Question

Direct tension indicators are sometimes used instead of torque wrenches to ensure that a bolt has a prescribed tension when used for connections. If a nut on the bolt is tightened so that the six 3 - mm high heads of the indicator are strained 0.1 mm/mm

Answer 1

Complete Question

The complete question is shown on the uploaded image

Answer:

The tension on the shank is  $T =8391.6 N$

Explanation:

From the question we are told that

       The strain on the strain on the head is $\Delta l = 0.1 mm/mm = \frac{0.1}{1000} = 0.1 *10^{-3} m/m$

         The contact area is  $A = 2.8 mm^2 = 2.8* (\frac{1}{1000} )^2 = 2.8*10^{-6} m^2$  

Looking at the first diagram

           At  600 MPa of stress

               The strain is  $0.3mm/mm$

          At   450 MPa of stress

                 The strain is   $0.0015 mm/mm$

 To find the stress at  $\Delta l$ we use the interpolation method

            $\frac{\sigma_{\Delta l} - \sigma_{0.0015} }{ \sigma _ {0.3} - \sigma_{0.0015} } = \frac{e_{\Delta l } - e_{0.0015}}{e_{0.3} - e_{ 0.0015}}$

Substituting values

              $\frac{\sigma _{\Delta l} - 450}{600 - 450} = \frac{0.1 -0.0015}{0.3 - 0.0015}$

            $\sigma _{\Delta l} -450 = 49.50$

             $\sigma _{\Delta l} = 499.50 MPa$

Generally the force on each head is mathematically represented as

              $F = \sigma_{\Delta l} * A$

Substituting values

             $F = 499.50*10^{6} * 2.8*10^{-6}$

                $=1398.6N$

Now the tension on the bolt shank is as a result of the force on the 6 head which is mathematically evaluated as

              $T = 6 * F$

                  $= 6* 1398.6$

              $T =8391.6 N$