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Bidang M

Bidang M. X Q A B L C Mx = Va.x – q x. ½ x

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Bidang M

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  1. Bidang M • X Q • A B L C Mx = Va.x – q x. ½ x = Va.x – ½ qx2 Mx = ½ q Lx – ½ qx2 lx Dmx=0 ½ ql – qx =0 dx ½ ql=qx =1/2 L Mmax = ½ ql x-1/2qx2 =1/2ql (1/2l)-1/2q(1/2l)2 =1/8 ql2

  2. Gambar bidang Momen • MA = ½ ql (0) – (1/2)2 = 0 • MB = ½ QL (L) – ½ Q (L)2 • =1/2 QL2 – ½ QL2 = 0 • A C B • 1/8 ql2

  3. Beban Terpusat • 5T 2T • A C D B • MB=0 • RA.8-5.4-2.1=0 • 8.RA-20-2=0 • 8.RA-22=0 • RA=22/8=2,75 T • MA=0 • -RB.8+2.7+5.4=0 • -8RB+14+20=0 • -8RB+34=0 • RB=34/8=4,25 T ( ) • CEK : 2,75+4,25=7 T

  4. Bidang D • D = RA =2,75 T • C = RA-P1 2,75-5 = 2,25 T • D = RA-P1-P2 2,75-5-2 = -0,25 T • B = RA-P1-P2+RB =2,75-5-2+4,25 T • 2,75 • -2,75 • -0,25

  5. Bidang Momen • MA = 0 • MC = 2,75 X 4= 11 tm (Mmax) • MD = (2,75 X 7 ) – (5 X 3) = 4,25 tm • MB = (2,75 X 8)-(5X4)-(2X1)=0 • 11tm • A C D B • 4,25 tm

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