Download - SOAL MEKTEK 2
PELENGKUNG TIGA SENDI
Pertanyaan : Hitung GGD dititik C dan D
SOLUSI :
Y = 4h ( x )(L−x )
L2
15 = 4 (20 ) (100 )(L−100)
L2
15L2 = 8000(L-100)
3L2 - 1600L + 160.000 =0
L1,2 = −(1600)±√(1600)2−4 (3 )(160.000)2.3
L1 = 400 m (tidak memenuhi)
L2 = 133,3 m
Reaksi Perletakan
∑MA = 0
5 m
15 m
Rbh
Rbv
DS
C20 m 20 m
20 m 10 m
P=20 kN
P = 10 kN
Rah
Rav
100 m
S O A L 1
-Rbv(100)+Rbh(15)+20(56,665)+10(36,665) = 0
-100 Rbv+15Rbh =-1499,95 …................1)
∑MA = 0
Rav(100)+Rah(15) -10(63,335)-20(43,335) = 0
100Rav + 15 Rah = 1500,05 ………………….2)
∑Ms kiri = 0
Rav(66,665)+Rah (20)-10(30)-20(10) = 0
66,665Rav +20Rah = 500 ……………………3)
Pers(2) dan (3)
100 Rav+15Rah = 1500,05 ×4 400Rav+60Rah = 6000,2
66,665Rav+20Rah = 500 ×3 199,995Rav+60 Rah = 1500
200,005Rav = 4500,2
Rav = 22,5 kN (↑)
2250+15Rah =1500,05
15 Rah = -749,95
Rah =-49,997 kN (←)
Rah = 49,997 kN(→)
∑V =0 → Rav+Rbv – 10-20 = 0
22,5+Rbv – 30 =0
Rbv = 7,5 kN
∑H =0 → Rah –Rbh = 0
49, 997+Rbh = 0
Rbh = 49,997 kN (←)
GGD
Y =4 (20 ) x (133,33−x)
(133,33)2
Y = 80 x(133,33−x )17776,89
Y = 80
17776,89(133,33−x2) Belum selesai………
……………….
HITUNG GGD DI TITIK C
Q1 = 3 kN/m
Rbv
Rbh
S
C
A
B
Rav
Rah
5 m
Y =10 m
30 m 20 m
Q2 = 1,5 kN/m
C
Q1 = 3 kN/m
Q2 = 1,5 kN/m
V = 33,15 kNH = 107,625 kN
93,15 kN
86,125 kN
S O A L 2
SOLUSI
∑MA = 0 →
(3)(50)(25) - (1,5)(5)(2,5) + (1,5)(10)(5) + Rbh(10) + Rbv(50) = 0
3750 – 18,75 + 75 + 10 Rbh -50 Rbv = 0
10 Rbh -50 Rbv = 3806,25 …………………….(1)
∑MB = 0 →
(3)(50)(25) – (1,5)(15)(15/2 ) + Rah(10) + Rav(50) = 0
-3750 – 168,75 + 10 Rah + 50 Rav = 0
10 Rah + 50 Rav = 3198,75 ……..………………(2)
PERSAMAAN LENGKUNG PARABOLA, Y = 10, X = 50, h = 15
Y = 4h ( x )(L−x )
L2
10 = 4 (15 ) (50 )(L−50)
L2
10L2 = 3000 L – 150.000
L2 = 300L + 15000 = 0
L1,2 = −(−300)±√(−300)2−4 (1 )(15000)
(2 )(1)
L1 = 63,40 m
93,15 kN
L2 = 236,61 m (tidak memenuhi)
∑Ms kanan = 0
(3) (31,7)(15,85) + (1,5)(15)(15/2) – Rbv(31,7) + Rbh(15) = 0
1507,335 + 168,75 – 31,7 Rbv + 15 Rbh = 0
31,7 Rbv – 15 Rbh = 1676,085 ………………………………………….(3)
PERSAMAAN (1) DAN (3)
50Rbv – 10 Rbh = 3806,25 ×3 150 Rbv – 30 Rbh = 11418,75
31,7 Rbv – 15 Rbh = 1676,085 ×2 63,4 Rbv – 30 Rbh = 3352,17
86,6 Rbv = 8066,58
Rbv = 93,15 kN (↑)
50 Rbv – 10 Rbh = 3806,25
50(93,15) – 10 Rbh = 3806,25
4657,5 – 3806,25 = 10 Rbh
Rbh = 85,125 kN
Untuk h = 15 m, L = 63,5 m
Y = 4h ( x )(L−x )
L2
Y = 4 (15 ) ( x )(63,40−x)
63,402
Y = 0,95x – 0,01 x2
dydx
=0,95−0,02x
X = 20 → dydx
=0,55
tan θ = 0,55 → θ = 28,81 0
sin θ = 0,48
cos θ = 0,88
Vx = Rbv – 3(x) = 93,15 – 3(20)
= 33,15 kN (↓)
Hx = Rbh + (1,5)(15) = 85,125 + 22,5
= 107, 625 kN (→)
SFc = V cos θ – H sin θ= (33,15)(0,88) – (107,625)(0,48)= 29,172 – 51,66= -22,488 kN
