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goldfiish [28.3K]
2 years ago
15

determine the position d of the 6- kn load so that the average normal stress in each rod is the same.

Engineering
1 answer:
Zinaida [17]2 years ago
7 0

The load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

<h3>What is meant by torque?</h3>

The force that can cause an object to rotate along an axis is measured as torque. Similar to how force accelerates an item in linear kinematics, torque accelerates an object in an angular direction. A vector quantity is torque.

Let the beam is of length L

Now the stress on both the end is the same now we can say that torque on the beam due to two forces must be zero

$N_1 * x=N_2 *(L-x)$

also, we know that stress at both ends are same

$\frac{N_1}{12}=\frac{N_2}{8}$

$2 * N_1=3 * N_2$

Now from two equations we have

$\frac{3}{2} N_2 * x=N_2 *(L-x)

solving the above equation we have

$x=\frac{2}{5} L

so the load is placed at distance 0.4 L from the end of $$12 \mathrm{~mm}^{2} $ area.

The complete question is:

47. the beam is supported by two rods ab and cd that have cross-sectional areas of $$12mm^2 and $$8mm^2, respectively. determine the position d of the 6-kn load so that the average normal stress in each rod is the same.

To learn more about torque refer to:

brainly.com/question/20691242

#SPJ4

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Water flows in a constant diameter pipe with the following conditions measured:
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Answer:

a) h_L=-3.331ft

b) The flow would be going from section (b) to section (a)

Explanation:

1) Notation

p_a =31.1psi=4478.4\frac{lb}{ft^2}

p_b =27.3psi=3931.2\frac{lb}{ft^2}

For above conversions we use the conversion factor 1psi=144\frac{lb}{ft^2}

z_a =56.7ft

z_a =68.8ft

h_L =? head loss from section

2) Formulas and definitions

For this case we can apply the Bernoulli equation between the sections given (a) and (b). Is important to remember that this equation allows en energy balance since represent the sum of all the energies in a fluid, and this sum need to be constant at any point selected.

The formula is given by:

\frac{p_a}{\gamma}+\frac{V_a^2}{2g}+z_a =\frac{p_b}{\gamma}+\frac{V_b^2}{2g}+z_b +h_L

Since we have a constant section on the piple we have the same area and flow, then the velocities at point (a) and (b) would be the same, and we have just this expression:

\frac{p_a}{\gamma}+z_a =\frac{p_b}{\gamma}+z_b +h_L

3)Part a

And on this case we have all the values in order to replace and solve for h_L

\frac{4478.4\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+56.7ft=\frac{3931.2\frac{lb}{ft^2}}{62.4\frac{lb}{ft^3}}+68.8ft +h_L

h_L=(71.769+56.7-63-68.8)ft=-3.331ft

4)Part b

Analyzing the value obtained for \h_L is a negative value, so on this case this means that the flow would be going from section (b) to section (a).

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