Thermal Effects and Stress on Inclined Planes

Introduction

Temperature changes cause materials to expand or contract. If deformation is unrestricted, no stress develops. If deformation is restrained, thermal strain produces internal stresses. Stress transformation equations are then used to determine normal and shear stresses on inclined planes.

Thermal Expansion

If a member is free to move, only deformation occurs. No thermal stress develops. Heating leads to expansion (+), while cooling leads to contraction (-).

Thermal strain: $\epsilon_{T} = \alpha\Delta T$

Elongation/Contraction: $\delta_{T} = \epsilon_{T}L = (\alpha\Delta T)L$

Restrained Thermal Expansion

If expansion is prevented by a fixed surface, thermal strain still occurs, but an equal and opposite mechanical strain ($\epsilon_{\sigma} = \frac{\sigma}{E})$ develops:

Total strain: $\epsilon = \epsilon_{T} + \epsilon_{\sigma}$

For a (.*?), the total deformation is zero, so $\epsilon = 0$. Therefore,

$0 = \alpha\Delta T + \frac{\sigma}{E}$

Thermal stress develops only (.*?) all free expansion (clearance/gap) has been used up. If there is a gap, no stress exists until the member contacts the restraint.

Compatibility

Compatibility ensures that the total deformation satisfies the physical constraints of the structure. The total deformation is the sum of the thermal and mechanical deformations:

$\delta_{total} = \delta_{T} + \delta_{P}$

where $\delta_{T} = (\alpha\Delta T)L$ and $\delta_{P} = \frac{PL}{AE}$

For problems with a gap:

if $\delta_{T} < gap \rightarrow$ no stress develops

if $\delta_{T} > gap \rightarrow$ the excess deformation creates force and stress

Excess deformation: $\delta_{P} = \delta_{T} - gap$

Stress on Inclined Planes

Given an axial stress $\sigma_{x} = \frac{P}{A_{cross}}$:

Normal stress: $\sigma_{\theta} = \sigma_{x}cos^{2}\theta$

Shear stress: $\tau_{\theta} = - \sigma_{x}\sin\theta\cos\theta$

Example 1:
Find the thermal stress in the beam of length 10 ft. The beam is held between immovable supports, has a modulus of elasticity $E = 29000\ ksi$ and coefficient of thermal expansion $\alpha = 6.5 \times 10^{- 6}/℉$, and the beam’s temperature is raised by $\Delta T = 20℉$.

Since the beam is held between immovable supports, it cannot expand.

The thermal strain of the beam is:

$\epsilon_{T} = \alpha\Delta T = (6.5 \times 10^{- 6}/℉)(20℉) = 0.00013$

Using Hooke’s Law:

$\sigma_{T} = E\epsilon_{T} = (29000\ ksi)(0.00013) = 3.77\ ksi$ (compression)

Example 2:
Find the compressive force N in the bar, and the maximum compressive stress. The diameters and lengths are given in the figure. The modulus of elasticity is 6 GPa, the coefficient of thermal expansion is $100 \times 10^{- 6}/℃$, and the bar is subjected to a temperature increase of 30 °C.

Since both ends are fixed and the force is constant through both sections, $N_{AC} = N_{CB} = N$.

Each section has mechanical and thermal deformation:

$\delta_{AC} = - \frac{NL_{1}}{EA_{1}} + \alpha L_{1}\Delta T$ $\delta_{CB} = - \frac{NL_{2}}{EA_{2}} + \alpha L_{2}\Delta T$

The total length of the bar cannot change, so we can use the compatibility equation:

$\delta_{AC} + \delta_{CB} = 0$

$( - \frac{NL_{1}}{EA_{1}} + \alpha L_{1}\Delta T) + ( - \frac{NL_{2}}{EA_{2}} + \alpha L_{2}\Delta T) = 0$

Plugging everything in:

$( - \frac{N(225\ mm)}{(6000\ N/mm^{2})(\frac{\pi}{4}(50\ mm)^{2})} + (100 \times 10^{- 6}/℃)(225\ mm)(30\ ℃)) +$

$( - \frac{N(300\ \ mm)}{(6000\ N/mm^{2})(\frac{\pi}{4}(75\ mm)^{2})} + (100 \times 10^{- 6}/℃)(300\ mm)(30\ ℃)) = 0$

Solving: $N = 51.8\ kN$

Maximum compressive stress takes place in the smaller cross-sectional area:

$\sigma_{c} = \frac{N}{A} = \frac{51.8\ kN}{\frac{\pi}{4}(0.05\ m)^{2}} = 26381\ kPa = 26.4\ MPa$