Chapter 3, Problem 3.48P

(a)

To determine

The deflection under load.

Answer to Problem 3.48P

The deflection under load is $0.02696\text{m}$ .

Explanation of Solution

Given:

The length of the beam is $L=2\text{m}$ .

The mass of the beam is $m=25\text{kg}$ .

The diameter of the shaft is $d=20\text{mm}$ .

Formula used:

The expression for the load is given as,

  $P=mg$

Here, $g$ is the acceleration due to gravity.

The expression for the moment of inertia of the shaft is given as,

  $I=\frac{\pi }{64}{d}^4$

The expression for the deflection is given as,

  $Y=\frac{1}{48}\frac{P{L}^3}{EI}$

Here, $E$ is the young’s modulus.

Calculation:

The value of load can be calculated as,

  $\begin{array}{c}P=mg\\ =25\text{kg}\times 9.8\text{m}/{\text{s}}^{\text{2}}\\ =245.25\text{N}\end{array}$

The value of moment of inertia of the shaft can be calculated as,

  $\begin{array}{c}I=\frac{\pi }{64}{d}^4\\ =\frac{\pi }{64}{\left(20\times {10}^{-3}\text{m}\right)}^4\\ =7.85\times {10}^{-9}{\text{m}}^{\text{4}}\end{array}$

The value of young’s modulus of AISI 303 steel is $193\text{GPa}$ .

The value of deflection can be calculated as,

  $\begin{array}{c}Y=\frac{1}{48}\frac{P{L}^3}{EI}\\ =\frac{1}{48}\frac{245.25\text{N}\times {\left(2\text{m}\right)}^3}{193\times {10}^9\text{Pa}\times 7.85\times {10}^{-9}{\text{m}}^{\text{4}}}\\ =0.02696\text{m}\end{array}$

Conclusion:

Therefore, the deflection under load is $0.02696\text{m}$ .

(b)

To determine

The diameter of the shaft made from 2024-T4 aluminum.

The diameter of the shaft made from architectural bronze.

The diameter of the shaft made from 99.5% titanium.

Answer to Problem 3.48P

The diameter of the shaft made from 2024-T4 aluminum is $25.495\text{mm}$ .

The diameter of the shaft made from architectural bronze is $\text{23}\text{.206}\text{mm}$ .

The diameter of the shaft made from 99.5% titanium is $23.288\text{mm}$ .

Explanation of Solution

Calculation:

The value of young’s modulus of 2024-T4 aluminum is $73.1\text{GPa}$ .

The value of moment of inertia of the shaft made from 2024-T4 aluminum can be calculated as,

  $\begin{array}{c}Y=\frac{1}{48}\frac{P{L}^3}{EI}\\ 0.02696\text{m}=\frac{1}{48}\frac{245.25\text{N}\times {\left(2\text{m}\right)}^3}{73.1\times {10}^9\text{Pa}\times I}\\ I=2.074\times {10}^{-8}{\text{m}}^{\text{4}}\end{array}$

The value of the diameter of the shaft made from 2024-T4 aluminum can be calculated as,

  $\begin{array}{c}I=\frac{\pi }{64}{d}^4\\ 2.074\times {10}^{-8}{\text{m}}^{\text{4}}=\frac{\pi }{64}{d}^4\\ d=0.025495\text{mm}\left(\frac{1000\text{m}}{1\text{mm}}\right)\\ d=25.495\text{mm}\end{array}$

The value of young’s modulus of architectural bronze is $97\text{GPa}$ .

The value of moment of inertia of the shaft made from architectural bronze can be calculated as,

  $\begin{array}{c}Y=\frac{1}{48}\frac{P{L}^3}{EI}\\ 0.02696\text{m}=\frac{1}{48}\frac{245.25\text{N}\times {\left(2\text{m}\right)}^3}{97\times {10}^9\text{Pa}\times I}\\ I=1.4236\times {10}^{-8}{\text{m}}^{\text{4}}\end{array}$

The value of the diameter of the shaft made from architectural bronze can be calculated as,

  $\begin{array}{c}I=\frac{\pi }{64}{d}^4\\ 1.4236\times {10}^{-8}{\text{m}}^{\text{4}}=\frac{\pi }{64}{d}^4\\ d=0.023206\text{mm}\left(\frac{1000\text{m}}{1\text{mm}}\right)\\ d=23.206\text{mm}\end{array}$

The value of young’s modulus of 99.5% titanium is $105\text{GPa}$

The value of moment of inertia of the shaft made from 99.5% titanium can be calculated as,

  $\begin{array}{c}Y=\frac{1}{48}\frac{P{L}^3}{EI}\\ 0.02696\text{m}=\frac{1}{48}\frac{245.25\text{N}\times {\left(2\text{m}\right)}^3}{105\times {10}^9\text{Pa}\times I}\\ I=1.4439\times {10}^{-8}{\text{m}}^{\text{4}}\end{array}$

The value of the diameter of the shaft made from 99.5% titanium can be calculated as,

  $\begin{array}{c}I=\frac{\pi }{64}{d}^4\\ 1.4439\times {10}^{-8}{\text{m}}^{\text{4}}=\frac{\pi }{64}{d}^4\\ d=0.023288\text{mm}\left(\frac{1000\text{m}}{1\text{mm}}\right)\\ d=23.288\text{mm}\end{array}$

Conclusion:

Therefore, the diameter of the shaft made from 2024-T4 aluminum is $25.495\text{mm}$ .

Therefore, the diameter of the shaft made from architectural bronze is $\text{23}\text{.206}\text{mm}$ .

Therefore, the diameter of the shaft made from 99.5% titanium is $23.288\text{mm}$ .