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2 years ago

8

The W16x50, steel beam below has a span of 26' and is subjected to a 2.3 k/ft uniform distributed loading. If a 8 kip load is al

so applied at the mid point, what is the maximum bending stress in the beam

1 answer:

MariettaO [177]2 years ago

5 0

Answer: hello the steel beam related to your question is missing attached below is the missing diagram

answer : max stress = -14.32 ksi

Explanation:

<em>For a W16 * 50 beam </em>

I = 659 in

d = 16.26 in

For steel : E = 29 * 10^3 ksi

Span ( length ) of steel beam = 26'

uniformly distributed load on the steel beam = 2.3 k/ft

load at middle = 8 kip , Distance from middle to either Ra or Rb = 13'

<em>First step : calculate the value of Rb and Ra </em>

∑Ma = 0

0 = ( 8 * 13 )+ (26 * 2.3 * 13 ) - Rb * 26

∴ 25 Rb = 881.4

Hence: Rb = 35.256 kip

∑fy = 0

0 = Ra + Rb - 2.3 * 26 - 8

Ra = 67.8 - Rb = 67.8 - 35.256

= 32.54 kip

second step :

Mmax = Mmidpoint

∴ Mmax = Ra * 13 - 2.3 ( 13 ) * (26/4 )

= 32.54 * 3 - 2.3 ( 13 ) * (26/4 )

= 97.62 - 194.35 = -96.73 kip fit

<u>Finally determine the maximum bending stress </u>

бmax = M * ( d/2 ) / I

= - 96.73 * ( 16.26 / 2 )*12 / 659

= -96.73 * 97.56 / 659 = -14.32 ksi

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