Chapter 3, Problem 3.8.18P

Two pipes {L, = 2.5 m and L, = 1.5 m) are joined al B by flange plales (thickness (, = 14 mm) with five bolts [dlt, = 13 mm] arranged in a circular pal tor n (see figure). Also, each pipe segment is atlaehed to a wall (at .1 and ( '. see figure! using a base plate Uh = 15 mm) and four bolts (dM, = 16 mm). All bolts are tightened until just snug. Assume £, = 110 GPa,E2 = 73 GPa,», = 0.33,andv, = 0.25. Neglect the self-weight of the pipes, and assume the pipes are in a stress-free stale initially. The cross-sectional areas of the pipes are At = 1500 mm: and A2 = (3/5)4. The ollter diameter of Pipe 1 is 60 mm. The outer diameter of Pipe 2 is equal to the inner diameter of Pipe 1. The bolt radius r = 64 mm for both base and flange plates.

(a)

If torque '/'is applied at .v = Lt. find an expression for reactive torques Iit and IL in terms of T.

(b)

Find the maximum load variable /'(i.e., Tmal) if allowable torsional stress in the two pipes is

Tall0* = 65 MPa-id Draw torsional moment iTMD i and torsional displacement (TDD) diagrams. Label all key ordinales. What is '/>.ll('.'

(d) Find mail, if allowable shear and bearing stresses in the base plate and flange bolts cannot be exceeded. Assume allowable stresses in shear an.: :vari:'.g I all bolls are r |Nill, = 45 MPa andtr MaK =90 MPa.

(e) Remove torque Tat x — L,. Now assume the flange-plate bolt holes are misaligned by some angle ß (see figure). Find the expressions for reactive torques Rx and R2 if the pipes are twisted to align the flange-plate bolt holes, bolts are then inserted, and the pipes released.

(f) What is the maximum permissible misalignment angle ß mix if allowable stresses in shear and bearing for all bolts [from part (d)] are not to be exceeded?

  

To determine

The reactive torque in terms of torque.

Answer to Problem 3.8.18P

The reactive torque at point (A) is = $-0.77T$.

The reactive torque at point (B) is = $-0.23T$.

Explanation of Solution

Given information:

       Figure (1) shows the systematic diagram of the system:

  

       Figure-(1)

The length of first pipe is $2.5\text{m}$, the length of second pipe is $1.5\text{m}$, thickness of flange is $14\text{mm}$, diameter of the flange bolt is $13\text{mm}$, thickness of the base plate attached to the wall is $15\text{mm}$, the diameter of the base plate bolt is $16\text{mm}$, young’s modulus of first pipe is $110\text{GPa}$, young’s modulus of second pipe is $73\text{GPa}$, poison ratio of first pipe is $0.33$, poison ratio of second pipe is $0.25$, cross sectional area of the first pipe is $1500{\text{mm}}^2$, cross sectional area of second pipe is $\left(3/5\right)A_1$, the outer diameter of first pipe is $60\text{mm}$, the outer diameter of second pipe is equal to inner diameter of first pipe and the bolt radius is $64\text{mm}$.

Write the expression for cross sectional area of the pipe (1).

  $A_1=\frac{\pi }{4}\left(d_1^2-d_2^2\right)$...... (I)

Here, the cross sectional area of the pipe (1) is $A_1$, outside diameter of the pipe is $d_1$and inner diameter is $d_2$.

Write the expression for cross sectional area of the pipe (2).

  $A_2=\frac{\pi }{4}\left(d_2^2-d_3^2\right)$...... (II)

Here, the cross sectional area of the pipe (2) is $A_2$, outside diameter of the pipe is $d_2$and inner diameter is $d_3$.

Write the expression for the thickness of the pipe (1).

  $t_1=\frac{d_1-d_2}{2}$...... (III)

Here, the thickness of the pipe (1) is $t_1$.

Write the expression for the thickness of the pipe (2).

  $t_2=\frac{d_2-d_3}{2}$...... (III)

Here, the thickness of the pipe (2) is $t_2$.

Write the expression for polar moment of inertia for pipe (1).

  $I_{p1}=\frac{\pi }{32}\left(d_1^4-d_2^4\right)$..... (IV)

Here, the polar moment of inertia of pipe (1) is $I_{P1}$.

Write the expression for polar moment of inertia for pipe (2).