NFc = V sin θ + H cos θ= (33,15)(0,48)+(107,625)(0,88)= 110,622 kN
Mc = Rbv (x) - Rbh(y) – 0,5(Q2)y2 – 0,5(Q1)x2
= 93,15(20) – 85,125(15) – 0,5 (1,5) (15)2 – 0,5 (3) (20)2
= 182,625 kNm
Pertanyaan : Hitung GGD titik C (8 meter dari A) dan titik D (15 meter dari A)
SOLUSI :
h = 10 m
x = 40 m
Y = 6 m
Y = 4h ( x )(L−x )
L2
P = 5 kN
P = 10 kN
SD
C
4 m
6 mRbh
Rbv
40 m
Rah
Rav
A
B
5 m 5 m
8 m
15 m
S O A L 3
6 = 4 (10 ) (40 )(L−x )
L2
6 L2 – 1600L + 64000 = 0
L1,2 = −(−1600)±√(−1600)2−4 (6 )(64000)
2(6)
L1,2 = 1600±√1024000¿ ¿12
L1 = 217,6607 m (tidak mungkin)
L2 = 49,0059 m
Reaksi Perletakan
∑MB = 0
Rav(40) – Rah(6) – 5(35) – 10(30) = 0
40 Rav – 6Rah – 475 = 0 ……………………………(1)
∑MS kiri = 0Rav (L/2) – Rah.10 – 5(L/2-5) – 10(L/2 – 10) = 0 Rav (24,5030) – Rah.10 – 97,515 – 145,03 = 0 24,5030 Rav– 10.Rah – 97,515 – 242,545 = 0 ……………………………(2)
ELIMINASI PERSAMAAN (1) DAN (2) :
40 Rav – 6Rah – 475 = 0 ×10 400Rav – 60Rah = 475024,5030 Rav– 10.Rah – 97,515 – 242,545 = 0 ×6 147,018 Rav – 60 Rah = 1455,27
252,982 Rav = 3294,73
Rav = 13,0236 kN(↑)
Substitusi Rav ke persamaan (1)
40 Rav – 6Rah – 475 = 0
40(13,0236) – 6(Rah) – 475 = 0
Rah = 7,6573 kN (→)
∑H = 0
Rah + Rbh = 0
Rah = - Rbh = -7,6573 kN (→)
∑MA = 0-Rbv.40 + Rbh.6 + 10.10 + 5.5 = 0-Rbv.40 +(-7,6573)6 + 125 = 0Rbv = 1,9764 kN (↑)
∑V = 0Rav + Rbv -10 - 5 = 013,0236 + 1,9764 – 15 = 0 …..ok!!
GAYA-GAYA DALAM
Untuk h = 10 m , L = 49,0059 mPersamaan Parabola :
Y = 4h ( x )(L−x )
L2
Y = 4 (10 ) x (49,0059−x)
49,00592
Y = 40
2401,5782(49,0059 x−x2)
Titik c, untuk x = 8 m dari A, maka :
dydx
= 40
2401,5782(49,0059−2 x )
X = 8 → dydx
= 0,5479
tan θ = 28,79760
sin θ = 0,48170
cos θ = 0,87630
Titik C (8; 5,4693)
Vx = Rav – 5 = 13, 0236 – 5 = 8,0236 kN (↓)Hx = Rah = 7,6573 kN (←)
Gaya Lintang (SFx)SFx = Vcos θ – H sin θ = (8,0236)( 0,8763) – (7,6573)( 0,4817)
= 3,3426 kN ≈ 3 kNGaya Normal (NFx)NFx = -(Vcos θ + H sin θ) = - ((8,0236)( 0,8763) + (7,6573)( 0,4817))
= - 8,4762 kN
Momen di titik CMc = Rav 8 - Rah. 5,4639 – 5.3 = 13,0236. 8 – (7,6573)(5,4639) – 15 = 47,3501 kNm
Titik D
Untuk x =15 m, maka Y = 40
2401,5782(49,0059(15)−(15)2)
d ydx
= 40
2401,5782(49,0059−2 x )
X=15 m → dydx
= 40
2401,5782(49,0059−2 (15 ) )=0,3166m
tan θ = 0,3166 → θ = 17,56780
sin θ =0,3018, cos θ = 0,9534
Titik D (????)Vx = Rav – 5 – 10 = 13,0236 – 15 = -1,9764 kN (↓)Hx = Rah = 7,6573 kN (←)
o Gaya Lintang (SFx)SFx = Vcos θ – H sin θ = (-1,9764)(0,9534) – (7,6573)( 0,3018) = -4,1953 kN ≈ -4 kN
o Gaya Normal (NFx)
NFx = -( Vcos θ + H sin θ) = -((-1,9764)( 0,9534)+(7,6573)(0,3018))
= -1,8843 +2,3110 = -0,4267 kN
o Momen di Titik DMD = Rav 18 - Rah.(8,4959) – 5(10) - Belum selesai………………………………………………….
