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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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bearhunter [10]

Answer:

A. Maple

Explanation:

Maple is a hardwood.

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2 years ago
Consider the base plate of an 800-W household iron with a thickness of L = 0.6 cm, base area of A =160 cm2, and thermal conducti
ratelena [41]

Given-

Power, P = 800W

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The heat conduction would be

\frac{d^2T}{dx^2} + \frac{d^2T}{dy^2} + \frac{d^2T}{dz^2} + \frac{e(gen)}{k}  = \frac{1}{\alpha } \frac{dT}{dt}

Except \frac{d^2T}{dx^2} all the values are 0.

Therefore,

\frac{d^2T}{dx^2} = 0

Thus, the boundary conditions here would be

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4 0
4 years ago
Liquid flows at steady state at a rate of 2 lb/s through a pump, which operates to raise the elevation of the liquid 100 ft from
Greeley [361]

Answer:

D) 1.04 Btu/s from the liquid to the surroundings.

Explanation:

Given that:

flow rate (m) = 2 lb/s

liquid specific enthalpy at the inlet (h_{1}=40.09 Btu/lb)

liquid specific enthalpy at the exit (h_{2}=40.94 Btu/lb)

initial elevation (z_1=0ft)

final elevation (z_2=100ft)

acceleration due to gravity (g) = 32.174 ft/s²

W_{cv} = 3 Btu/s

The energy balance equation is given as:

Q_{cv}-W{cv}+m[(h_1-h_2)+(\frac{V_1^2-V_2^2}{2})+g(z_1-z_2)]=0

Since  kinetic energy effects are negligible, the equation becomes:

Q_{cv}-W{cv}+m[(h_1-h_2)+g(z_1-z_2)]=0

Substituting values:

Q_{cv}-(-3)+2[(40.09-40.94)+\frac{32.174(0-100)}{778*32.174} ]=0\\Q_{cv}+3+2[-0.85-0.1285 ]=0\\Q_{cv}+3+2(-0.9785)=0\\Q_{cv}+3-1.957=0\\Q_{cv}+1.04=0\\Q_{cv}=-1.04\\

The heat transfer rate is 1.04 Btu/s from the liquid to the surroundings.

8 0
3 years ago
In a full duplex communication by using UTP cable, Pt =140 w, Pr =10 w and NEXTdb =14,47. According to this information which an
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The answer that is correct is c
4 0
3 years ago
A company specification calls for a steel component to have a minimum tensile strength of 1240 MPa. Tension tests are conducted
polet [3.4K]

Answer:

HV = 372.4 KGf/mm^2

HR_C = 39.445

Explanation:

given data:

minimum tensile strength = 1240 MPa

we knwo that \sigma_{vrs}  and brinell hardness number (HB} relation given as

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H.B = \frac{1240}{3.4} = 364.7

1) relation betwen H.B and Rockwell is

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2) relation between HB  and Vickers hardnes HV is given as

 HB \times 1.05 = HV + 10.5

HV = HB\times 1.05 - 10.5

HV = 372.4 KGf/mm^2

4 0
4 years ago
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