Question

The rigid beam is supported by a pin at A and wires BD and CE. If the load P on the beam causes the end C to be displaced 10 mm downward, determine the normal strain developed in wires CE and BD.

Answer 1

Concepts and reason

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 normal strain developed in wires is calculated by using the similar triangle rule.

Deformation is assumed to be linear as they are very small compared to the original length of the wire.

Fundamentals

Stress: A body offers some internal resistant force under the action of a load.

This internal resistant force per unit cross-sectional area is termed as stress.

The stress can be calculated by,

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

Here, the force applied on the material is P and the cross-sectional area on which the force acts is A.

The strain can be calculated by,

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

Here, the change in length of the material is $\delta L$ and its original length is L.

Draw the deformation diagram.

Calculate the change in length of the wire BD.

$\frac{{\Delta {L_{BD}}}}{3} = \frac{{\Delta {L_{CE}}}}{7}$

Here, $\Delta {L_{BD}}$ is the change in length of the wire BD and $\Delta {L_{CE}}$ is the change in length of the wire CE.

Substitute 10 mm for $\Delta {L_{CE}}$ .

$\begin{array}{l}\\\frac{{\Delta {L_{BD}}}}{3} = \frac{{10}}{7}\\\\\Delta {L_{BD}} = 4.3\,{\rm{mm}}\\\end{array}$

Calculate the normal strain developed in the wire CE.

${\varepsilon _{CE}} = \frac{{\Delta {L_{CE}}}}{{4000}}$

Substitute 10 mm for $\Delta {L_{CE}}$ .

$\begin{array}{l}\\{\varepsilon _{CE}} = \frac{{10}}{{4000}}\\\\{\varepsilon _{CE}} = 0.0025\,{\rm{mm/mm}}\\\end{array}$

Calculate the normal strain developed in the wire BD.

${\varepsilon _{BD}} = \frac{{\Delta {L_{BD}}}}{{4000}}$

Substitute 4.3 mm for $\Delta {L_{BD}}$ .

$\begin{array}{l}\\{\varepsilon _{BD}} = \frac{{4.3}}{{4000}}\\\\{\varepsilon _{BD}} = 0.00107\,{\rm{mm/mm}}\\\end{array}$

Ans:

Therefore, the normal strain developed in the wires CE and BD are ${\bf{0}}{\bf{.0025}}\,{\bf{mm/mm}}$ and ${\bf{0}}{\bf{.00107}}\,{\bf{mm/mm}}$ respectively.

Answer 2

Concepts and reason

Free body diagram:

It is a graphical or symbolic representation used to visualize the external forces, moments and the support reactions on a body for a condition.

Normal stress:

It refers to the stress induced in the member that is loaded along longitudinal or axial direction of the member. It is calculated mathematically from the ratio of the axial or longitudinal force to the cross-sectional area of the member.

Normal strain:

It is the ratio of deformation in length to the original length when a member is subjected to axial load.

Deflection:

It refers to the displacement of any point in a structure due to applied load. It is denoted by .

Hooke’s law:

Hooke’s law states that the normal strain in a member is directly proportional to the normal stress.

Moment:

It refers to the propensity of the force to cause rotation in a body about any fixed point. The moment’s magnitude can be obtained by multiplying force’s magnitude with the perpendicular distance at which the force acts. The moment is denoted by and its unit is.

Apply similar triangle method to determine the deflection in the steel rod , then determine the axial strain in the steel rod . Use deflection equation to calculate the force acting in the wooden post and steel rod. Finally apply moment equilibrium condition about point to calculate the vertical force .

Fundamentals

The formula to calculate the strain of an object is as follows:

Here, change in length of the specimen is and original length of the specimen is.

The equation to calculate the modulus of elasticity is as follows:

Here, stress in the specimen is and modulus of elasticity is

Unit Conversions:

(b)

The free body diagram and deflection diagram of bar is shown below:

Use similar triangle method to determine the deflection in the steel rod B.

Calculate the normal strain in the steel rod as follows:

Here, length of the steel rod is and deflection in the steel rod is .

Substitute for and for. .

(a)

Calculate the normal strain in the steel rod as follows:

Here, length of the steel rod is and deflection in the steel rod is .

Substitute for and for. .

Ans: Part b

The axial strain in the steel rod BD is .

Part a

The axial strain in the steel rod CE is .