Answer
$1.2\times10^{11}N/m^{2}$
Work Step by Step
Let's apply equation 10.17 $F=Y(\frac{\Delta L}{L_{0}})A$ to the composite rod,
$F=Y_{COMP}(\frac{\Delta L_{COMP}}{L_{COMP}})A-(1)$
We know that,
$\Delta L_{COMP}=\Delta L_{ALU}+\Delta L_{TUNG}=\frac{FL_{ALU}}{Y_{ALU}A}+\frac{FL_{TUNG}}{Y_{TUNG}A}-(2)$
(2)=>(1),
$F=(\frac{Y_{COMP}A}{L_{COMP}})(\frac{FL_{ALU}}{Y_{ALU}A}+\frac{FL_{TUNG}}{Y_{TUNG}A})$
$1=Y_{COMP}(\frac{L_{ALU}}{L_{COMP}Y_{ALU}}+\frac{L_{TUNG}}{L_{COMP}Y_{TUNG}})-(3)$
Given that,
$\frac{L_{ALU}}{L_{COMP}}=\frac{L_{TUNG}}{L_{COMP}}=\frac{1}{2}$
(3)=>
$1=Y_{COMP}(\frac{1}{2Y_{ALU}}+\frac{1}{2Y_{TUNG}})=>Y_{COMP}=\frac{2Y_{ALU}Y_{TUNG}}{Y_{ALU}+Y_{TUNG}}$
Let's plug the values from table 10.1and we get,
$Y_{COMP}=\frac{2(3.6\times10^{11}N/m^{2})(6.9\times10^{10}N/m^{2})}{(3.6\times10^{11}N/m^{2})+(6.9\times10^{10}N/m^{2})}=1.2\times10^{11}N/m^{2}$
