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Ketikan Mekrek III(1)
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Ketikan Mekrek III(1)
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Alina Dua
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1.
Carilah ΔB dan θB Bagian Balok a. Titik Awal B
C
Batas- Batas M
A L/2
m →ΔB
L/2
m →θB P=1 x
x
M=1 C
B x
x A
=∫24_^▒(1)(2) =∫24_^▒(1)(2)
$=∫24_^▒(
_0^(/2)▒ 〖 (0)(− )" − " 1/2 4_0^(/2)▒ 〖 ( 〖− 1/2 〗 ^2)(−1(1/2+) )+∫=∫24_0^(/2)▒ (〖 〗 =∫24_0^(/2)▒ 〖〗
" =∫24_0^(/2)▒ 〖 ( 〖 =∫24_0^(/2)▒ 〖 ( 〖− 1/2 〗 ^2)(−1/2−) 〗 +"0 =∫24_0^(/2)▒
∫24_0^(/2)▒ 〖〖 1/4 〗 ^2+ 〖 1/2 〗 ^3 〗 " "
= = 〖 1/(2(2+1))
∫24_0^(/2)▒ 〖〖 1/4 〗 ^2+ 〖 1/2 〗 ^3 〗 " "
= = 〖 1/ 〗 ^3
= = 〖 1/(4(2+1)) 〗 ^3+ 〖 1/(2(3+1)) 〗 ^4 {█(∕2@0)┤ = = 〖 1/ (/2 = = 〖 1/12 〗 ^3+ 〖 1/ 〗 ^4 {█(∕2@0)┤ = = 〖 1/12 (/2) 〗 ^3+ 〖 1/ (/2) 〗 ^4 =1/12 =1/12 !(^3/)+1/ (^4/1) = = 〖 〖 〗 ^4/#+ 〖 〖
^4/12 〗
=1/ =1/ !(^ = = 〖 〖 〗 ^3/
AC
BC
C
B
0 - L/2
0 - L/2
-1/2#!2
0
-(1/2!"
-!
-1
-1
)(3) 〗 ^2)(−1) 〗 )+∫_0^(/2)▒ 〖 (0)(− 1/2 〗 ^2) 〗 +"0 "
〗 ^3 {█(∕2@0)┤
{█(∕2@0)┤ 〗 ^3
/)
P=1
$.
Bagian Balok Titik Awal
B
C
Batas- Batas A L/2
M
L/2
m →ΔB m →θB
P=1 x
x
M=1 C
B x
x A
=∫24_^▒(1)(2)
$=∫24_^▒(
=∫24_0^(/2)▒ 〖 (−%)(−/2−) 〗 )+∫_0^(/2)▒ − =∫24_0^(/2)▒ =∫24_0^(/2)▒ 〖 (−% 〖 (0)( 〗 )&" " 〖 〖 ^2 =∫24_0^(/2)▒ =∫24_0^(/2)▒ 〖 〖 /2+ 〗 〗 +"0 " = 〖 〖
^2/(2(1+1)) 〗 + 〖 〖
= = 〖 〖
^2/ 〗 4+ 〖 〖
^3/(2+1) {█(∕2@0)┤ 〗
〗 {█(∕2@0)┤ ^3/3
= = 〖 /4 (/2) 〗 ^2+%/3 (/2)^3
=∫24_0^(/2) =∫24_0^(/2)
= = 〖 〖 ^2/2 〗 { =%/2 =%/2 (/2)^2 { = = /2 (^2/4
=%/4 =%/4 (^2/4)+ %/3 (^3/) = = 〖 〖 〗 ^2/ ^3/24 = = 〖 〖 〗 ^3/1+ 〖 〖 〗 ^3)/4 = = 〖 3% 〗 ^3/4+ (2 〖 〖 〗
= = 〖 '% 〗 ^3/4
AC
BC
C
B
0 - L/2
0 - L/2
-%!2
0
-(L/2!"
-!
-1
-1
)(3) )(−1) 〗 )+∫_0^(/2)▒ 〖 (0)(−1 〗 )" %
(∕2@0)┤ (∕2@0)┤ )
q
&.
Bagian Balok Titik Awal
B
C
Batas- Batas A L/2
M
L/2
m →ΔB m →θB x
x
M=1 C
B x
x A
=∫24_^▒(1)(2)
$=∫24_^▒(
=∫24_0^(/2)▒ =∫24_0^(/2)▒ !^2 )(−1/2− 〗 ) 〖 (0)(−1) 〗 =∫24_0^(/2)▒ 〖 (0)(−) 〗 +∫_0^(/2)▒ 〖 (−1/2 24_0^(/2)▒ 〖 1/4 = =1/(4(2+1)) = =1/12
〖
〖
〖
〖 ^2+ 〗 1/2
〖 ^3 〗+1/(2(3+1))
〖 ^3 〗+1/
〖
〖
3 〗 " " =∫24_0^(/2)▒ 〖 ^〗 =∫24_0^(/2)▒ 〖 1/
〖
〖 ^4 〗 {█(∕2@0)┤ =1/(2(2+1)) =1/(2(2+1))
〖 ^4 〗 {█(∕2@0)┤
〗 ^3+1/ 〖 =1 =1/12 〖 〖 (/2) 〖(/2) 〗 ^4
=1 =1/12 !(^3/)+1/ !(^4/1) =1 =1/# !^4+1/12 !^4 =(!^4)/#+ =(!^4)/#+ (^4)/12 =(4!^4)/34+ =(4!^4)/34+(3^4)/34 =(!^4)/34 =(!^4)/34
=1/ =1/
〖
〖
〖 ^〗
= =1/ ! 〖 〖(/2 = =1/ !(^ = = 〖 〖 〗 ^4
AC
BC
C
B
0 - L/2
0 - L/2
0
-1/2l#!2
-!
-1(1/2!"
-1
-1
1)(3) +∫_0^(/2)▒ 〖 (−1/2 〖
〖
〖 ^2 〗〗
^3 〗 {█(∕2@0)┤
{█(∕2@0)┤ ) 〗 ^3 /)
4
〖
〖 ^2 〗)(
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