What is the Stress in a Bolt and Spacer System?

[hatchelhoff]

Homework Statement

A Steel bolt having nominal Diameter of 20 mm and a pitch of 2.5 mm
(Pitch is the distance from thread to thread in the axial direction), and an aliminium tube,
40 mm OD by 22 mm ID, act as a spacer for two plates. The distance between the two plates is 0.35 m. The nut is pulled snug and then given a one-third additional turn. Find the resulting stress in the bolt and in the tube, neglecting the deformation of the plates

(The correct answers are 241 MPa and -86.3 MPa

Homework Equations

$\tau$ = Shear Stress, $\lambda$ = Shear Strain
σ = F/A, ε = Elongation/Origional Length, σ = Modulus of Elasticity x ε
Elongation = (Force x Length) / (AE), v = -Lateral Strain/axial Strain
E = 2G(1 + v), ε1 = σ1/E - vσ2/E, σ1 = (E(ε1 + vε2))/(1-vxv)
$\tau$ = G$\lambda$

For Uniaxial Stress
ϵ1 = (σ1/E), ε2 = -v*ε1, ε2 = -v*ε1,

For Biaxial stress
ϵ1 = (σ1/E) -(v.σ2/E)

ϵ2 = (σ2/E) -(v.σ1/E)

ϵ3 = v.(σ1/E) - v.(σ2/E)

For Triaxial Stress
ϵ1 = (σ1/E) -(v.σ2/E) -(v.σ3/E)
ϵ2 = (σ2/E) -(v.σ1/E) -(v.σ3/E)
ϵ3 = (σ3/E) -(v.σ1/E) -(v.σ2/E)

The Attempt at a Solution

Part A
The steel bolt is in a state of uniaxial stress
ϵ1 = (σ1/E)
ϵ1 = Elongation/ Origional Length
Elongation = 2.5 mm / 3 = 0.83333 mm
ε1 = 0.83333/ 350 mm = 0.002381

I am saying that since the bolt is steel, E = 200 GPa

Therefore
σ1 = ε1 * E = 0.002381 * 200 GPA = 476 MPa

 

[nvn]

hatchelhoff: (1) What is the given value of E for the aluminum? (2) Is the thickness of each plate given? If so, what is the thickness of each plate? (3) Is the bolt fully threaded, all the way to the bolt head? Or is the bolt only partially threaded, with a solid shank? If partially threaded, what is the length of the solid shank?

 

[hatchelhoff]

NVN: 1) There is no value give for E of Aluminium. There was no value given of E for the steel bolt either, I just felt it was correct to asume it is 200 GPa. 2) The thickness of each plate is not given. The bolt seems to be partially threaded. The only dimension given for the bolt is the distance between the bolt head and the nut which is 0.35 m.
This 0.35 m includes the shank and some of the thread.

 

[nvn]

hatchelhoff: OK. Near the top of post 1, did you round the 241 and -86.3 MPa stresses? I.e., did these two stresses have a few more significant digits after the decimal point, which you did not tell us about?

 

[hatchelhoff]

nvn: The answers are exactly as I have given them.

 

[nvn]

hatchelhoff: Could you post a diagram? If the distance between the bolt head and nut is 350 mm, then the distance between the two plates cannot be 350 mm, as stated in post 1. Or else I do not understand. (The dimensions do not seem to make sense yet.) Could you post a dimensioned diagram?

 

[hatchelhoff]

Please see attached crude painting of the bolt and the spacer.

 

[nvn]

hatchelhoff: Is this a test question? If not, is it a school assignment worth a lot of points?

You can see from your diagram, the distance between the plates is not 350 mm. Do you agree? Did you make a typographic mistake in post 1, where you said, "The distance between the two plates is 0.35 m"? If we call the distance between the two plates L2, should L2 be L2 = 330 mm, instead of 350 mm? I think L2 should be L2 = 330 mm. We can see, the bolt initial length is L1 = 350 mm.

