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Dmitriy789 [7]
3 years ago
13

A 35 ft simply supported beam is loaded with concentrated loads 15 ft in from each support. On one end, the dead load is 8 kips

and the live load is 18 kips. At the other end, the dead load is 4 kips and the live load is 9 kips. Include the self-weight of the beam in the design. Lateral supports are provided at the supports and the load points. Determine the least-weight W-shape to carry the load. Use A992 Steel and Cb = 1.0.

Engineering
1 answer:
agasfer [191]3 years ago
4 0

Answer: ASD = 306 kips-ft

               LRSD = 1387.5 k-ft

Explanation:

To begin, we will take a step by step process to solving this problem.

Attached below is a picture to guide us to solving this.

To begin, we have that to reaction of the support

ΣMд = 0

where;

RB * 35 - (8+18)15 - (4+9)20 = 0

RB = 18.57k

also Ey = 0;

RA + RB = 18 + 8 + 9 +4 = 20.43 k

taking the maximum moment at mid point;

Mc = RA * 35/2 - (8 +18) (35/2 -15)

Mc = 292.525

therefore, MD = RA * 15 = 20.43 * 15 = 306.45 k.ft

MD = 306.45 k.ft

ME = 279 k.ft i.e 18.57 * 15

considering the unsupported  length; 35 - (15*2 = 5ft

now we have that;

Lb = Lp = 5ft

where Lp = 1.76 ry(√e/fy)

Lp = 1.76 ry √29000/50 ......

ry = 1.4 inch

so we have that Mr = Mp for Lb = Lp where

Mp = 2 Fy ≤ 1.5 sx Fy

Recall from the expression,

RA + RB = (8+4) * 1.2 + (18+9) * 1.6 = 57.6

RA * 35 = 4 * 1.2 * 15 + 9 *1.6 * 15 + 8 * 1.2 * 20 + 18 * 1.6 * 20

RA = 30.17 k

the maximum moment at D = 30.17 * 15 = 452.55 k.ft

Zrequired = MD/Fy = 452.55 * 12 / 50 = 108.61 inch³

so we have Sx = 452.55 * 12 / 1.5 * 50 = 72.4 inch³

also r = 1.41 in

Taking LRFD solution:

where the design strength ∅Mn = 0.9 * Zx * Fy

given r = 2.97

Zx = 370 and Sx = 81.5, we have

∅Mn = 0.9 * 370 * 50 = 16650 k-inch = 1387.5 k-ft

this tells us it is safe.

ASD solution:

for Lb = Lp, and where Mn = Mp = Fcr Sx

we already have value for Sx as 81.5 so

Fcr = ZxFy/Sx

Fcr = 370 * 50 / 81.5 = 227 ksi

considering the strength;

Strength = Mn / Ωb = (0.6 * 81.5 * 50) * (1.5) / 12 = 306 kips-ft

This justifies that it is safe because is less than 306

cheers i hope this helps.

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7 0
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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

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Radius, r = 0.25/2 = 0.125 m

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Exit temperature, T₂ = 284 + 273 = 557 K

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

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

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\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
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