The cross-section shown is used to support the loads onthe beam below. The moment of inertia of the section is |=1384 in4 and distance of the centroid of the section fromthe bottom is y = 5.8 in.[MA=4800 lb-ft, w=600 lb/ft, P=1400 lb, a=5ft,b=7ft,c=2 ft,d=5ft]Mabཅ་d15 in1.5 in1.5 in.2 in.k11 in.Cross section of the beama) Construct the complete shear-force andbending-moment diagrams for the beam anddetermine the maximum positive bendingmoment in the beam.b) determine the maximum negative bendingmoment in the beam.c) Determine the maximum tension bendingstress at any location along the beam.d) Determine the maximum compressionbending stress at any location along the beam.

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Answered: The cross-section shown is used to…

Structural Analysis

6th Edition

ISBN:9781337630931

Author:KASSIMALI, Aslam.

Publisher:KASSIMALI, Aslam.

Chapter2: Loads On Structures

Section: Chapter Questions

Problem 1P

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hi im stuck in this please help me out A wooden beam is loaded by a concentrated load ?P = 5000 N and supported by boundary conditions as shown in Figure Q5(a). The cross-section of the beam is shown in Figure Q5(b). Given that ???LAB = 3000 mm, ???LBC = 2000 mm, ?0B0=10 mm, ?1B1 = 30 mm, and ?H = 240 mm, ℎ0h0=200 mm, ℎ1h1=20 mm, determine the following quantities below: last two sub parts of question didnt fit in so i will type here- theres no question d e) Determine the maximum magnitude of shear force and bending moment, neglecting the sign. The maximum magnitude of internal shear force, positive value: N The maximum magnitude of internal bending moment, positive value: N*mm f) Determine the magnitude of the maximum bending stress in the beam, neglecting the sign. Maximum bending stress, positive value: MPa
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CIVIL ENGINEERING -REINFORCED Problem#2: Determine the permissible ultimate moment capacity of the beam shown in figure. fc'= 27.5 MPa, fy = 345 MPa. 350 mm As' = 775mm² O O As =3625 mm2 O O f Questions: the green How they get the value of circle below: 96, 4, 3042-4 and 2 to arrive 121-12 mm. Please how answers that answer you get and show your formula. Solution to #2 As, As - As'=3625-775 = 2850 mm ² Afy 2850(345) 0.85 fc'b 0.85(27.5)350 a= a 63 mm 600(a−ß,d')_600(120.18-[0.85(63)]) a = B₁c 129.12=0.85c c=14249mm 600 mm a= Compression steel does not yields at failure [Fx=0 0.85fc'ab + A, 'fs' = Afy 0.85(27.5)350a + 775fs' = 3625(345) 10.56a+fs'= 1613.7 с 120.18 d-c 96+√(96)² + 4(30426) 2 = 120.18mm fs'=1613.7-10.56 a f=334.7 MPa
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