Page 59 - eEISBN FINALIZE

MK = MK

56 . 67 ( ) F4 = CD ( ) 203 + ( ) 2
226.68 = 3 F CD + 40
3 F CD + 40 = 226.68

3 F CD = 226.68 - 40
186 . 68
F =
CD
3
= 62 . 23 kN (T)

3
=
tan 
2
1 - 
 = tan  3 

 2 
= 56.3 0

 F =  F 

56.67 = F sin 56.3+ 30 + 40
KD
56.67 = F KD sin 56.3+ 70

56.67 - 70 = F KD sin 56.3
F KD sin 56.3 = 56.67 - 70
−
F KD sin 56.3 = 13 . 33
13 . 33
=
F KD −
sin 56.3
= -16.02 kN (C)

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