile or shear strengths of bolts, threads, joint members, etc. Those, like us, who are also interested in the response of bolts and joints to service loads, temperature cycles, etc. will find the modulus a very important factor. Table 5.1 lists the modulus for many bolt and joint materials.

5.1.2 CHANGE IN LENGTH OF THEBOLT

5.1.2.1 Effective Length

Tensile loads are not applied to bolts ‘‘from end to end’’; they’re applied between the inner face of the nut and the undersurface of the head. The entire bolt is not loaded, therefore, the way the test rods are. There is zero tensile stress in the free ends, for example.

There is, however, some stress in portions of both the head and the threads (see Figure 3.5). We cannot assume that the bolt is merely a cylinder equal in length to the grip length.

Instead, we have to make some assumption concerning the stress levels which will allow us to estimate an ‘‘effective length’’ for the bolt that is somewhere between the true overall length and the grip length.

Change in length (∆L)

Force (F)

FIGURE 5.2 Elastic curves for a short, stubby bolt and a long, thin bolt cut from the same material.

We know from Chapter 3 that tensile stress in a bolt is maximum near the inner faces of the head and nut, and that tensile stress is zero at the outboard faces of the nut and head.

Assuming that there is a uniform decrease in stress from inboard to outboard faces of the head, as suggested by Figure 3.4, we can make the assumption that the average stress level in the head of the bolt is one-half the body stress; or we can make a mathematically equivalent assumption and say that one-half of the head is uniformly loaded at the body stress level and that the rest of the head sees zero stress. Similarly, we can say that one-half of the threads engaged by the nut are loaded at the ‘‘exposed thread’’ stress level. We are now in a position to say that the effective length (LE) of the fastener is equal to the length of the body (LB) plus one-half the thickness of the head (TH) added to the length of the exposed threads (LT) plus one-half the thickness of the nut (TN), as suggested by Figure 5.3 or by

LE¼ Lð Bþ TH=2Þ þ Lð Tþ TN=2Þ (5:7a) Compare the actual stress levels sketched in Figure 5.3 with those shown in Figures 3.5, 3.6, and 3.7. We have taken the simplest case for estimating the effective length of our bolt.

There’s really no simple way we could deal with the ‘‘true’’ stress distribution, which would involve a finite-element analysis or the like and would require more information about the geometry of a particular bolt and joint than we’ll ever have in practice. We’ll find, however, that the assumptions we have made give us reasonable predictions in many applications,

TABLE 5.1

Modulus of Elasticity at Room Temperature (310⁶psi)

Bolt Materials Modulus Ref.

because the bulk of the fastener is stressed at, or near, the levels we have assumed. It is only the surfaces of the fastener that exhibit the maximum deviations from these averages.

At least that’s true as far as long, thin bolts are concerned. As the length-to-diameter ratio of the bolt decreases, and the bolt becomes more and more short and stubby, our assumption of effective length becomes more and more suspect. More about this in Chapter 9.

I have suggested that we use one-half the thickness of the head and one-half the thickness of the nut to compute the effective length of the equivalent fastener. I should mention in passing that other sources recommend slightly different correction factors, such as 0.4
nominal diameter for the head, and another 0.4
D for the nut [1]; or 0.3
D for each [2]; or nothing for the head and 0.6
D for the nut [3]. The thickness of a standard heavy hex nut, incidentally, is equal to the nominal diameter, so the correction I have suggested is equal to 0.5
D for a heavy hex nut and a little more for a light nut.

5.1.2.2 Cross-Sectional Areas of the Bolt

We also have to make some assumptions concerning cross-sectional areas of the bolt when computing change in length. The body area is no mystery; it’s merely equal to p
D²=4, where D is the nominal diameter of the fastener.

For the cross-sectional area of the threads, however, we must use the effective or ‘‘stress area’’ discussed in Chapter 3.

5.1.3 COMPUTING CHANGE INLENGTH OF THEBOLT

We can now compute the approximate change in length of the bolt under load:

DLC¼ FP

Lbe

EAB

þ Lse

EAS

 

(5:8)

where

Lbe ¼ the effective length of the body (true body length plus one-half the thickness of the head of the bolt) (in., mm) (see Appendix F)

Lse ¼ the effective length of the threads (length of exposed threads plus one-half the thickness of the nut) (in., mm) (see Appendix F)

DLC¼ combined change in length of all portions (in., mm)

AS ¼ the effective stress area of the threads (see Chapter 3) (in.², mm²) (see Appendix E) AB ¼ the cross-sectional area of the body (in.², mm²)

LE

TH LS LT TN

F F F F

(A) (B)

FIGURE 5.3 Illustration of actual bolt configuration and average tensile stress levels (A), and the equivalent configuration and stress distribution assumed for calculation purposes (B).

E ¼ the modulus of elasticity (psi, N=mm²) (Table 5.1) FP ¼ tension in bolt (lb, N)

If the fastener has a more complex shape, as shown in Figure 5.4, then additional sections must be computed, but there is otherwise no change in the procedure. The change in length for the fastener shown in Figure 5.4, for example, would be

DLC¼ FP

Once we know how to compute the change in length of the fastener, we can also estimate the spring constant or stiffness, using the relationship

KB¼ FP=DLC (5:10)

Before using this equation let me mention that Equation 5.12 gives us an alternative—and perhaps more convenient—way to estimate bolt stiffness. But for now, let’s continue with Equation 5.10.

5.2.2 EXAMPLE

Let’s compute the stiffness and change in length of a ³⁼⁸–16
¹⁼² SAE, Grade 8 hex bolt (shown in Figure 5.5) in a joint having a 1 in. grip length (Lg). We can get the nominal dimensions we’ll need by measuring a sample bolt, or, more safely, by referring to the pertinent specifications, or to the data in Appendices E and F. These tell us the following:

Specification Dimension

SAE J 104 (nut) Height of nut (TN)¼ 0.3285 in.

SAE J 105 (bolt) Height of head (TH)¼ 0.2345 in.

ANSI Bl.1-1974 (thread) Thread length (LT)¼ 1.000 in.

Tensile stress area of threads (AS)¼ 0.0775 in.²

From the description of the bolt, we already know, of course, that nominal body diameter (D)¼ 0.375 in. and nominal shank length (L) ¼ 1.500 in.

FIGURE 5.4 Each cross section of a complex fastener must be computed separately.