Question

The C83400-red-brass (E=14.6*10^6) rod AB and 2014-T6-aluminum rod BC are joined at the collar B and fixed connected at their ends. The cross-sectional area of each member is 1.75in2 .

If there is no load in the members when T1 = 50?F, determine the average normal stress in the member BC when T2 = 150 ?F.

Also, how far will the collar be displaced?

The C83400-red-brass (E=14.6*10^6) rod AB and 2014-T6-aluminum rod BC are joined at the collar B and fixed connected at their ends. The cross-sectional area of each member is 1.75in2 . If there is no load in the members when T1 = 50?F, determine the average normal stress in the member BC when T2 = 150 ?F. Also, how far will the collar be displaced?

Answer 1

Concepts and reason

Stress: When an alloy is loaded with a force, it produces a stress which makes the alloy to deform.

Strain: When the alloy is stressed in specific direction the response of the system for this stress is strain. The deformation of an alloy in specified direction divided by the original length gives the strain the alloy.

The deformation in a rigid body is represented by deformation diagram.

When load is applied in a specific direction the body deforms in that direction.

The displacement constraints that occur at the supports or other points of a member are compatibility conditions.

Free body diagram is the representation of all forces acting on the body.

When there is temperature change thermal stresses are included in the body.

Hooke’s law states that stress is directly proportional to strain.

Fundamentals

Calculate the stress using the following relation:

$\sigma = \frac{F}{A}$

Here, the force applied on the body is $F$ and the cross sectional area is $A$ .

Write the expression for deflection due to temperature change.

${\delta _T} = \alpha \Delta T$

Calculate the normal strain in a body using following relation:

$\varepsilon = \frac{{\delta l}}{l}$

Here, the change in length is $\delta l$ and the original length is $l$ .

Write the equilibrium equations.

$\begin{array}{l}\\{F_R} = \sum {\bf{F}} = 0\\\\{\left( {{M_R}} \right)_O} = \sum {{{\bf{M}}_O}} = 0\\\end{array}$

Here, the resultant force is ${F_R}$ and the resultant moment about any arbitrary point is ${\left( {{M_R}} \right)_O}$ .

Write the load displacement relation.

$\delta = \frac{{PL}}{{AE}}$

Here, the load applied is $P$ , the length of the member is $L$ , the cross sectional area of the member is $A$ , and the modulus of elasticity is $E$ .

Draw the free body diagram of the member.

Apply equilibrium equation in x-direction.

$\begin{array}{l}\\\sum {{F_x}} = 0\\\\{F_{Br}} - {F_{Al}} = 0\\\\{F_{Br}} = {F_{Al}} = F\\\end{array}$

As the rod assembly is fixed the deflection of A with respect to C is zero.

‎

Write the compatibility equation.

$\begin{array}{l}\\\sum {{\delta _T}} - \sum {{\delta _P}} = 0\\\\\left( {{\alpha _{Br}}\Delta T{L_{Br}} + {\alpha _{Al}}\Delta T{L_{Al}}} \right) - \left( {\frac{{{P_{AB}}{L_{AB}}}}{{{A_{AB}}{E_{AB}}}} + \frac{{{P_{BC}}{L_{BC}}}}{{{A_{BC}}{E_{BC}}}}} \right) = 0\\\\\left( {{\alpha _{Br}}\left( {{T_2} - {T_1}} \right){L_{Br}} + {\alpha _{Al}}\left( {{T_2} - {T_1}} \right){L_{Al}}} \right) - \left( {\frac{{{P_{AB}}{L_{AB}}}}{{{A_{AB}}{E_{AB}}}} + \frac{{{P_{BC}}{L_{BC}}}}{{{A_{BC}}{E_{BC}}}}} \right) = 0\\\\\left\{ \begin{array}{l}\\9.8 \times {10^{ - 6}}\left( {150 - 50} \right) \times \left( {3\,{\rm{ft}} \times 12\,\frac{{{\rm{in}}{\rm{.}}}}{{{\rm{ft}}}}} \right)\\\\ + 12.8 \times {10^{ - 6}} \times \left( {150 - 50} \right) \times \left( {2\,{\rm{ft}} \times 12\,\frac{{{\rm{in}}{\rm{.}}}}{{{\rm{ft}}}}} \right)\\\end{array} \right\} = \frac{{F \times \left( {3\,{\rm{ft}} \times 12\,\frac{{{\rm{in}}{\rm{.}}}}{{{\rm{ft}}}}} \right)}}{{1.75 \times 14.6 \times {{10}^6}}} + \frac{{F \times \left( {2\,{\rm{ft}} \times 12\,\frac{{{\rm{in}}{\rm{.}}}}{{{\rm{ft}}}}} \right)}}{{1.75 \times 10.6 \times {{10}^6}}}\\\\F = 24419\,{\rm{lb}}\\\end{array}$

Calculate the average normal stress.

$\begin{array}{c}\\\sigma = \frac{F}{A}\\\\ = \frac{{24419}}{{1.75}}\\\\ = 13953.71\,{\rm{psi}}\\\\{\rm{ = 13}}{\rm{.953}}\,{\rm{ksi}}\\\end{array}$

From the “average mechanical properties of typical engineering materials”,

Yield stress of C83400-Red brass is ${\left( {{\sigma _y}} \right)_{Br}} = 10\,{\rm{ksi = 10000}}\,{\rm{psi}}$

Yield stress of 2014-T6-Aluminum is ${\left( {{\sigma _y}} \right)_{Al}} = 60\,{\rm{ksi = 60000}}\,{\rm{psi}}$

The obtained stress is less than the yield stress hence it is safe stress.

Write the expression for deflection at B.

$\begin{array}{l}\\\sum {{\delta _T}} - \sum {{\delta _P}} = {\delta _B}\\\\{\delta _B} = \alpha \Delta TL - \frac{{{P_{AB}}{L_{AB}}}}{{{A_{AB}}{E_{AB}}}}\\\\ = 9.8 \times {10^{ - 6}}\left( {150 - 50} \right) \times \left( {3 \times 12} \right) - \frac{{24419 \times 3 \times 12}}{{1.75 \times 14.6 \times {{10}^6}}}\\\\ = 0.000873\,{\rm{in}}{\rm{.}}\\\end{array}$

Ans:

Therefore, the average normal stress in the member is $13.953\,{\rm{ksi}}$ .

The amount of collar displacement is $0.000873\,{\rm{in}}{\rm{.}}$ .