Draw the shear and moment diagrams for the loaded beam. After you have the diagrams, answer the questions as a check on yourwork.13 kN3 kN/m5 kN/mBto mtam5 m9 mQuestions:When x = 1.1 m,V =ikN M=kN-mWhen x = 6.4 m,V=kN M=kN-mWhen x = 10.3 m, V=kN M=ikN-mWhen x = 14.1 m, V=kN M=ikN-mThe maximum (absolute value) shear force in the beam is:|Vmaxl =ikNThe maximum (absolute value) bending moment in the beam is:|Mmaxl =kN-m

Given: The beam with loads is shown below:Calculation: Determine the reactions as follows: Take the moment about point A as follows: By15+3552=133+12952×93+6By=18.1 kN Take the summation of force in y-direction as follows: By+Ay=35+13+129518.1+Ay=50.5Ay=32.4 kNTake the section on each segment at distance x as shown below: For segment PA, For shear force, Vx=-3x=-3xAt P, x=0VP=-30=0 kNAt x=1.1mVx=1.1=-31.1=-3.3 kNAt A, x=5 mVA=-35=-15 kN For bending moment, Mx=-3xx2=-3x22At P, x=0 mMP=-3022=0 kN-mAt x=1.1 mMx=1.1=-31.122=-1.815 kN-mAt A, x=5 mMA=-3522=-37.5 kN-m Thus, the shear force and bending moment at distance 1.1 m are -3.3 kN and -1.815 kN-m respectively. For segment AQ, For shear force, Vx=-35+32.4=17.4At A, x=5VA=17.4 kNAt x=6.4mVx=6.4=17.4 kNAt Q, x=8 mVQ=17.4 kN For bending moment, Mx=-35x-2.5+32.4x-5=-15x-2.5+32.4x-5At x=6.4 mMx=6.4=-156.4-2.5+32.46.4-5=-13.14 kN-mAt Q, x=8 mMQ=-158-2.5+32.48-5=14.7 kN-m Thus, the shear force and bending moment at distance 6.4 m are 17.4 kN and -13.14 kN-m respectively. For segment QR, For shear force, Vx=-35+32.4-13=4.4At Q, x=8VQ=4.4 kNAt x=10.3mVx=10.3=4.4 kNAt R, x=11 mVR=4.4 kN For bending moment, Mx=-35x-2.5+32.4x-5-13x-8=-15x-2.5+32.4x-5-13x-8At x=10.3 mMx=10.3=-1510.3-2.5+32.410.3-5-1310.3-8=24.82 kN-mAt R, x=11 mMR=-1511-2.5+32.411-5-1311-8=27.9 kN-m Thus, the shear force and bending moment at distance 10.3 m are 4.4 kN and 24.82 kN-m respectively. For segment RB, For shear force, Vx=-35+32.4-13-12x-11x-11×59=4.4-5x-11218At R, x=11VR=4.4-511-11218=4.4 kNAt x=14.1mVx=14.1=4.4-514.1-11218=1.73056 kNAt B, x=20 mVB=4.4-520-11218=-18.1 kN For bending moment, Mx=-35x-2.5+32.4x-5-13x-8-12x-11x-11×59x-113=-15x-2.5+32.4x-5-13x-8-5x-11354At x=14.1 mMx=14.1=-1514.1-2.5+32.414.1-5-1314.1-8-514.1-11354=38.78 kN-mAt B, x=20 mMR=-1520-2.5+32.420-5-1320-8-520-11354=0 kN-m Thus, the shear force and bending moment at distance 14.1 m are 1.73056 kN and 38.78 kN-m respectively. The shear force and bending moment diagram is shown below: From above diagrams, Maximum absolute shear force is 18.1 kN. Maximum absolute bending moment is 39.57 KN-m.

Answered: Draw the shear and moment diagrams for…

Engineering

Mechanical Engineering

Draw the shear and moment diagrams for the loaded beam. After you have the diagrams, answer the questions as a check on your work. 13 kN 3 kN/m 5 kN/m tom 5 m 9 m

Elements Of Electromagnetics

7th Edition

ISBN:9780190698614

Author:Sadiku, Matthew N. O.

Publisher:Sadiku, Matthew N. O.

ChapterMA: Math Assessment

Section: Chapter Questions

Problem 1.1MA

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