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The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take : A = C kN/m m 16 B = 12 C = 16 kN/ m B B_ m A m 1 m Solution : Equation of Equilibrium: C(A+1) kN C KNlm 0.5(A+1) m Ax AM Im Fec A meter (a) (b)

The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take : A = C kN/m m 16 B = 12 C = 16 kN/ m B B_ m A m 1 m Solution : Equation of Equilibrium: C(A+1) kN C KNlm 0.5(A+1) m Ax AM Im Fec A meter (a) (b)

Elements Of Electromagnetics
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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Ум -0; Fвc(3 /5)4 - 0.5C (4 +1)* - о EF, = 0; 4, + F3c (3/5)– C(4+1)=0 F3c kN 240.83 Ay = kN 127.5 v,(kN)= x, (m)= |0.913333 x 127.5 M(KNIM) v3 M1 v · v,(kN)= -Lin) v,(kN)= 128 16 х2 x2 M,(kNm) M,(kNm)= M3 508 v2 (C) (d) -8
Transcribed Image Text:Ум -0; Fвc(3 /5)4 - 0.5C (4 +1)* - о EF, = 0; 4, + F3c (3/5)– C(4+1)=0 F3c kN 240.83 Ay = kN 127.5 v,(kN)= x, (m)= |0.913333 x 127.5 M(KNIM) v3 M1 v · v,(kN)= -Lin) v,(kN)= 128 16 х2 x2 M,(kNm) M,(kNm)= M3 508 v2 (C) (d) -8
The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take : A = C kN/m m 16 B = 12 C = 16 kN/ m B B_ m A m 1 m Solution : Equation of Equilibrium: C(A+1) kN C KNlm 0.5(A+1) m Ax AM Im Fec A meter (a) (b)
Transcribed Image Text:The beam is subjected to the uniform distributed load shown. Draw the shear and moment diagrams for the beam. Take : A = C kN/m m 16 B = 12 C = 16 kN/ m B B_ m A m 1 m Solution : Equation of Equilibrium: C(A+1) kN C KNlm 0.5(A+1) m Ax AM Im Fec A meter (a) (b)
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