Question

IR 2 2. (20 pts.) The below beam is loaded by three different loads as shown. The cross-section of the beams consists of a t-shped member whose dimensions are shown below. The centroid of the cross-section is located at 8 inches from the base of the cross- section and the moment of inertia of the cross-section is 814 in 4 a) Draw the shear diagram for the beam. Submit that information through ELC. Label all points. b) determine the maximum shear stress in the beam P = 15 (kips). X = 3 01:04:18 Cross-sectonal dimensions a = 3.2 (in.), b = 10 (in.). c-2 (in.) and d = 14 (in.) The maximum value of Qis (in3 The maximum shear stress in the beam is (ksi).

Answer 1

2 Given that INA = AMI abook Centroid INA=8iu in. P = 15 (kip) x=3. P = 15 kip xp=315 kip XP = 315 kip D RBD B RE 3ftsi k 3ft >k 6 ft RBy = 130.72 Ro=514.28 kip kip 315 Kip 115.72 Kip B E J -is kie 1-199128 kip from the eavilibrium Conditions EMBzo (Ro) (u) = (315)(17) +815)(6) - (15) (3) Ro=514.28 kip EFnzo -> RBn=0

Efyzo RByt Ro- 15+315 +315 Røy= 130.72 kip Considen a section between ARB 15 kip Sfa SFn = 15 kip. STA = ~15 kip at n=oft atn=3ft - SFB= -15 kip section blo B2C SFX = 130-72- 15 15 eip Sf = 115.72 kip e 3ft 1 x= 3ft -> SFR = 115.72 kip at na x=qft - SF = 115.72 kip. 13072 k Section blw can 15 kip 315 kip. SEX 347 6fb SFn = 130.72-15- 315 13092 kip K - 199.28 kip q at a= =aft - Spa =-199.28 kip at na 17 ft - Sto: -19928 kip. section blw DRE 15 kip 315 kip SF 3 st 6 ft 8 ft 오 130.72 sly.28 x 키

SFx = = 130.72 + 514.28-15-315 = 315 kip. at x = 17 ft - SFD=315 kip at n = 2016- = 315 kip SFE maximum Shear fore is Sfman-315 kip at point D-E - The Given Cross-section is 14 in It - 2 in ↑ ain .in 2.2- lo 7.8 y=8 Eman in. 3.2 in koescon Centroid je (14)C) (104) + (10)(3-2)(5) (14)(2) + (10) (3.2) = 7.8 in forom bottom s I ca (14) (2)3 +(14) (2) (1-7.8)² + (3.2) (10) + 12 (10) (3:2). (7.8-5) 2 Ica= 813.6 inu

- maximum shear stress always induces at the Centroidal anis. (If y=7.8 in & INA 813,6 in) QaG = Q man = (14) (2) (4.2) + (3-2) (2.2) (2:2) Qman = 125.34u ing -) marimom shear stress b=3.2 in. Znan = $fman) (Oman) (Ich) (b) = 15.16 ksi Eman = (315) (125.344) (813.6) (3.2) Eman = 15.16 ksi If y = 8 in & INA=84 inh (as given in proble) Qman = (14) (2) (4) +(3-2) (2) (1) 18. y in 3. ruan = (315) Cusiw rman (814) (3.2) 14.318 ksi Zmax = 14-318 ksi