8/12/2019 Mekanika Struktur III

1/4

1. Diketahui :

Ditanya : Carilah reaksi peletakan, momen tumpuan, momen lapangan dan gaya geser.

Serta gambarkan Free Body Diagram (FBD), Shear Forced Diagram (SFD),

Bending moment Diagram (BMD) dan Normal Forced Diagram (NFD) dengan

Metode Distribusi Moment!

Jawaban

Segmen I Segmen II

Fixed End Moment(FEM)

Pada , = 4m dan = 4m

MPa = tidak ada

MPb =

=

= -30,705 Tm

Pada , = 3m dan = 6 m

, = 6m dan = 3 m

MPb =

+

=

+

= 31,41 Tm

MPc = Tidak ada

Stiffness Factor(SF)

Ketentuan :

= = =

=

= 1,5EI

Ketentuan :

= = =

=

= 0,667EI

Distribution Factor

(DF)

= Tidak ada

=

=

= 0,692

=

=

= 0,308

= Tidak ada

8/12/2019 Mekanika Struktur III

2/4

Check!

+ = 10,692 + 0,308 = 1 oke!

Corry Over Factor

(COF)= = 0,5

Tabel Distribusi Momen

Maka, didapat Mb = 31,193 Tm

a. PerhitunganFree Body Diagram(FBD)

Segmen I Segmen II

Reaksi ujung

akibat beban

yang bekerja

+10,235 +10,235 +10,47 +10,47

Reaksi ujung

akibatmomenujung

-3,899 +3,899 +3,466 -3,466

Reaksi ujung

total+6,336 +14,134 +13,936 +7,004

= 6,336 = 28,07 =7,004

8/12/2019 Mekanika Struktur III

3/4

Cek nilai reaksi peletakan :

= 0

( + + )(P1 + P2 + P3) = 0

(6,336 + 28,07 + 7,004)(20,47 + 10,47 + 10,47) = 0

41,4141,41 = 0 . OKE!

b. Perhitungan Shear Forced Diagram(SFD)SFa = Ra = 6,336 T

SFP1 = SFaP1 = 6,33620,47 = -14,134 T

SFb = SFP1Rb = -14,134 + 28,07 = 13,936 T

SFP2 = SFbP2 = 13,93610,47 = 3,466 T

SFP2 = SFF2P3 = 3,46610,47 = -7,004 T

SFc = SFP2 + Rc = -7,004 + 7,004 = 0 T

c. PerhitunganBending Moment Diagram(BMD)Ma = Ra . 0 = 6,336 x 0 = 0 Tm

MP1 = Ra . 4 = 6,336 x 4 = 15,476 Tm

Mb = Ra . 8 x P1.4 = 6,336 x 8 . 20,47 x 4 = -31,193 Tm

MP2 = Rb. 3 + Mb = 13,936 x 3

31,193 = 10,615TmMP3 = Rb. 6P2. 3 + Mb = 13,936 x 610,47 x 331,193 = 21,013 Tm

Mc = Rb. 9P2. 6 + MbP3.3 = 13,936. 910,47. 631,19310,47.3

= 0 Tm

8/12/2019 Mekanika Struktur III

4/4

d. Gambar diagram masing-masing