- #1

hatchelhoff

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## 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

[itex]\tau[/itex] = Shear Stress, [itex]\lambda[/itex] = 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)

[itex]\tau[/itex] = G[itex]\lambda[/itex]

**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