Question

The force applied at the handle of the rigid lever causes the lever to rotate clockwise about the pin B through an angle of 2 degrees. Determine the average normal strain developed in each wire. The wires are unstretched when the lever is in the horizontal position. Please show work.

  

Answer 1

Concepts and Reason

Angular displacement: It the as the angle through which an object moves on a circular path. It is the angle, in radians, between the initial and final positions.

Strain: The change in dimension over the original dimension caused due to the applied load is called strain.

Fundamentals

Strain is the ratio of change in length over the original length.

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

The relation for angular displacement is,

$\delta = L\theta$

Conversion of degree to radians,

$\begin{array}{l}\\\theta = 2^\circ \left( {\frac{\pi }{{180}}} \right)\;{\rm{rad}}\\\\\;\;\;{\rm{ = 0}}{\rm{.03491}}\;{\rm{rad}}\\\end{array}$

Calculate the displacement of point A.

${\delta _A} = {L_{AB}}\theta$

Here length of lever AB is ${L_{AB}}$ and angle of tilt about the pin B is $\theta$ .

Substitute, $200\;{\rm{mm}}$ for ${L_{AB}}$ , ${\rm{0}}{\rm{.03491}}\;{\rm{rad}}$ for $\theta$ in relation,

$\begin{array}{l}\\{\delta _A} = \left( {200 \times 0.03491} \right)\\\\\;\;\;\; = 6.9813\;{\rm{mm}}\\\end{array}$

Calculate the displacement of point C.

${\delta _C} = {L_{BC}}\theta$

Here length of lever BC is ${L_{BC}}$ and angle of tilt about the pin B is $\theta$ .

Substitute, $300\;{\rm{mm}}$ for ${L_{BC}}$ , ${\rm{0}}{\rm{.03491}}\;{\rm{rad}}$ for $\theta$ in relation,

$\begin{array}{l}\\{\delta _C} = \left( {300 \times 0.03491} \right)\\\\\;\;\;\; = 10.4720\;{\rm{mm}}\\\end{array}$

Calculate the displacement of point D.

${\delta _D} = {L_{BD}}\theta$

Here length of lever BD is ${L_{BD}}$ and angle of tilt about the pin B is $\theta$

Substitute $300\;{\rm{mm}}$ for ${L_{BD}}$ , ${\rm{0}}{\rm{.03491}}\;{\rm{rad}}$ for $\theta$ in relation,

$\begin{array}{l}\\{\delta _D} = \left( {500 \times 0.03491} \right)\\\\\;\;\;\; = 17.4533\;{\rm{mm}}\\\end{array}$

Calculate the average normal strain developed in wire AH.

${\left( {{ \in _{avg}}} \right)_{AH}} = \frac{{{\delta _A}}}{{{L_{AH}}}}$

Here displacement of point A of the lever is ${\delta _A}$ , unstretched length of the wire AH is ${L_{AH}}$ .

Substitute $6.9813\;{\rm{mm}}$ for ${\delta _A}$ , $200{\rm{ mm}}$ for ${L_{AH}}$ in relation,

$\begin{array}{l}\\{\left( {{ \in _{avg}}} \right)_{AH}}\; = \frac{{6.9813}}{{200}}\\\\\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.0349\\\end{array}$

Calculate the average normal strain developed in wire CG.

${\left( {{ \in _{avg}}} \right)_{CG}} = \frac{{{\delta _C}}}{{{L_{CG}}}}$

Here displacement of point C of the lever is ${\delta _C}$ , unstretched length of the wire CG is ${L_{CG}}$

Substitute $10.4720\;{\rm{mm}}$ for ${\delta _C}$ and $300{\rm{ mm}}$ for ${L_{CG}}$ in relation,

$\begin{array}{l}\\{\left( {{ \in _{avg}}} \right)_{CG}}\; = \frac{{10.4720}}{{300}}\\\\\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.0349\\\end{array}$

‎

Calculate the average normal strain developed in wire DF.

${\left( {{ \in _{avg}}} \right)_{DF}} = \frac{{{\delta _D}}}{{{L_{DF}}}}$

Here displacement of point D of the lever is ${\delta _D}$ , unstretched length of the wire DF is ${L_{DF}}$ .

Substitute, $17.4533\;{\rm{mm}}$ for ${\delta _D}$ and $300{\rm{ mm}}$ for ${L_{DF}}$ in relation,

$\begin{array}{l}\\{\left( {{ \in _{avg}}} \right)_{DF}} = \frac{{17.4533}}{{300}}\\\\\;\;\;\;\;\;\;\;\;\;\;\;\; = 0.0582\\\end{array}$

Ans:

Therefore, the average normal strain developed in wire AH is $0.0349$ .