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Debora [2.8K]
3 years ago
7

A 35 ft simply supported beam is loaded with concentrated loads 15 ft in from each support. On one end, the dead load is 8.0 kip

s and the live load is 18.0 kips. At the other end, the dead load is 4.0 kips and the live load is 9.0 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
Engineering
2 answers:
Dafna11 [192]3 years ago
4 0

Answer:

From the load equation

F=stress*Area

Given stresses are 8 kips and 9 kips.

Hence the minimum weight supported=6.695 lbs.

Anastaziya [24]3 years ago
4 0

Answer:

ASD = 306 kips-feet

LRSD = 1387.5 kips-feet

Explanation:

a step by step process to solving this problem.

ΣM at A = 0

where;

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

RB = 18.57k

also E y = 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 kips-feet

MD = 306.45 kip-feet

ME = 279 kip-feet .IE 18.57 * 15

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

now we have that;

L b = L p = 5ft

where L p = 1.76 r y(√e/f y)

L p = 1.76 r y √29000/50

r y = 1.4 inch

so we have that M r = M p  for L b = L p where

M p = 2 F y ≤ 1.5 s x   F y

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 kips-feet

Z required = MD / F y = 452.55 * 12 / 50 = 108.61 inch³

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

also r = 1.41 in

Taking LRFD solution:

where the design strength ∅ M n = 0.9 * Z x * F y

given r = 2.97

Z x = 370 and S x = 81.5, we have

∅ M n = 0.9 * 370 * 50 = 16650 k-inch = 1387.5 kips-feet

this tells us it is safe.

ASD solution:

for L b = L p, and where M n = M p = F c r    S x

we already have value for S x as 81.5 so

F c r = Z x times F y divide by  S x

F c r = 370 * 50 / 81.5 = 227 kips per sq.in

considering the strength;

Strength = M n / Ωb = (0.6 * 81.5 * 50) * (1.5) / 12 = 306 kips-feet  

This justifies that it is safe because is less than 306

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Answer:

Part a

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<em>The value of cpk cannot be more than cp. It can at maximum equal to the value of the cp.</em>

Explanation:

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As per the given data

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c_p_{J}=\frac{UTL-LTL}{6\sigma}\\c_p_{J}=\frac{47-30}{6\times 2}\\c_p_{J}=1.42

c_p_k_{J}=min(\frac{UTL-\mu}{3\sigma},\frac{\mu-LTL}{3\sigma})\\c_p_k_{J}=min(\frac{47-38}{3\times 2},\frac{38-30}{2\times 3})\\c_p_k_{J}=min(1.5,1.33)\\c_p_k_{J}=1.33

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<em></em>

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