Question

A Review The allowable bending stress is allow = 24 ksi and the allowable shear stress is Tollow = 14 ksi 8 kip/It 6 ft Part A Select the lightest-weight wide-lange beam with the shortest depth from Appendix B that will safely support the loading shown

Answer 1

8 Kiplift Q Solution → allow = 24 klin? Callow 14 Kiplin² toft Maximum Bending moment, Momax = (1/2 x6x8) + (1 - 2 x 6) or; max = 48 kib-ft According to the Moment - flexure relationship; Amox where, I min = minm required allow Imin Section Modules Also; we have 24 ins. (h) Thickness Thickness 5.3 02 بليييع 24 = (48X 12) => Zmin = (22348) = 24 in3 Zemin = 24 ing Zmin te. The given lightest weight wide flange Beam should have minimum Section modulus of Now; from the steel table corresponding to American wide flange Beams ak to ASTM6 ; we have Designation Depth Width weight Web Imperial (W) Flange Moment of Inertial Section Modules (inx lb/ft) on) (8174) lin) welcin Up (in) Ixlint) Iy lind) Zxliny) Zylin) W12X 26 12-22 6.490 26 0.23 0.380 204 17:3 33.4 W12 X 22 12.31 4.03 22 0.26 0425 156 4.7. 25.4 2.3 W12 x 16 11.99 3.990 16 022 0.265 103 2-8 17.1 1.4 W 12 X 14 11.91 3.970 14 0.225 88.6 2.4 14.9 1-2 for W 12x22; (2x = 25.4 in3) > (Zmrin = 24 in3). Hence; The lightest weight wide flange beam with the shortest depth that safely support the given leading is (W 12422). Port B We have : IT = VAMP or; mox Imaxx A Ib Ixb where; Iman mann Shear stresss ( x 6X 8) = 24 kips: We know that maximum Shear I-section at neutral anes Moment of the shaded area above Cross 1231 the neutal axis of the Section about the centroid of the Section or; (A xğ”) = (5-73 X0:26 x 5.73 +4.03X 0:425 x (5•7340-429)) or; (Ağ') 14.44629388 Also; = bu=tw = 0.26 I= 156 int They, morem possible 24x 14.446294 8.5481 Kiplina Shear Stress in; Tmax 1 56 x 0.26 the section i.e. (mor= 8.5481) < (Callow = 14) Hence; The required value of shear stress, I is a IT8.5481 Ksil the beam is since (ELT allow); adequeate for Also; Shear. 4.03 stress in the an 5.73 0:21 N- --- --Aoo (Ay) 0-425 k ز