Question

Problem 5.4 J1: No.2 grade Douglas fir-Larch is to be used for a series of floor beams 8ft on center, spanning 10ft. If the total uniformly distributed load on each beam, including the beam weight, is 3050 lb, select the section with the least cross-sectional area based on bending stress. Problem 5.4 J2: A simple beam of Hem Fir, No. 1 grade, has a span of 14ft with two concentrated loads of 5 kips each placed at the third points of the span. Neglecting its own weight, determine the size of the beam with the least cross-sectional area based on bending stress. Problem 5.5 J3: An 8x14 beam of Douglas fir-larch No. 2 grade is 12' long and has a concentrated load of 6 kips located 4ft from one end. Investigate the beam for shear. Problem 5.5 J4: What should be the nominal cross-sectional dimensions for the beam of least weight that supports a total uniformly distributed load of 16 kips on a simple span and consists of Hem Fir No. 1 grade? Consider only the limiting shear stress. Problem 5.6 J5: A 2x14 rafter cantilevers over a 4x12 support beam. If both members are of Hem Fir, No. 2 grade, is the situation adequate for bearing? The rafter load on the support beam is 3500#. Problem 5.7 J6: Find the least weight section that can be used for a simple span of 14 ft with a total uniformly distributed load of 15 kips. Check bending, shear, deflection and find the required bearing length to support the beam. Only check deflection for TL, ATL = L/240

Answer 1

*305025710 = 30501147 1. Now for determing Least Cross-section area Based on Ben eling Stress:- we use & M = 0 . where m= homent on the beam 1 I = Moment of Inentia os Bending stress Note Stene we assume this Beam of retangular Beam as por Indian standards So I = lis and m= were 305x10" , sof y = 6/2 Here fe is the force due to Load . . A+ Cross-section Area I bds bowidth 8 de depth so m= q XI - a bol = axbd? and or fu so m. f xbd = fxd/6 brd Here M= 30 Saloola = 3050d/6 da so se doreste 15 = 7.567

So according to it Minimum width sequised combe mury if I de Bos Minimom mor 3050 x 16 305xlox10x12 62. dx 305x10x10x12 3050x16 5 - 7.5% 306x10x10x12 (7:5)? 3059 x 16 then b = 7.5 ft Canbe the width So cross-sectional area = bxd = 7.5x7.5 = 56.25ft² - x Skips Skips is -* skips - 1827. s 1467 'e un momement KU13 1413 1473 A =AS B (Bending 1913 e 1413 1413 77 137 Diagram) - RA * - RB RA+FB = lokipe and EMA=0, 5*14/3 + 5x 2014 - RBaly - RB = 5(1/2 + 2 ) = 5x3 = 5 SORA= 5 Now Max - Rax14/3 = 5* 14 = 23.33 kipsft then using M G 1. M = Pxd/6= 23.33 af p = Skips d. 23:33 - d: 284+,

Now calculatoa interpolation criteria ☆ - m xy at I = base b2- 284 23.33 % 12 5 5-10.5822467 6 kips 5x 14 16kipe ا د ГА - 4 x 6 و 2 -2 kips ge 12° TRR =8x6 +22 = 4 kipe so for shear just at right of eft : 6-4 2 kips and shear force just at Left of 4ftif : 4kips .