 

[hatchelhoff]

nvn: It is just a course work question from the book which we are using.
it Is neither a test question or a school assignment worth a lot of points?

I agree that the distance brtween the two plates is not 350mm. I am not sure how you Calculated L2 = 330mm.

 

[nvn]

hatchelhoff: Always leave a space between a numeric value and its following unit symbol. E.g., 330 mm, not 330mm.

Your answer at the end of post 1 is currently incorrect. Try again.

First, forget about Poisson's ratio, eps2, eps3, and stress in the y and z directions. In this problem, you have only axial stress, no transverse stress. Therefore, hereafter, let us use the number "1" to denote the steel bolt, "2" to denote the aluminum tube, and "3" to denote each of the two steel plates. Therefore, we have the following.

(1) E1 = 200 000 MPa, E2 = 69 000 MPa, E3 = infinity.
(2) t3 = initial thickness of each steel plate = 10 mm.
(3) tf3 = final thickness of each steel plate = 10 mm.
(4) M20 bolt; i.e., M20 x 2.5.
(5) p = bolt thread pitch = 2.5 mm.
(6) L1 = bolt initial length = (350 mm) - p/3.
(7) L2 = aluminum tube initial length = (350 mm) - 2*t3 = 330 mm.
(8) Lf1 = bolt final length = L1 + eps1*L1.
(9) Lf2 = aluminum tube final length = L2 + eps2*L2.
(10) Lf2 = Lf1 - 2*tf3.
(11) P1 = bolt axial force.
(12) P2 = aluminum tube axial force.
(13) delta1 = bolt elongation = Lf1 - L1.
(14) delta2 = aluminum tube elongation = Lf2 - L2.
(15) eps1 = delta1/L1.
(16) eps2 = delta2/L2.
(17) The correct answers are sigma1 = 240.85 MPa, sigma2 = -86.325 MPa.​

Hint 1: P2 = -P1. Hint 2: Assume bolt is a solid shank for its full length, L1; i.e., ignore threads. (Footnote: Normally, you do not ignore threads. Ignore threads only for this particular homework question.)

 

[hatchelhoff]

How did you know to allow 10 mm for the steel plate.

 

[nvn]

How did you know to allow 10 mm for the steel plate.

Only because the stress answers are given, E1 = 200 GPa, and, for Al 6061-T6, E2 = 69 GPa.

 

[hatchelhoff]

There must be another way of solving it as we are not supposed to know the final answers for stress. I only gave the answer which i got from the back of the book for reference.

 

[nvn]

hatchelhoff: If the stress answers were not given, then another approach (the typical approach) is, assume a value for t3, then solve for the stresses.

 

[nvn]

hatchelhoff: I want to clarify that all of the following quantities, from the list in post 10, are in the bolt axial (longitudinal) direction. I clarified this below.

(1) E1 = 200 000 MPa, E2 = 69 000 MPa, E3 = infinity.
(2) t3 = initial thickness of each steel plate = 10 mm.
(3) tf3 = final thickness of each steel plate = 10 mm.
(4) M20 bolt; i.e., M20 x 2.5.
(5) p = bolt thread pitch = 2.5 mm.
(6) L1 = bolt initial length = (350 mm) - p/3.
(7) L2 = aluminum tube initial length = (350 mm) - 2*t3 = 330 mm.
(8) Lf1 = bolt final length = L1 + eps1*L1.
(9) Lf2 = aluminum tube final length = L2 + eps2*L2.
(10) Lf2 = Lf1 - 2*tf3.
(11) P1 = bolt axial force.
(12) P2 = aluminum tube axial force.
(13) delta1 = bolt axial elongation = Lf1 - L1.
(14) delta2 = aluminum tube axial elongation = Lf2 - L2.
(15) eps1 = bolt axial strain = delta1/L1.
(16) eps2 = aluminum tube axial strain = delta2/L2.
(17) The correct answers are sigma1 = 240.85 MPa, sigma2 = -86.325 MPa.​

Hint 1: P2 = -P1. Hint 2: Assume bolt is a solid shank for its full length, L1; i.e., ignore threads. (Footnote: Normally, you do not ignore threads. Ignore threads only for this particular homework question.) Hint 3: You have an equation relating delta1 and P1 (and delta2 and P2) listed in post 1.