  $I_{p2}=\frac{\pi }{32}\left(d_2^4-d_3^4\right)$..... (V)

Here, the polar moment of inertia of pipe (2) is $I_{P2}$.

Write the expression for modulus of rigidity for pipe (1).

  $G_1=\frac{E_1}{2\left(1+v_1\right)}$...... (VI)

Here, modulus of rigidity for pipe (1) is $G_1$, young’s modulus of pipe is $E_1$and poison ratio is $v_1$.

Write the expression for modulus of rigidity for pipe (2).

  $G_2=\frac{E_2}{2\left(1+v_2\right)}$...... (VII)

Here, modulus of rigidity for pipe (2) is $G_2$, young’s modulus of pipe is $E_2$and poison ratio is $v_2$.

       Figure-(2) shows free body diagram of the system with internal torque and reactions at the end:

  

       Figure-(2)

Write the expression of equilibrium condition for figure (2)

  $R_1+R_2=-T$...... (VIII)

Here, the reaction torque acting at the point (A) is $R_1$, reaction torque acting at the point (B) is $R_2$and the torque applied at $x=L$is $-T$.

Write the expression for torsional flexibility of pipe (1)

  $f_{T_1}=\frac{L_1}{G_1{I}_{p_1}}$...... (IX)

Here, the torsional rigidity of pipe (1) is $f_{T_1}$.

Write the expression for torsional flexibility of pipe (2)

  $f_{T_2}=\frac{L_2}{G_2{I}_{p_2}}$...... (X)

Here, the torsional rigidity of pipe (1) is $f_{T_2}$.

Write the expression for angle of twist for pipe (1)

  ${\varphi }_1=T\times f_{T_1}$

Here, the angle of twist for pipe (1) is ${\varphi }_1$.

Write the expression for angle of twist for pipe (2)

  ${\varphi }_2=R_2\left(f_{T_1}+f_{T_2}\right)f_{T_2}{}_{}^{}$

Here, the angle of twist for pipe (1) is ${\varphi }_2$.

Since, the angle of twist in pipe (1) and pipe (2) are opposite in direction

  ${\varphi }_1=-{\varphi }_2$..... (XI)

Substitute, $T\times f_{T_1}$for ${\varphi }_1$and $R_2\left(f_{T_1}+f_{T_2}\right)f_{T_2}$for ${\varphi }_2$in Equation (XI)

  $\begin{array}{l}T\times f_{T_1}=-\left(R_2\left(f_{T_1}+f_{T_2}\right)f_{T_2}\right)\\ R_2=\frac{-T{f}_{T_1}}{f_{T_1}+f_{T_2}}\end{array}$...... (XII)

Calculation:

Substitute $1500{\text{mm}}^2$for $A_1$and $60\text{mm}$for $d_1$in Equation (I).

  $\begin{array}{r}1500{\text{mm}}^2=\frac{\pi }{4}\left({\left(60\text{mm}\right)}^2-d_2^2\right)\\ 1909.85{\text{mm}}^2=\left({\left(60\text{mm}\right)}^2-d_2^2\right)\\ 1909.85{\text{mm}}^2-{\left(60\text{mm}\right)}^2=-d_2^2\\ 41.11\text{mm}=d_2\end{array}$

Substitute $\left(3/5\right)A_1$for $A_2$and $41.11\text{mm}$for $d_2$in Equation (II).

  $\begin{array}{r}\left(\frac{3}{5}\right)1500{\text{mm}}^2=\frac{\pi }{4}\left({\left(41.11\text{mm}\right)}^2-d_3^2\right)\\ 1145.9{\text{mm}}^2=\left({\left(41.11\text{mm}\right)}^2-d_3^2\right)\\ 1145.9{\text{mm}}^2-{\left(41.11\text{mm}\right)}^2=-d_3^2\\ 23.33\text{mm}=d_3\end{array}$

Substitute $60\text{mm}$for $d_1$and $41.11\text{mm}$for $d_2$in Equation (III).

  $\begin{array}{c}{t}_1=\frac{\left(60\text{mm}\right)-\left(41.11\text{mm}\right)}{2}\\ =9.945\text{mm}\end{array}$

Substitute $41.11\text{mm}$for $d_2$and $23.33\text{mm}$for $d_3$in Equation (IV).

  $\begin{array}{c}{t}_1=\frac{\left(41.11\text{mm}\right)-\left(23.33\text{mm}\right)}{2}\\ =8.89\text{mm}\end{array}$

Substitute $60\text{mm}$for $d_1$, $41.11\text{mm}$for $d_2$in Equation (V)

  $\begin{array}{c}{I}_{p1}=\frac{\pi }{32}\left({\left(60\text{mm}\right)}^4-{\left(41.11\text{mm}\right)}^4\right)\\ =0.098\left(\left(12960000{\text{mm}}^4\right)-\left(2856208.4{\text{mm}}^4\right)\right)\\ =991.91\times {10}^3{\text{mm}}^4\end{array}$

Substitute $41.11\text{mm}$for $d_2$, $23.33\text{mm}$for $d_3$in Equation (V)

  $\begin{array}{c}{I}_{p2}=\frac{\pi }{32}\left({\left(41.11\text{mm}\right)}^4-{\left(23.33\text{mm}\right)}^4\right)\\ =0.098\left(\left(2856208.4{\text{mm}}^4\right)-\left(296250.4{\text{mm}}^4\right)\right)\\ =251.36\times {10}^3{\text{mm}}^4\end{array}$

Substitute $110\text{GPa}$for $E_1$, $0.33$for $v_1$in Equation (VI)

  $\begin{array}{c}{G}_1=\frac{110\text{GPa}}{2\left(1+0.33\right)}\\ =\frac{110\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)}{2.66}\\ =0.414\times {10}^5\text{N}/{\text{mm}}^2\end{array}$

Substitute $73\text{GPa}$for $E_2$, $0.25$for $v_2$in Equation (VII)

  $\begin{array}{c}{G}_2=\frac{73\text{GPa}}{2\left(1+0.25\right)}\\ =\frac{73\text{GPa}\left(\frac{{10}^9\text{Pa}}{1\text{GPa}}\right)}{2.5}\\ =0.292\times {10}^5\text{N}/{\text{mm}}^2\end{array}$

Substitute $2500\text{mm}$for $L_1$, $0.414\times {10}^5\text{N}/{\text{mm}}^2$for $G_1$and $991.91\times {10}^3{\text{mm}}^4$for $I_{P_1}$in Equation (IX)

  $\begin{array}{c}{f}_{T_1}=\frac{2500\text{mm}}{0.414\times {10}^5\text{N}/{\text{mm}}^2\left(991.91\times {10}^3{\text{mm}}^4\right)}\\ =\frac{2500\text{mm}}{4.106\times {10}^{10}}\\ =6.088\times {10}^{-8}\end{array}$

Substitute $1500\text{mm}$for $L_1$, $0.292\times {10}^5\text{N}/{\text{mm}}^2$for $G_1$and $251.36\times {10}^3{\text{mm}}^4$for $I_{P_2}$in Equation (X)

  $\begin{array}{c}{f}_{T_2}=\frac{1500\text{mm}}{0.292\times {10}^5\text{N}/{\text{mm}}^2\left(251.36\times {10}^3{\text{mm}}^4\right)}\\ =\frac{2500\text{mm}}{7.339\times {10}^{10}}\\ =20.437\times {10}^{-8}\end{array}$

Substitute $6.088\times {10}^{-8}$for $f_{T_1}$and $20.437\times {10}^{-8}$for $f_{T_2}$in equation (XII).

  $\begin{array}{l}{R}_2=\frac{-T\left(6.088\times {10}^{-8}\right)}{\left(6.088\times {10}^{-8}\right)+\left(20.437\times {10}^{-8}\right)}\\ R_2=-0.23T\end{array}$

Substitute $-0.23T$for $R_2$in Equation (VIII).

  $\begin{array}{l}{R}_1+-0.23T=-T\\ R_1=-T+0.23T\\ R_1=-0.77T\end{array}$

Conclusion:

The reactive torque at point (A) is $-0.77T$.

The reactive torque at point (B) is $-0.23T$.

To determine

The maximum load variable

Answer to Problem 3.8.18P

The maximum load variable

Explanation of Solution

Given information:

Permissible torsional stress in two pipes is $65\text{MPa}$.

Write the expression for maximum load in pipe (1).

  $T_1=\frac{{\tau }_a{I}_{p_1}}{\left(\frac{d_1}{2}\right)}$...... (XIII)

Here, maximum load in the pipe (1) is $T_1$and the permissible stress is ${\tau }_a$.

