Mekanika Rekayasa II

Mekanika Rekayasa II (Merek II) Kali ini kami menulis rumus pada bangunan berlantai 2, dimana rumus tersebut kita gunakan dan perletakan bahan tersebut.

Perletakan :

semntara di gambar

∑MB = 0

RAV.9 – P1 . 7 – P2 . 4 – Q3 (1/3 . 9) – Q1 (1/2 . 9) = 0

RAV . 9 – 2 . 7 – 3 . 4 – 19 ( 1/3 . 9) – 18 (1/2 . 9) = 0

RAV . 9 – 161 = 0

RAV = 161/9 = 17 8/9……………………………………………….( ↑ )

∑MA = 0

-          RBV . 9 + P2 . 5 + P1 . 2 + Q3 (2/3 . 9) + Q1 (1/2 . 9)

-          RBV . 9 + 3 . 5 + 2 . 2 + 18 (6) + 18 (4 ½) = 0

-          RBV . 9 + 15 + 4 + 108 + 81 = 0

RBV . 9 + 208/9 23 1/9 ………………………………(  ↑ )

KONTROL  : ∑M = 0

RAV . + RBV – P1 – P2 – Q1 – Q2 – Q3 = 0

17  8/9 + 23 1/9 – 2 – 3 – 18 -18 = 0

41  - 41 = 0 ………………………………………………………….ok….!!!!!!

Bentang A – P1 0 ≤ X ≤ 2

q31 =  =  =  

q31x =  =  =  x                 

Q31x = ½ . x q31x = 1/2x  .  4/9x = 2/9x²

Q11x = q1 . x = 2 . x = 2x

MX         = RAV . X – Q11X . 1/2X – Q31X . 1/3 X

= 17 - 2X . ½ X – 2/9X . 1/3X

= 17 – X² – 2/27X³

DX          = 17   - 2X – 2/9X²

X	0	1	2
MX	0	16	31
DX			13

Bantang p1 – p2 = 0 ≤ x ≤ 3

q32 =  =  =  

q32x =  =  =  x

Q31x = ½ . x q31x = 1/2x  .  4/9x = 2/9x²