 

[hatchelhoff]

nvn: I am using the equation which I feel you were pointing me to
which is
Delta1 = (P1*L1)/A1*E1
This however creates a circular reference which I cannot solve
for Example
In order to find P1 i need to find Delta1
in order to find Delta1 I need to find LF1
in order to find LF1 I need to find Eps1
in order to find Eps1 I need to find Delta1
Which leads me back to the first unknown.
What am I doing wrong?

 

[nvn]

hatchelhoff: The equation you cited in post 16 is good. And, there are two more equations, which will get you out of a circular reference. See the hints at the bottom of post 15.

 

[hatchelhoff]

nvn: Can I say that delta1 = delta2

 

[nvn]

No. Keep trying.

 

[hatchelhoff]

nvn: Using your hint P2 = -P1 I get
P2= delta2(A2*E2)/L2 = -P1 = - (delta1(A1*E1)/L1)
I am still missing somthing as I can't figure out how to get out of the circular reference.

 

[nvn]

hatchelhoff: Nice work. Hint 4: delta1 and delta2 are given in post 15.

 

[hatchelhoff]

P2= delta2(A2*E2)/L2 = -P1 = - (delta1(A1*E1)/L1)
Therefore
P2 = (LF2-L2)(A2*E2)/L2 = - (LF1 - L1)(A1*E1)/L1)

 

[nvn]

OK, continue.

 

[hatchelhoff]

P2 = ((L2 + eps2*L2)-L2)(A2*E2)/L2 = - ((L1 + eps1*L1) - L1)(A1*E1)/L1)

 

[nvn]

hatchelhoff: No, go back to post 22. But only use one equal sign, not two equal signs.

 

[hatchelhoff]

(lf2-l2)(a2*e2)/l2 = - (lf1 - l1)(a1*e1)/l1)

 

[nvn]

Yes, nice work. Continue.

 

[hatchelhoff]

(L2 + eps2*L2 -L2)(A2*E2)/L2 = - (L1 + eps1*L1 - L1)(A1*E1)/L1)

 

[nvn]

No, go back to post 26, and try again.

 

[hatchelhoff]

Should I try to get LF2 and LF1 on the same side of the equation

 

[nvn]

Yes.

 

[hatchelhoff]

((lf2-1)/l2)(l1/(lf1+1)) = (a1*e1)/(a2*e2)

 

[nvn]

hatchelhoff: No, that algebra in post 32 is no good. Go back to post 26, and expand the equation, which means multiply all terms. Afterwards, get Lf1 and Lf2 on the same side of the equation. (Also, use uppercase L, not lowercase L, so we do not get confused.) Try again.

 

[hatchelhoff]

(Lf2-l2)(a2*e2)/L2 = - (Lf1 - l1)(a1*e1)/L1)
(LF2 * a2 * e2 -L2* a2 * e2)/L2 = -(Lf1*a1*e1 + L1 * a1 * e1)/L1
((LF2*a2*e2)/L2) - L2/L2 * a2*e2 = ((-Lf1*a1*e1)/L1) + L1/L1 * a1*e1
((LF2*a2*e2)/L2) = ((-Lf1*a1*e1)/L1) + a1e1 + a2e2

((LF2*a2*e2)/L2)+((Lf1*a1*e1)/L1) = a1e1 + a2e2

 

[nvn]

hatchelhoff: Nice work. Hint 5: Now look for an expression for Lf2 in post 15.