Question

e = 2.1×10^5 mpa, determine the deformation in the length of the bar caused by these loads. a_1 = 250 mm^2 a_2 = 200 mm^2 50kn 30kn 20kn a b c d e 1m 2m 1m 3m

Explanation:

Step1: Calculate internal forces in each segment

• Segment AB: $F_{AB}=50\times10^{3}\ N$
• Segment BC: $F_{BC}=50\times10^{3}\ N$
• Segment CD: $F_{CD}=50\times10^{3}- 30\times10^{3}=20\times10^{3}\ N$
• Segment DE: $F_{DE}=20\times10^{3}\ N$

Step2: Use the formula for axial - deformation $\Delta L=\frac{FL}{AE}$

• For segment AB: $\Delta L_{AB}=\frac{F_{AB}L_{AB}}{A_{1}E}$, where $L_{AB} = 1\ m=1000\ mm$, $A_{1}=250\ mm^{2}$, $E = 2.1\times10^{5}\ MPa=2.1\times10^{5}\ N/mm^{2}$, $F_{AB}=50\times10^{3}\ N$. So $\Delta L_{AB}=\frac{50\times10^{3}\times1000}{250\times2.1\times10^{5}}=\frac{50\times10^{6}}{250\times2.1\times10^{5}}=\frac{500}{25\times2.1}=\frac{20}{2.1}\ mm$
• For segment BC: $\Delta L_{BC}=\frac{F_{BC}L_{BC}}{A_{1}E}$, $L_{BC}=2000\ mm$, $F_{BC}=50\times10^{3}\ N$. So $\Delta L_{BC}=\frac{50\times10^{3}\times2000}{250\times2.1\times10^{5}}=\frac{100\times10^{6}}{250\times2.1\times10^{5}}=\frac{1000}{25\times2.1}=\frac{40}{2.1}\ mm$
• For segment CD: $\Delta L_{CD}=\frac{F_{CD}L_{CD}}{A_{2}E}$, $L_{CD}=1000\ mm$, $A_{2}=200\ mm^{2}$, $F_{CD}=20\times10^{3}\ N$. So $\Delta L_{CD}=\frac{20\times10^{3}\times1000}{200\times2.1\times10^{5}}=\frac{20\times10^{6}}{200\times2.1\times10^{5}}=\frac{200}{2\times2.1}=\frac{100}{2.1}\ mm$
• For segment DE: $\Delta L_{DE}=\frac{F_{DE}L_{DE}}{A_{2}E}$, $L_{DE}=3000\ mm$, $F_{DE}=20\times10^{3}\ N$. So $\Delta L_{DE}=\frac{20\times10^{3}\times3000}{200\times2.1\times10^{5}}=\frac{60\times10^{6}}{200\times2.1\times10^{5}}=\frac{600}{2\times2.1}=\frac{300}{2.1}\ mm$

Step3: Calculate total deformation

$\Delta L=\Delta L_{AB}+\Delta L_{BC}+\Delta L_{CD}+\Delta L_{DE}=\frac{20 + 40+100 + 300}{2.1}=\frac{460}{2.1}\approx219.05\ mm$

Answer:

$\approx219.05\ mm$