Page 80 - KAJIDAYA BAHAN

KAJIDAYA BAHAN

Penyelesaian:

a) Kedudukan paksi neutral bagi keratan tersebut.

120 mm

30 mm 30mm
80 mm

30 mm

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

120mm 80
= t x t = A x y1
y1 = ------ + 0
= 120 x 80 = 9600 x 40
2
80mm = 9600 = 40 = 384000

( - )
60 mm
50
= t x t = A x y2
50mm y2 = ------ + 30
= 60 x 50 = 3000 x 55
2
= 3000 = 165000
= 55
30 mm

A = 6600 Ay = 219000

Ay
¯
y = yteg max
A
46.82 mm
219000 P N
= ------------
6600

= 33.18 mm 0

y = 33.18 mm
¯

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