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TRANSCRIPT
Tugas Besar Analisa Struktur 2
Diketahui λ = 1m, p = 5 ton, Hitung besar Gaya di setiap Simpulnya!
Jawab :
Reaksi Perletakan :
Soal 1 1
∑MA=0−RB .6+ p . 6+ p . 5+p . 4+2 p . 3+ p .2+ p . 1=0−6 RB+5 .6+5 .5+5 .4+10 .3+5 .2+5 .1=0−6 RB+120=0
RB=1206
=20 t (↑ )
∑MB=0RA . 6−p . 6−p . 5−p . 4−2 p. 3−p . 2−p . 1=06 RA−5 .6−5 .5−5 .4−10 .3−5 .2−5.1=06 RA−120=0
RA=1206
=20 t (↑)
αα α
α
Tugas Besar Analisa Struktur 2
CEK :
SIMPUL A
SIMPUL C
SIMPUL H
Soal 1 2
∑ KV=0RA+RB−8 p=020+20−40=00=0(OKE )
∑ Kx=0S 7−S 1cosα=0S 7=S 1cosα
S 7=15√5×2√5S 7=30 t
∑ Ky=0RA−p−S1. sinα=020−5−0,5S1=00,5 S1=15S1=30 t
S9=0S 7−S 8=0S 8=S 7S 8=30 t
∑ Kx=0S 1cos α−S 2cos α−S 10cos α=0
15√5×2√5−2
√5S 2−2
√5S 10=0
2√5
(S 2+S 10 )=30
S 2+S 10=15√5 .. .. . .. .. . .. ..(1 )
α
Tugas Besar Analisa Struktur 2
SIMPUL D
Soal 1 3
∑ Ky=0p+S 2sinα−S 10 sinα−S 1sin α=0
5+1
√5S 2−1
√5S 10−15√5×1√5
=0
1√5
(S 2−S 10 )=10
S 2+S 10=10√5 .. .. . .. .. . .. .(2)
S 2=15√5−S10S 2=15√5−5
2√5
S 2=252
√5 t
S 2+S 10=15√5S 2−S 10=10 √5−2S 10=5√5S 10=5
2√5 t
S 10cos α−S 8+S 12=052
√5×2√5−30+S 12=0
S 12=30−25S12=25 t
S 10sinα−S 11=0S 11=S 10 sinα
S 11=52
√5×1√5
S 11=52t
αα
45
α α
Tugas Besar Analisa Struktur 2
SIMPUL I
SIMPUL J
Soal 1 4
∑ Kx=0S 2cos α−S 3 cosα+S 13 cos 45=0252
√5⋅2√5−2
√5⋅S 3−1
√2⋅S 13=0
2
√5⋅S 3+
1
√2⋅S 13=25 .. . .. .. .. . .. .. . ..(3 )
2
√5⋅S 3+
1
√2⋅S 13=25
−1√5S 3+1
√2S 13=−5
3√5S 3=30
S 3=10√5 t1
√2S 13=25−10√5⋅2√5
S 13=5√2 t
∑ Ky=0S 2sinα+S 13 sin 45−S 11−S 3sinα−p=0252
√5⋅1√5+1
√2S 13−5
2−1
√5S 3−5=0
−1√5S 3+1
√2S 13=−5. . .. .. . .. .. . .. .(4 )
∑ Kx=0S 3cos α−S 4cos α=0S 3=S 4S 4=10 √5 t
4545
Tugas Besar Analisa Struktur 2
SIMPUL E
Soal 1 5
∑ Ky=0S 3sinα+S 4 sinα−2 p−S 14=0
10√5⋅1√5+10√5⋅1√5
−10−S 14=0
S14=10−10+10S 14=10 t
∑ Ky=0S 14−S 13 sin 45−S 15 sin 45=0
10−5√2⋅1√2−S 15⋅1
√2=0
1√2
⋅S 15=10−5
S 15=5√2t
∑ Kx=0S 13 cos45+S 16−S 12−S 15 cos45=0
5√2⋅1√2+S 16−25−5=0
S 16=25 t
α
45α
α
Tugas Besar Analisa Struktur 2
SIMPUL K
SIMPUL F
Soal 1 6
∑ Kx=0S 4 cosα+S 15 sin 45−S 5 cosα=0
10√5⋅2√5+5√2⋅1√2
−2
√5S 5=0
−2
√5S 5=−25
S 5=252
√5 t
∑ Ky=0S 5sinα+S 15cos 45−S 17−p−S 4 sinα=0252
√5⋅1√5+5 √2⋅2√5
−S 17−5−10√5⋅1√5=0
S 17=25+10−10−202
S 17=52t
∑ Ky=0S 17−S 18 sinα=0S 18 sinα=S 171√2S 18=5
2
S 18=52
√2 t
∑ Kx=0S 19−S 16−S 18 cosα=0
S 19−25−52
√5⋅2√5=0
S 19=30 t
ααα
α
Tugas Besar Analisa Struktur 2
SIMPUL G
SIMPUL L
SIMPUL B
Soal 1 7
S 20=0∑ Kx=0S 21−S 19=0S 21=S 19S 21=30 t
∑ Ky=0S 6sin α+S 18sin α−p−S 5sinα=01
√5S 6+5
2√5⋅1√5
−5−252
√5⋅1√5=0
1√5S 6=15
S 6=15√5t
∑ Kx=0S 6cos α−S 21=0S 21=S 6cos α
S 21=15√5⋅2√5S 21=30 t
Tugas Besar Analisa Struktur 2
TABEL :
GayaBesar (TON)
GayaBesar (TON)
Tarik (+) Tekan (-) Tarik (+) Tekan (-)
RA - 20 S13 - 5 √2=7,07
RB - 20 S14 10 -
S1 - 15√5=33,54 S15 - 5 √2=7,07
S2 - 25/2 √5 = 27,95 S16 25 -
S3 - 10 √5 = 22,36 S17 2,5 -
S4 - 10 √5=22,36 S18 - 5/2 √5 = 5,59
S5 - 25/2 √5 = 27,95 S19 30 -
S6 - -15√5=33,54 S20 - -
S7 30 - S21 30
S8 30 -
S9 - -
S10 - 5/2 √5 = 5,59
S11 2,5 -
S12 25 -
Soal 1 8
∑ Ky=0R B−p−S 6sinα=0
20−5−15 √5 .1√5=0
20−5−15=00=0 (Oke )