Page 83 - KAJIDAYA BAHAN

KAJIDAYA BAHAN

penyelesaian
a) Kedudukan paksi neutral bagi keratan tersebut.
Ay
y =
¯
A

2
3
BENTUK A (mm ) y (mm) Ay (mm )

100mm 200
= t x t = A x y1
y1 = ------ + 0
= 100 x 200 = 20000 x 100
2
200mm = 20000 = 100 = 2000000

( - )
80 mm
100
= t x t = A x y2
100mm y2 = ------ + 95
= 80 x 100 = 8000 x 145
= 8000 2 = 1160000
= 145
95 mm

¯
A = 12000 Ay = 840000 y

Ay
=
A
840000
= --------------
12000
= 70 mm

b) Jumlah moment luas kedua sekitar paksi neutral.

IPN = IPN 1 + IPN 2

3
3
bd + Ah 2 bd + Ah 2
= - ------------ + --------------
12 1 12 2

100(200) + 20000( 100-70) 2 80(100) + 8000(145-70) 2
3
3
= ----------------------------------- + -----------------------------------
12 1 12 2

-5
4
= 3.300 x 10 m
74