An inverted tee lintel is made of two 8" x 1/2" steel plates. Calculate the maximum bending stress in tension and compression when the lintel carries a total uniformly distributed load of 10000 lb on a simple span of 6 ft. Also, calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

hello your question lacks some information attached is the complete question

A) (i)maximum bending stress in tension = 0.287 * 10^6 Ib-in

    (ii) maximum bending stress in compression =  0.7413*10^6 Ib-in

B) (i)  The average shear stress at the neutral axis = 0.7904 *10 ^5 psi

    (ii)  Average shear stress at the web = 18.289 * 10^5 psi

    (iii) Average shear stress at the Flange = 1.143 *10^5 psi

Explanation:

First we calculate the centroid of the section,then we calculate the moment of inertia and maximum moment of the beam( find attached the calculation)

A) Calculate the maximum bending stress in tension and compression

lintel load = 10000 Ib

simple span = 6 ft

( (moment of inertia*Y)/ I ) = MAXIMUM BENDING STRESS

I = 53.54

i) The maximum bending stress (fb) in tension=

=  = =  0.287 * 10^6 Ib-in

ii) The maximum bending stress (fb) in compression

= = = 0.7413*10^6 Ib-in

B) calculate the average shear stress at the neutral axis and the average shear stresses at the web and the flange

i) The average shear stress at the neutral axis

V = = = 3.6*10^5 Ib

Ay = 8 * 0.5 * (2.375 - 0.5 ) + 0.5 * (2.375 - ) *

= 5.878 in^3

t = VQ / Ib  = ( 3.6*10^5 * 5.878 ) / (53.54 8 0.5) = 0.7904 *10 ^5 psi

ii) Average shear stress at the web ( value gotten from the shear stress at the flange )

t = 1.143 * 10^5 * (8 / 0.5 )  psi

  = 18.289 * 10^5 psi

iii) Average shear stress at the Flange

t = VQ / Ib =

= 1.143 *10^5

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explanation: