Chapter 6, Problem 6.2.6P

A r o lukI f/frm f «m t ub e of ou t sid e d ia met er ^ and a copper core of diameter d_xare bonded to form a composite beam, as shown in the figure,

(a) Derive formulas for the allowable bending moment M that can be carried by the beam based upon an allowable stress <7_Ti in the titanium and an allowable stress ( _u in the

copper (Assume that the moduli of elasticity for the titanium and copper are E_r- and £_Cu, respectively.)

(b)

If d₁= 40 mm, d_{= 36 mm, E_Tl= 120 GPa, E_Cu= 110 GPa, o-_Ti = 840 MPa, and

ctqj = 700 MPa, what is the maximum bending moment Ml (c)

What new value of copper diameter d_twill result

in a balanced design? (i.e., a balanced design is

that in which titanium and copper reach allow-

able stress values at the same time).

  

To determine

The formula for the allowable bending moment M for titanium tube and copper core of the composite beam

Answer to Problem 6.2.6P

  $\begin{array}{l}Theallowablebendingmomentfortitianium,M_{allowableTi}=\frac{{\sigma }_{Ti}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Ti}*d_2}\\ Theallowablebendingmomentforcopper,M_{allowableCu}=\frac{{\sigma }_{Cu}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Cu}*d_1}\end{array}$

Explanation of Solution

Given:

Allowable stress for titanium is s_ti

Allowable stress for titanium is s_cu

Diameter of the copper rod = d₁

Outer Diameter of the titanium rod = d₂

Concept Used:

  $\begin{array}{l}Themoment\text{}ofinertiaofhollowtube=\frac{\pi \left(d_2{}^4-d_1{}^4\right)}{64}\\ Themoment\text{}ofinertiaofarod=\frac{\pi \left(d_1{}^4\right)}{64}\\ Allowablebendingmoment=\frac{{\sigma }_1\left(E_1{I}_1+E_2{I}_2\right)}{E_1*\left(\frac{d_2}{2}\right)}\\ \end{array}$

Calculation:

  $Themoment\text{}ofinertiaofti\tan iumtube=\frac{\pi \left(d_2{}^4-d_1{}^4\right)}{64}$

  $Themoment\text{}ofinertiaofthecopperrod=\frac{\pi \left(d_1{}^4\right)}{64}$

  $Allowablebendingmomentofti\tan ium=\frac{{\sigma }_{Ti}\left(E_{Ti}{I}_{Ti}+E_{Cu}{I}_{Cu}\right)}{E_{Ti}*\left(\frac{d_2}{2}\right)}$

  $M_{allowableTi}=\frac{{\sigma }_{Ti}\left(E_{Ti}*\left(\frac{\pi \left(d_2{}^4-d_1{}^4\right)}{64}\right)+E_{Cu}\left(\frac{\pi \left(d_1{}^4\right)}{64}\right)\right)}{E_{Ti}*\left(\frac{d_2}{2}\right)}$ $M_{allowableTi}=\frac{{\sigma }_{Ti}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Ti}*d_2}$

  $Allowablebendingmomentofcopper=\frac{{\sigma }_{Cu}\left(E_{Ti}{I}_{Ti}+E_{Cu}{I}_{Cu}\right)}{E_{Cu}*\left(\frac{d_2}{2}\right)}$ $M_{allowableCu}=\frac{{\sigma }_{Cu}\left(E_{Ti}*\left(\frac{\pi \left(d_2{}^4-d_1{}^4\right)}{64}\right)+E_{Cu}\left(\frac{\pi \left(d_1{}^4\right)}{64}\right)\right)}{E_{Cu}*\left(\frac{d_2}{2}\right)}$

  $M_{allowableCu}=\frac{{\sigma }_{Cu}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Cu}*d_1}$

Conclusion:

  $\begin{array}{l}Theallowablebendingmomentfortitianium,M_{allowableTi}=\frac{{\sigma }_{Ti}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Ti}*d_2}\\ Theallowablebendingmomentforcopper,M_{allowableCu}=\frac{{\sigma }_{Cu}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Cu}*d_1}\end{array}$

To determine

The allowable bending moment for titanium and copper.

Answer to Problem 6.2.6P

Allowable bending moment for titanium, M_allowableti = 4989 N-m

Allowable bending moment for Copper, M_allowableCu =5039.6 N-m

Explanation of Solution

Given:

Allowable stress for titanium, s_ti = 840 MPa

Allowable stress for titanium, s_cu= 700 MPa

Diameter of the copper rod, d₁= 36 mm

Outer Diameter of the titanium rod, d₂ = 40 mm

E_ti= 110 GPa

E_cu= 120 GPa

Concept Used:

  $\begin{array}{l}Theallowablebendingmomentfortitianium,M_{allowableTi}=\frac{{\sigma }_{Ti}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Ti}*d_2}\\ Theallowablebendingmomentforcopper,M_{allowableCu}=\frac{{\sigma }_{Cu}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Cu}*d_1}\end{array}$

Calculation:

  $\begin{array}{l}Theallowablebendingmomentfortitianium,M_{allowableTi}=\frac{{\sigma }_{Ti}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Ti}*d_2}\\ M_{allowableTi}=\frac{840\pi \left(110*{10}^3*\left({40}^4-{36}^4\right)+120*{10}^3*{36}^4\right)}{32*110*{10}^3*36}\\ M_{allowableTi}=4989N-m\\ Theallowablebendingmomentforcopper,M_{allowableCu}=\frac{{\sigma }_{Cu}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Ti}*d_2}\\ M_{allowableTi}=\frac{700\pi \left(110*{10}^3*\left({40}^4-{36}^4\right)+120*{10}^3*{36}^4\right)}{32*120*{10}^3*36}\\ M_{allowableTi}=5039.6N-m\end{array}$Conclusion:

Allowable bending moment for titanium, M_allowableti = 4989 N-m

Allowable bending moment for Copper, M_allowableCu =5039.6 N-m

To determine

The value of the diameter of the copper rod for a balanced design

Answer to Problem 6.2.6P

The diameter of the copper for a balanced design is 36.4 mm

Explanation of Solution

Given:

Allowable stress for titanium, s_ti = 840 MPa

Allowable stress for titanium, s_cu= 700 MPa

Outer Diameter of the titanium rod, d₂ = 40 mm

E_ti= 110 GPa

E_cu= 120 GPa

Concept Used:

  $\begin{array}{l}Allowablebendingmomentofti\tan ium=Allowablebendingmomentofcopper\\ \frac{{\sigma }_{Ti}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Ti}*d_2}=\frac{{\sigma }_{Cu}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Cu}*d_1}\end{array}$

Calculation:

  $\begin{array}{l}Allowablebendingmomentofti\tan ium=Allowablebendingmomentofcopper\\ \frac{{\sigma }_{Ti}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Ti}*d_2}=\frac{{\sigma }_{Cu}\pi \left(E_{Ti}*\left(d_2{}^4-d_1{}^4\right)+E_{Cu}{d}_1{}^4\right)}{32*E_{Cu}*d_1}\\ \frac{{\sigma }_{Ti}}{E_{Ti}*d_2}=\frac{{\sigma }_{Cu}}{E_{Cu}*d_1}\\ \frac{840}{120*40}=\frac{700}{110*d_1}\\ d_1=36.4mm\end{array}$

Conclusion:

The diameter of the copper for a balanced design is 36.4 mm