Question

QUESTION 4 A steel beam (W12 x 30) extends horizontally out of a solid concrete wall as shown. The beam is subjected to a horizontal loading as shown and the weight of the beam is 360 lb per ft. 20 ft Aft Side view 6 ft H= 15000 lb H psi Considering the weight of the beam, calculate the following: a). the maximum combined tensile stress at the wall. Olen b). the maximum combined compressive stress at the wall. O comp psi

Answer 1

since a part of beam is inside the wall hence we can take it as fixed beam at left end.

also for W12 30 beam section we have all standard values listed in databook.

see below the complete solution.

solution :- Given beam s section. I section for beam W 12 X 30 We have #tfl ca bf = 6.52 in Af=ooyyin HH n ५ d=12.34 in section. beam W 12x30 of Area 2 A=807g in? Inna 20.3 inu Iyy = 238 inu How find stresses. lil stress due to HE 15000 lb compressive. H 15 000 - o come = 1706-48 eblina A 8.79 epsi)

(ii) Binding stress due to self weight. w МА Inn Ү. оъ – Bending moment W x 2 м= 36 от 2 - - Equivalent Beam, M А 2. L - -o 54 Mr Звох 20- * 4-5.48 000E -

Now, Bending stress at wall due to M. ba MY Ixn 72000x (bs/2) x12 - 20.3 oba 72000 X 12 X Go S2 2 2003 b= 13875067389 eblin2 or(psi) Now part (a) Maximum combined tensile stresa at wall. occurs at top surface. =) - ob. - Comp 138750.7389 - 1706-48 a = 1370 44.25 lbline](psi) parlb) Maximum combined compressive stress. at lower point of beam at wall, bi - bt o comp = 138750.7+1706.5 = 140457.2 eblina