Ketikan Mekrek III(1)

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)/34+ =(4!^4)/34+(3^4)/34  =(!^4)/34 =(!^4)/34

 =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 〗)(