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irina1246 [14]
3 years ago
9

The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the av

erage normal stress in each bar if the rigid beam is subjected to a force of P
Engineering
1 answer:
GenaCL600 [577]3 years ago
5 0

Answer:

hello a diagram attached to your question is missing attached below is the missing diagram

The three suspender bars AB, CD, and EF are made of A-36 steel and have equal cross-sectional areas of 500 mm2. Determine the average normal stress in each bar if the rigid beam is subjected to a force of P = 70kN , let d = 2.4 m , L = 4m  

Answer :

Stress = force / area

for Bar 1 = (40.03 * 1000) / 500  = 80.06 MPa

Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa

for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa

Explanation:

Given data:

Type of steel = A-36

cross-sectional area = 500 mm^2

Calculate the average normal stress in each bar

we have to  make some assumptions

assume forces in AB, CD, EF  to be p1,p2,p3  respectively

∑ Fy = 0 ; p1 + p2 + p3 = 70kN  ----------  ( 1 )

∑ Mc = 0 ; P1 * d - p * d/2 - p3 * d = 0

where d = 2.4 hence ; p1 - p3 = 35 -------- ( 2 )

Take ; Tan∅

Tan∅  = MN / 2d = OP/d

i.e. s1 - 2s2 - s3 = 0

\frac{P1L}{AE} - \frac{2P2}{AE} + \frac{P3L}{AE}  = 0

L , E and A are the same hence

P1 - 2p2 + p3 = 0 ----- ( 3 )

Next resolve the following equations

p1 = 40.03 kN,  p2 = 23.33 kN, p3 = 5.33 kN

Stress = force / area

for Bar 1 = (40.03 * 1000) / 500  = 80.06 MPa

Bar 2 = ( 23.33 * 1000 ) / 500 = 46.66 MPa

for Bar 3 = ( 5.33 * 1000 ) / 500 = 10.66 MPa

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Answer:

Plans; blueprints.

Explanation:

In Engineering, it is a common and standard practice to use drawings and models in the design and development of various tools or systems that are being used for proffering solutions to specific problems in different fields such as engineering, medicine, telecommunications and industries.

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5 0
3 years ago
Ordan has _ 5 8 can of green paint and _ 3 6 can of blue paint. If the cans are the same size, does Jordan have more green paint
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Answer:

Jordan has more green paints

Explanation:

Given

Green = \frac{5}{8}

Blue = \frac{3}{6}

Required

Which paint does he have more?

For better understanding, it's better to convert both measurements to decimal.

For the green paint:

Green = \frac{5}{8}

Green = 0.625

For the blue paint:

Blue = \frac{3}{6}

Blue = 0.5

By comparison:

0.625 > 0.5

<em>This means that Jordan has more green paints</em>

3 0
3 years ago
Steam enters a turbine at 120 bar, 508oC. At the exit, the pressure and quality are 50 kPa and 0.912, respectively. Determine th
Archy [21]

Answer:

The turbine produces <u>955.53 KW</u> power.

Explanation:

Taking the turbine as a system. Applying Law of Conservation of Energy:

m(h₁ - h₂) - Heat Loss = P

where,

m = mass flow rate of steam = 1.31 kg/s

h₁ = enthalpy at state 1 (120 bar and 508°C)

h₂ = enthalpy at state 2 (50 KPa and x = 0.912)

Heat Loss = 225 KW

P = Power generated by turbine = ?

First, we find h₁ from super steam tables:

At,

T = 508°C

P = 120 bar = 12000 KPa = 12 MPa

we find that steam is in super-heated state with enthalpy:

Due to unavailibility of values in chart we approximate the state to 500° C and 12.5 MPa:

h₁ = 3343.6 KJ/kg

Now, for state 2, we have:

P = 50 KPa and x = 0.912

From saturated steam table:

h₂ = hf₂ + x(hfg₂) = 340.54 KJ/kg + (0.912)(2304.7 KJ/kg)

h₂ = 2442.4 KJ/kg

Now, using values in the conservation equation:

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<u>P = 955.53 KW</u>

5 0
3 years ago
A total of 245 kip force is applied to a set of 10 similar bolts. If the spring constant of each bolt is 0.4 Mlb/in and that of
zubka84 [21]

Answer: The net force in every bolt is 44.9 kip

Explanation:

Given that;

External load applied = 245 kip

number of bolts n = 10

External Load shared by each bolt (P_E) = 245/10 = 24.5 kip

spring constant of the bolt Kb = 0.4 Mlb/in

spring constant of members Kc = 1.6 Mlb/in

combined stiffness factor C = Kb / (kb+kc) = 0.4 / ( 0.4 + 1.6)  = 0.4 / 2 = 0.2 Mlb/in

Initial pre load Pi = 40 kip

now for Bolts; both pre load Pi and external load P_E are tensile in nature, therefore we add both of them

External Load on each bolt P_Eb = C × PE = 0.2 × 24.5 = 4.9 kip

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Fb = 4.9 kip + 40 kip

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