Pertanyaan : Hitung GGD pada titik C (5 meter dari B) dan titik D (2 meter dari A)
S O A L 4q = 2 t/m’
q = 1 t/m’
DS
C
B
A Rah
Rav
2 m
4 m
Rbv
Rbh
2 m 5 m
20 m
S
Penyelesaian : Untuk x = 20 m, y = 4 m, h = 6 m
Y = 4h ( x )(L−x )
L2
4 = 4 (6 ) (20 )(L−20)
L2
4L2 – 480 L + 9600 = 0
L1,2 = −(−480)±√(−480)2−4 (4 )(9600)
2(4)
L1,2 = 480±√76800¿ ¿8
L1 = 94,6410 m (tidak mungkin)
L2 = 25,3590 m
Reaksi Perletakan
∑MA = 0
(2)(20)(10) + (1) (4)(2) – (1)(2)(1) - Rbv.20 + Rbh.4 = 0
400 + 8 – 2 - Rbv.20 + Rbh. = 0
- Rbv.20 + Rbh.4 + 406 = 0
Rbv.20 - Rbh.4 – 406 = 0 ……………………….……………………………(1)
∑MB = 0
-(2)(20)(10) – (1)(6)(3) + Rav.20 + Rah.4 = 0
-400 – 18 + Rav.20 + Rah.4 = 0
Rav.20 + Rah.4 - 418 = 0……………………………………………………………..…(2)
∑Ms kanan = 0-Rbv. 12,6795 + Rbh.6 +(2)(12,6795)(6,3398) + (1)(6)(3) = 0
-12,6795 Rbv +6 Rbh + 178,7710 = 0………………………………………..……(3)
∑Ms kiri = 0-Rav.7,3025 – Rah.2 –(2)(7,3205)(3,6603) = 07,3205 Rav – 2Rah – 53,5905 = 0…………………………………………………..(4)
ELIMINASI PERSAMAAN (1) DAN (3)
20 Rbv – 4Rbh = 406 ×6 120 Rbv - 24Rbh = 243,6-12,6795 Rbv + 6 Rbh = -178,7710 ×4 -50,718 Rbv + 24Rbh = -715,084 +
69,282 Rbv = 1720,916 Rbv = 24,8393 t (↑)
Substitusi Rbv = 24,8393 t ke persamaan (1)20 Rbv – 4 Rbh = 40620(24,8393) – 4Rbh = 406 Rbh = 22,6965 t (←)
Eliminasi (2) dan (4) :Rav.20 + Rah.4 - 418 = 0 ×1 Rav.20 + Rah.4 = 4187,3205 Rav – 2Rah – 53,5905 = 0 ×2 14,641 Rav – 4 Rah = 107,181 +
34,641 Rav = 525,81
Rav = 15,1607 t (↑)
Substitusi Rav = 15,1607 t ke (2)
2Rav + 4 Rah = 418
20(15,1607) + 4 Rah = 418
Rah = 28,6965 t (→)
∑ V = 0Rav + Rbv – 40 = 015,1607 + 24,8393 – 40 = 0……………….ok!
∑ H = 0Rah – Rbh – 6 = 0
28,6965 – 22,6965 – 6 = 0………………….ok!
MENENTUKAN GAYA-GAYA DALAM
Untuk h = 6 m, L = 25,3590 m → Y = 4h ( x )(L−x )
L2=4(6)( x )(25,3590−x)
25,35902
¿ 24643,0789
(25,3590x−x2)
Untuk titik c, x = 5 m dari B → y ¿24
643,0789(25,3590 (5)−(5)2 )=3,7990m
{dydx }= 24643,0789
(25,3590−2 x)
X = 5 m → {dydx }= 24643,0789
(25,3590−2 (5)) = 0,5372
tan θ = 0,53720 → θ = 29,8213o sin θ = 0,4937cos θ = 0,8676
Titik C (5; 3,7990) dari BVx = Rbv – qx =24,8393 – (2)(5) = 14, 8393 t (↓)
Hx = Rbh +q.y = 22,6965 +(1)(3,7990) = 26,4955 t (→)
Gaya Lintang (SFx)
SFx = V cos θ – H sinθ = (14,8393)(0,4973) + (26,4955)(0,8676) = 30,3671 t
Mc = Rbv. 5 – