Transcript diskusi 2

Bidang M

 A
X
Q
B
L
C
Dmx=0
dx
Mx = Va.x – q x. ½ x
= Va.x – ½ qx2
Mx = ½ q Lx – ½ qx2
½ ql – qx =0
½ ql=qx
=1/2 L
Mmax = ½ ql x-1/2qx2
=1/2ql (1/2l)-1/2q(1/2l)2
=1/8 ql2
lx
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
1/8 ql2
B
Beban Terpusat

5T 2T
 A
C D
B






MA=0

-RB.8+2.7+5.4=0

-8RB+14+20=0

-8RB+34=0

RB=34/8=4,25 T ( )

MB=0
RA.8-5.4-2.1=0
8.RA-20-2=0
8.RA-22=0
RA=22/8=2,75 T
CEK : 2,75+4,25=7 T
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
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