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

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

answer : max stress = -14.32 ksi

Explanation:

For a W16 * 50 beam

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'

First step : calculate the value of  Rb and Ra

∑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

Finally determine the maximum bending stress

бmax = M * ( d/2 ) / I

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

         = -96.73 * 97.56 / 659 = -14.32 ksi

explanation:

i don't understand really? can u provide more information

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answer: 500 wph.

explanation: