Question

a. A member LMNP is subjected to point loads as shown in Figure 3. Calculate: (i) Force P necessary
for equilibrium (ii) Total elongation of the bar (iii) stresses in each portion. Take modulus of
Elasticity is equal to 210 GN/m².
[C01]
2400 mm?
600 mm
1200 mm
50 KN
P 500 KN
2
L
200 KN
P
M
IN
1000 mm
1000 mm
-600 mm
my
Figure 3​

Answer 1

Explanation:

1000m

I hope it may help to you

Answer 2

Answer:

Concept:

Chemical equilibrium refers to the situation in a chemical reaction where both the reactants and products are present in concentrations that have no further tendency to change over time, preventing any discernible change in the system's properties.

Explanation:

1) The force P is necessary for equilibrium so, resolving the force on the rod horizontally,

                      $\begin{aligned}&\sum P_{x}=0 \\&-P_{1}+P_{2}-P_{3}+P_{4}=0 \\&-50+P-500+200=0 \\&P=350 k N\end{aligned}$

Hence, the force 350kN is necessary for equilibrium.

2) Refer to the image

   $Let \ the \ \delta_{L M}, \delta_{M N}, \delta_{N P} be change in the lengths LM, MN, NP. \begin{aligned}\delta_{L M} &=\frac{P_{1} L_{1}}{A_{1} E} \\\delta_{L M} &=\frac{50 \times 10^{3} \times 1000 \times 10^{-3}}{600 \times 10^{-6} \times 210 \times 10^{9}} \\\delta_{L M} &=0.0003968 \\\delta_{L M} &=3.97 \times 10^{-4} \mathrm{~m}\end{aligned}$

The elongation will be increased for length LM. Because the forces on LM is tension or elongated. And elongation will be positive.

                                      $\begin{aligned}\delta_{M N} &=\frac{P_{2} L_{2}}{A_{2} E} \\\delta_{M N} &=\frac{300 \times 10^{3} \times 1000 \times 10^{-3}}{2400 \times 10^{-6} \times 210 \times 10^{9}} \\\delta_{M N} &=0.000595 \\\delta_{M N} &=5.95 \times 10^{-4} \mathrm{~m}\end{aligned}$

The elongation will be decreased for length MN. Because the forces on MN is compressive.

The elongation for MN will be negative.

                                    $\begin{aligned}&\delta_{M N}=\frac{P_{2} L_{2}}{A_{2} E} \\&\delta_{M N}=\frac{300 \times 10^{3} \times 1000 \times 10^{-3}}{2400 \times 10^{-6} \times 210 \times 10^{9}} \\&\delta_{M N}=0.000595 \\&\delta_{M N}=5.95 \times 10^{-4} \mathrm{~m}\end{aligned}$

The elongation will be increased for length NP. Because the forces on NP is tensile.

The elongation for NP will be positive.

                                          $\begin{aligned}\delta_{N P} &=\frac{P_{3} L_{3}}{A_{3} E} \\\delta_{N P} &=\frac{200 \times 10^{3} \times 600 \times 10^{-3}}{1200 \times 10^{-6} \times 210 \times 10^{9}} \\\delta_{N P} &=0.000476 \\\delta_{N P} &=4.76 \times 10^{-4} \mathrm{~m}\end{aligned}$

Total elongation,

                                       $\begin{aligned}&\delta=\delta_{L M}+\delta_{M N}+\delta_{N P} \\&\delta=3.97 \times 10^{-4}-5.95 \times 10^{-4}+4.76 \times 10^{-4} \\&\delta=2.78 \times 10^{-4} m\end{aligned}$

Hence the total elongation is   $&\delta=2.78 \times 10^{-4} m$

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