Write the expression for maximum load in pipe (1).

  $T_2=\frac{{\tau }_a{I}_{p_2}}{\left(\frac{d_2}{2}\right)}$...... (XIV)

Here, maximum load in the pipe (2) is $T_2$.

Write the expression for maximum load variable in pipe (1).

  $T_{\max }=\frac{T_1}{\left(\frac{f_{T_2}}{f_{T_1}+f_{T_2}}\right)}$...... (XV)

Here, the maximum load variable is $T_{\max }$.

Calculation:

Substitute $65\text{MPa}$for ${\tau }_a$, $991.91\times {10}^3{\text{mm}}^4$for $I_{P_1}$, and $60\text{mm}$for $d_1$in equation (XIII).

  $\begin{array}{c}{T}_1=\frac{65\text{MPa}\left(991.91\times {10}^3{\text{mm}}^4\right)}{\left(\frac{60\text{mm}}{2}\right)}\\ =\frac{65\text{MPa}\left(991.91\times {10}^3{\text{mm}}^4\right)}{\left(30\text{mm}\right)}\\ =2149\times {10}^3\text{N}\cdot \text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =2.149\text{kN}\cdot \text{m}\end{array}$

Substitute $65\text{MPa}$for ${\tau }_a$, $251.36\times {10}^3{\text{mm}}^4$for $I_{P_1}$, and $41.11\text{mm}$for $d_1$in equation (XIII).

  $\begin{array}{c}{T}_2=\frac{65\text{MPa}\left(251.36\times {10}^3{\text{mm}}^4\right)}{\left(\frac{41.11\text{mm}}{2}\right)}\\ =\frac{65\text{MPa}\left(991.91\times {10}^3{\text{mm}}^4\right)}{\left(20.55\text{mm}\right)}\\ =795\times {10}^3\text{N}\cdot \text{mm}\left(\frac{1\text{m}}{1000\text{mm}}\right)\left(\frac{1\text{kN}}{1000\text{N}}\right)\\ =0.795\text{kN}\cdot \text{m}\end{array}$

Since, the pipe (1) is greater, the maximum load variable in pipe (1).

Substitute $2.149\text{kN}\cdot \text{m}$for $T_1$, $6.088\times {10}^{-8}$for $f_{T_1}$and $20.437\times {10}^{-8}$for $f_{T_2}$in Equation (XV).

  $\begin{array}{c}{T}_{\max }=\frac{2.149\text{kN}\cdot \text{m}}{\left(\frac{6.088\times {10}^{-8}}{6.088\times {10}^{-8}+20.437\times {10}^{-8}}\right)}\\ =\frac{2.149\text{kN}\cdot \text{m}}{\left(0.77\right)}\\ =2.79\text{kN}\cdot \text{m}\end{array}$

Conclusion:

The maximum load variable is = $2.79\text{kN}\cdot \text{m}$.

To determine

The torsional moment diagrams.

The torsional displacement diagrams.

The maximum angle of twist.

Answer to Problem 3.8.18P

The maximum angle of twist is $7.49°$.

Explanation of Solution

Given information:

       Figure (3) shows torsional moment diagram and torsional displacement diagram for the given system.

  

       Figure-(3)

Write the expression for maximum angle of twist

  ${\varphi }_{\max }=\left[\frac{\left(f_{T_1}\times f_{T_2}\right)}{\left(f_{T_1}+f_{T_2}\right)}\right]\times T_{\max }$...... (XVI)

Here, maximum angle of twist is = ${\varphi }_{\max }$.

Calculation:

Substitute $2.79\text{kN}\cdot \text{m}$for $T_{\max }$, $6.088\times {10}^{-8}$for $f_{T_1}$and $20.437\times {10}^{-8}$for $f_{T_2}$in Equation (XVI).

  $\begin{array}{c}{\varphi }_{\max }=\left[\frac{\left(\left(6.088\times {10}^{-8}\right)\times \left(20.437\times {10}^{-8}\right)\right)}{\left(\left(6.088\times {10}^{-8}\right)+\left(20.437\times {10}^{-8}\right)\right)}\right]\times 2.79\text{kN}\cdot \text{m}\\ =\left[\frac{\left(1.2442\times {10}^{-14}\right)}{\left(\left(26.525\times {10}^{-8}\right)\right)}\right]\times 2.79\text{kN}\cdot \text{m}\left(\frac{1000\text{N}}{1\text{kN}}\right)\\ \text{=0}\text{.1308}\left(\frac{180}{\pi }\right)\\ =7.49°\end{array}$

Conclusion:

The maximum angle of twist is = $7.49°$.

