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TitleMekanika Bahan Bab 6
File Size217.9 KB
Total Pages14
Document Text Contents
Page 7

FAKULTAS TEKNIK JURUSAN TEKNIK SIPIL

UNIVERSITAS LAMPUNG





http://mahasiswasipilunila.wordpress.com 57





q

25 mm




2,5 m 225 mm





25 mm 150 mm 25 mm



Penyelesaian :



y





1 25 mm

99,04

x

2 2

225 mm

150,96









25 mm 150 mm 25 mm





Ix = Ix
o + Ay’2

Ix1 =
23

12
1 54,86.25.20025.200.  = 37706274,67 mm4

Ix2 =
23

12
1 46,38.225.25.2225.25..2  = 64101618,00 mm4 +

Ix = 101807892,67 mm
4

Mmaks = ½ q.l
2

= ½ .q.2,52


x

b
b

I

yM .


(dipakai yb supaya tegangan pada serat bawah akan maksimum 140 MPa)

140 =
7,101807892

96,150.10.5,2..
62

2
1 q



q = 30,21 kN/m



Mmaks = ½ .30,21.2,5
2 = 94,40625 kNm



84,91
7,101807892

04,99.10.40625,94.
6


x

a

a
I

yM
 MPa

140
7,101807892

96,150.10.40625,94.
6


x

b

b
I

yM
 MPa

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