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Ierofanga [76]
3 years ago
15

A simply supported beam spans 25 ft and carries a uniformly distributed dead load of 0.6 kip/ft, including the beam self-weight,

and a live load of 2.1 kip/ft. Determine the minimum required plastic section modulus and select the lightest-weight W-shape to carry the moment. Consider only the limit state of yielding and use A992 steel. Design by (a) LRFD and (b) ASD
Engineering
1 answer:
USPshnik [31]3 years ago
5 0

Answer:

Check the explanation

Explanation:

beam span = 25 ft

dead load = 0.6 kip/ft

live load = 2.1 kip/ft

factored load = 1.2*0.6 +1.6*2.1=4.08 kip/ft

moment in beam = 4.08*252/8=318.75 kip-ft = 3825 kip-in

design strength =0.9* 50 = 45 ksi

plastic section modulus required = 3825/45=85 in3

Moment in beam in ASD = (0.6+2.1)*252/8 = 210.9 kip-ft

lighest W section from LRFD = W21x44

lightest W section from ASD = W21x44

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Radioactive wastes generating heat at a rate of 3 x 106 W/m3 are contained in a spherical shell of inner radius 0.25 m and outsi
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Inner surface temperature= 783K.

Outer surface temperature= 873K

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Outside radius= 0.30 m

Temperature at infinity, T(¶)= 10°c = 273. + 10 = 283K.

Convection coefficient,h = 500 W/m^2 . K

Temperature of the surface= T(s) = ?

Temperature of the inner= T(I) =?

STEP 1: Calculate for heat flux at the outer sphere.

q= r × e/3

This equation satisfy energy balance.

q= 1/3 ×3000000(W/m^3) × 0.30 m

= 3× 10^5 W/m^2.

STEP 2: calculus the temperature for the surface.

T(s) = T(¶) + q/h

T(s) = 283 + 300000( W/m^2)/500(W/m^2.K)

T(s) = 283+600

T(s)= 873K.

TEMPERATURE FOR THE OUTER SURFACE is 873 kelvin.

The same TWO STEPS are use for the calculation of inner temperature, T(I).

STEP 1: calculate for the heat flux.

q= r × e/3

q= 1/3 × 3000000(W/m^3) × 0.25 m

q= 250,000 W/m^2

STEP 2:

calculate the inner temperature

T(I) = T(¶) + q/h

T(I) = 283K + 250,000(W/m^2)/500(W/m^2)

T(I) = 283K + 500

T(I) = 783K

INNER TEMPERATURE IS 783 KELVIN

5 0
3 years ago
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