To determine

The maximum torque for limiting value shear and bearing stress of base plate and flange bolts.

Answer to Problem 3.8.18P

The maximum torque for limiting value shear and bearing stress of base plate and flange bolts is = $2480727.27\text{N}$.

Explanation of Solution

Given information:

The permissible shear stress for all bolts is $45\text{MPa}$, and permissible bearing stress is $90\text{MPa}$.

Write the expression for cross -sectional area of the base plate bolt.

  $A_{cbb}=\frac{\pi }{4}{d}_{bb}^2$...... (XVII)

Here, the cross-sectional area of the base plate bolt is $A_{cbb}$, the diameter of the base plate $d_{bb}$.

Write the expression for cross sectional area of the flange plate bolt.

  $A_{cbf}=\frac{\pi }{4}{d}_{bf}^2$...... (XVII)

Here, the cross-sectional area of the flange plate bolt is $A_{bf}$, the diameter of the flange plate $d_{bf}$.

Write the expression for maximum shear load in base plate bolt.

  $T_{\max ,bb}=\frac{n_b\times r\left({\tau }_{allow}\times A_{cbb}\right)}{\left(\frac{f_{T_2}}{f_{T_1}+f_{T_2}}\right)}$...... (XVIII)

Here, the maximum shear load in base plate is $T_{\max ,bb}$and number of base plate bolts is $n_b$Write the expression for maximum shear load in flange plate bolt.

  $T_{\max ,bf}=\frac{n_f\times r\left({\tau }_{allow}\times A_{cbf}\right)}{\left(\frac{f_{T_2}}{f_{T_1}+f_{T_2}}\right)}$...... (XIX)

Here, the maximum shear load in flange plate is $T_{\max ,bf}$and number of base plate bolts is $n_f$.

Write the expression for area of the base plate bolt.

  $A_{bb}=d_{bb}\times t_b$...... (XX)

Here, the area of base plate bolt is $A_{bb}$and the thickness of the is $t_b$.

Write the expression for area of the flange plate bolt.

  $A_{bf}=d_{bf}\times t_f$...... (XXI)

Here, the area of base plate bolt is $A_{bf}$and the thickness of the is $t_f$.

Write the expression for bearing load on base plate

  $T_{\max ,bb}=\frac{n_b\times r\left({\sigma }_{allow}\times A_{bb}\right)}{\left(\frac{f_{T_2}}{f_{T_1}+f_{T_2}}\right)}$...... (XXII)

Here, the maximum shear load in base plate is $T_{\max ,bb}$and number of base plate bolts is $n_b$Write the expression for maximum shear load in flange plate bolt.

  $T_{\max ,bb}=\frac{n_f\times r\left({\sigma }_{allow}\times A_{bf}\right)}{\left(\frac{f_{T_2}}{f_{T_1}+f_{T_2}}\right)}$...... (XXIII)

Here, the maximum shear load in flange plate is $T_{\max ,bf}$and number of base plate bolts is = $n_f$.

Calculation:

Substitute $16\text{mm}$for $d_{bb}$in Equation (XVII).

  $\begin{array}{c}{A}_{bb}=\frac{\pi }{4}{\left(16\text{mm}\right)}^2\\ =\left(0.785\right)\left(256{\text{mm}}^2\right)\\ =200.96{\text{mm}}^{\text{2}}\end{array}$

Substitute $13\text{mm}$for $d_{bf}$in Equation (XVIII).

  $\begin{array}{c}{A}_{bf}=\frac{\pi }{4}{\left(13\text{mm}\right)}^2\\ =\left(0.785\right)\left(169{\text{mm}}^2\right)\\ =132.65{\text{mm}}^{\text{2}}\end{array}$

Substitute $200.96{\text{mm}}^{\text{2}}$for $A_{bb}$, $64\text{mm}$for $r$, $4$for $n_b$, $45\text{MPa}$for ${\tau }_{allow}$, $6.088\times {10}^{-8}$for $f_{T_1}$and $20.437\times {10}^{-8}$for $f_{T_2}$in Equation (XVIII)

