Page 107 - KAJIDAYA BAHAN

KAJIDAYA BAHAN

Bahagian AB

5.5 kNm

TAB

x
A

Mx = 0 5.5 – TAB = 0
TAB = 5.5 kNm

Bahagian BC

5.5 kNm

2 kNm

TBC

A
x
B
Mx = 0 5.5 – 2 – TBC = 0

TBC = 3.5 kNm

Bahagian CD

5.5 kNm
2 kNm

0.5 kNm
TCD

A
x
B
C
Mx = 0 5.5 – 2 + 0.5 – TCD = 0
TCD = 4 kNm
b) Kirakan tegasan maksimum bagi aci tersebut.

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