Question

Problem 3: A simply supported wooden beam, 12' long, is acted on by P= 10 kip at the mid-span and a uniformly distributed load w = 1.5 kip/feet over the entire length. The nominal size of beam cross-section is 8 x 12, however, actual cross-section is 7.5 x 11.5. Calculate the maximum tensile & compressive bending stresses at the top and bottom fibers of the section. ANS: Maximum tensile & compressive stresses (o) are = . -

Answer 1

# Solution:- LE 12' | P= 10 kip AK Ra= RB = ( + ) (Beam is symmetric) — Maximum bending moment at centre :- Mmax = RAX 1 - (WL) ( 2x ) = (1 + w Mmax = P + WL with

Mmax = (10 kip) (12 ft) + (liskielce) (12442 4 Mmax * 57 kip-ft = 57x12 kip-in (Mmax = 684 kip-in. Calculating moment of Inertiq :- I= 7.5X11.53 Tu = 5.75" 12 lis E - - - ----------- - - I= 950:546875 int =5.751 - Maximum tensile stress:- CAt bottom) Emax = M max lt - (684 kip-in) (5175 in) 950:546875 int 4 ( E, max = 4.137618147 ksi) Ans Maximum compressive bending styes:- Cat top) a - Maman c ( Tension → tive, compression>-ive) c max I Tamar = -(684 kip-in)(5.75 in.). 950:546875 int ( Formax = – 4:137618147 ksi) Ans