  $\begin{array}{c}{T}_{\max ,bb}=\frac{4\times 64\text{mm}\left(\left(45\text{MPa}\right)\times \left(200.96{\text{mm}}^{\text{2}}\right)\right)}{\left(\frac{20.437\times {10}^{-8}}{\left(6.088\times {10}^{-8}\right)+\left(20.437\times {10}^{-8}\right)}\right)}\\ =\frac{4\times 64\text{mm}\left(\left(45\text{MPa}\right)\times \left(200.96{\text{mm}}^{\text{2}}\right)\right)}{0.77}\\ =\frac{2.315\times {10}^6\text{N}}{0.77}\\ =3006493.5\text{N}\end{array}$

Substitute $132.65{\text{mm}}^{\text{2}}$for $A_{bf}$, $64\text{mm}$for $r$, $5$for $n_b$, $45\text{MPa}$for ${\tau }_{allow}$, $6.088\times {10}^{-8}$for $f_{T_1}$and $20.437\times {10}^{-8}$for $f_{T_2}$in Equation (XVIII)

  $\begin{array}{c}{T}_{\max ,bb}=\frac{5\times 64\text{mm}\left(\left(45\text{MPa}\right)\times \left(132.65{\text{mm}}^{\text{2}}\right)\right)}{\left(\frac{20.437\times {10}^{-8}}{\left(6.088\times {10}^{-8}\right)+\left(20.437\times {10}^{-8}\right)}\right)}\\ =\frac{5\times 64\text{mm}\left(\left(45\text{MPa}\right)\times \left(132.65{\text{mm}}^{\text{2}}\right)\right)}{0.77}\\ =\frac{1.910\times {10}^6\text{N}}{0.77}\\ =2480727.27\text{N}\end{array}$

Substitute $16\text{mm}$for $d_{bb}$and $15\text{mm}$for $t$in Equation (XX).

  $\begin{array}{c}{A}_{bb}=16\text{mm}\times 15\text{mm}\\ =240{\text{mm}}^2\end{array}$

Substitute $13\text{mm}$for $d_{bb}$and $14\text{mm}$for $t$in Equation (XXI).

  $\begin{array}{c}{A}_{bf}=13\text{mm}\times 14\text{mm}\\ =182{\text{mm}}^2\end{array}$

Substitute $240{\text{mm}}^2$for $A_{bb}$, $64\text{mm}$for $r$, $4$for $n_b$, $90\text{MPa}$for ${\tau }_{allow}$, $6.088\times {10}^{-8}$for $f_{T_1}$and $20.437\times {10}^{-8}$for $f_{T_2}$in Equation (XVIII)

  $\begin{array}{c}{T}_{\max ,bb}=\frac{4\times 64\text{mm}\left(\left(90\text{MPa}\right)\times \left(240{\text{mm}}^2\right)\right)}{\left(\frac{20.437\times {10}^{-8}}{\left(6.088\times {10}^{-8}\right)+\left(20.437\times {10}^{-8}\right)}\right)}\\ =\frac{4\times 64\text{mm}\left(\left(90\text{MPa}\right)\times \left(240{\text{mm}}^2\right)\right)}{0.77}\\ =\frac{5.529\times {10}^6\text{N}}{0.77}\\ =7181298.701\text{N}\end{array}$

Substitute $182{\text{mm}}^2$for $A_{bf}$, $64\text{mm}$for $r$, $5$for $n_b$, $90\text{MPa}$for ${\tau }_{allow}$, $6.088\times {10}^{-8}$for $f_{T_1}$and $20.437\times {10}^{-8}$for $f_{T_2}$in Equation (XVIII)

  $\begin{array}{c}{T}_{\max ,bf}=\frac{5\times 64\text{mm}\left(\left(90\text{MPa}\right)\times \left(182{\text{mm}}^2\right)\right)}{\left(\frac{20.437\times {10}^{-8}}{\left(6.088\times {10}^{-8}\right)+\left(20.437\times {10}^{-8}\right)}\right)}\\ =\frac{5\times 64\text{mm}\left(\left(90\text{MPa}\right)\times \left(182{\text{mm}}^2\right)\right)}{0.77}\\ =\frac{5.184\times {10}^6\text{N}}{0.77}\\ =6732467.532\text{N}\end{array}$

Conclusion:

The maximum torque for limiting value shear and bearing stress of base plate and flange bolts is = $2480727.27\text{N}$.