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Arte-miy333 [17]
4 years ago
15

Assume you have a beam that is 15 feet long and carries a concentrated load of 900 pounds at a distance of 4 feet from R . React

ion (R ) is pinned 10 feet from the load. Reaction R = 300 lbs and R = 600. What is the maximum shear in lbs? Please show your work and provide answer.

Engineering
1 answer:
skelet666 [1.2K]4 years ago
8 0

Answer:

maximum shear force is 600 lb

Explanation:

given data

long = 15 feet

load = 900 pounds

distance from R  = 5 feet

pinned from the load =  10 feet

Reaction R = 300 lbs

Reaction R = 600

solution

we get here moment at point 2 that is

∑M2 = 0

that is express as

R1(15) - 900 (5)  = 0

so R1 will be = 300 lb

and

when ∑Fy = 0

so here 300 + R2 - 900 = 0

so R2  = 600 lb

so here maximum shear force is 600 lb

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Answer / Explanation:

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3 years ago
25°C is flowing in a covered irrigation ditch below ground. Every 100 ft, there is a vent line with a 1.0 in. inside diameter an
Gnesinka [82]

The total evaporation loss of water is 87.873 ×10^{-5}  lbm /day.

<u>Explanation:</u>

Assume A is the water and B is air.

A is diffusing to non diffusing B.

N_{A}=\frac{D_{A B} P}{R T Z P_{B/ M}}\left(P_{A_{1}} \ - P_{B_{2}}\right)

By the table 6.2 - 1 at 25°C, the diffusivity of air and water system is 0.260 \times 10^{-4} \mathrm{m}^{2} / \mathrm{s}.

Total pressure P = 1 atm = 101.325 KPa

P_{A_{1} } = 23.76 mm Hg

P_{A_{1} } = \frac{23.76}{760}

P_{A_{1} } = 0.03126 atm

P_{A_{1} } = 3.167 K Pa

When air surrounded is dry air, then PA_{2} = 0 mm Hg

R = 8.314 \frac{K P a \cdot m^{3}}{mol \cdot k}

P_{B/M} = \frac{P_{B_{1}}-P_{B_{2}}}{\ln \left(P_{B_{1}} / P_{B_{2}}\right)}

P_{B_{1}}=P_{T}-P_{A1_{}}

= 101.325 - 3.167

P_{B_{1} } = 98.158 K Pa

P_{B_{2}}=P_{T}-P_{A_{2}}

P_{B_{2} } = 101.325 - 0

P_{B_{2} } = 101.325 K Pa

P_{B / M}=\frac{98.158-101.325}{\ln (98.158 / 101 \cdot 325)}

P_{B/M} = 99.733 K Pa

Z = 1 ft = 0.3048 m

T = 298 K

N_{A} =  \frac{(0.206*10^{-4})(101.325)(3.167 - 0) }{(8.314) ( 298 )( 0.3048 )( 99.733 ) }

N_{A} = 0.11077 × 10^{-6} mol/ m^{2}.s

N_{A} = 0.11077 × 10^{-6} × 18 × (60×60×24)

N_{A} = 0.1723 lb_{m} / m^{2}.day

\tilde{N}_{A}=N_{A} \times Area

Area of individual pipe is

A=\frac{\pi}{4}(0.0254)^{2}

A = 0.00051 m^{2}

\bar{N}_{\boldsymbol{A}} = 0.1723 × 0.00051

\bar{N}_{\boldsymbol{A}} = 0.000087873 lbm/day

In 1000 ft length of ditch,there will be a 10 pipes. The amount of evaporation water is

= 10 × 0.000087873 = 0.00087873 lbm /day

The total evaporation loss of water is 87.873 ×10^{-5}  lbm /day.

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3 years ago
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3 years ago
Read 2 more answers
Python Codio Question:
tester [92]

Answer:

# -*- coding: utf-8 -*-

# Get N from the command line

import sys

N = int(sys.argv[1])

if N > 0: #N is positive  

   positve = list(range(N,0,-1))

   print(positve)

elif N < 0: #N is negative  

   negative = list(range(N,0,1))

   print(negative)

else:

   print("Invalid in input")

Explanation:

First, you need to identify if the number entered is positive, negative or none of them, for that we use one if, one elif and one else statement:

  • If the number entered (N) is greater than zero (is positive) we print a list in the range N to 0 in steps of minus one, the zero is not printed because the function range by default omits the last value
  • If the previous statement was False and the number entered is smaller than zero (is negative) we print a list in the range N to 0 in steps of one, the zero is not printed because the function range by default omits the last value
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3 years ago
An adiabatic air compressor compresses 10 L/s of air at 120 kPa and 20 degree C to 1000 kPa and 300 degree C.
Oksana_A [137]

Answer:

work=281.4KJ/kg

Power=4Kw

Explanation:

Hi!

To solve follow the steps below!

1. Find the density of the air at the entrance using the equation for ideal gases

density=\frac{P}{RT}

where

P=pressure=120kPa

T=20C=293k

R= 0.287 kJ/(kg*K)= gas constant ideal for air

density=\frac{120}{(0.287)(293)}=1.43kg/m^3

2.find the mass flow by finding the product between the flow rate and the density

m=(density)(flow rate)

flow rate=10L/s=0.01m^3/s

m=(1.43kg/m^3)(0.01m^3/s)=0.0143kg/s

3. Please use the equation the first law of thermodynamics that states that the energy that enters is the same as the one that must come out, we infer the following equation, note = remember that power is the product of work and mass flow

Work

w=Cp(T1-T2)

Where

Cp= specific heat for air=1.005KJ/kgK

w=work

T1=inlet temperature=20C

T2=outlet temperature=300C

w=1.005(300-20)=281.4KJ/kg

Power

W=mw

W=(0.0143)(281.4KJ/kg)=4Kw

5 0
3 years ago
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