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nadya68 [22]
2 years ago
9

In case of damaged prestressed concrete I girders which are used for restoring strength?

Engineering
1 answer:
Ahat [919]2 years ago
7 0

Answer: The use of post-tensioning rods is useful in this situation.

Explanation:

The prestressed concrete has not been accepted as a good building material. The high tensile strength of steel is combined with the concrete to give the compressive strength to the concrete.

Post-tensioning rods of steel along with jacking can be used to restore the strength of the prestressed girder and it can begin with the calculated preload. This way the damaged concrete can be repaired and restored.  

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The results of a _________ test will determine if there is suitable drainage and the size of the drain field that will be requir
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The results of a percolation test will determine if there is suitable drainage and the size of the drain field that will be required for a septic system.

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2 years ago
Una frase de: ama la vida quien___________________________________
TEA [102]

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A phrase from: who loves life

Explanation:

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2 years ago
Horizontal shear forces and, consequently, horizontal shear stresses are caused in a flexural member at those locations where th
jek_recluse [69]

Answer:

False

Explanation:

When the horizontal shear forces act on the surface there is transverse shear stress at a particular point which is equal in magnitude. Pure bending is less common than a non uniform bending because the beam is not in equilibrium.

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3 years ago
Methane gas is 304 C with 4.5 tons of mass flow per hour to an uninsulated horizontal pipe with a diameter of 25 cm. It enters a
Arada [10]

Answer:

a) h_c = 0.1599 W/m^2-K

b) H_{loss} = 5.02 W

c) T_s = 302 K

d) \dot{Q} = 25.125 W

Explanation:

Non horizontal pipe diameter, d = 25 cm = 0.25 m

Radius, r = 0.25/2 = 0.125 m

Entry temperature, T₁ = 304 + 273 = 577 K

Exit temperature, T₂ = 284 + 273 = 557 K

Ambient temperature, T_a = 25^0 C = 298 K

Pipe length, L = 10 m

Area, A = 2πrL

A = 2π * 0.125 * 10

A = 7.855 m²

Mass flow rate,

\dot{ m} = 4.5 tons/hr\\\dot{m} = \frac{4.5*1000}{3600}  = 1.25 kg/sec

Rate of heat transfer,

\dot{Q} = \dot{m} c_p ( T_1 - T_2)\\\dot{Q} = 1.25 * 1.005 * (577 - 557)\\\dot{Q} = 25.125 W

a) To calculate the convection coefficient relationship for heat transfer by convection:

\dot{Q} = h_c A (T_1 - T_2)\\25.125 = h_c * 7.855 * (577 - 557)\\h_c = 0.1599 W/m^2 - K

Note that we cannot calculate the heat loss by the pipe to the environment without first calculating the surface temperature of the pipe.

c) The surface temperature of the pipe:

Smear coefficient of the pipe, k_c = 0.8

\dot{Q} = k_c A (T_s - T_a)\\25.125 = 0.8 * 7.855 * (T_s - 298)\\T_s = 302 K

b) Heat loss from the pipe to the environment:

H_{loss} = h_c A(T_s - T_a)\\H_{loss} = 0.1599 * 7.855( 302 - 298)\\H_{loss} = 5.02 W

d) The required fan control power is 25.125 W as calculated earlier above

5 0
3 years ago
**Please Help ASAP**
RoseWind [281]

Answer:

look up the assignment number. its in the left side of the screen. its what i did when i had problems.

Explanation:

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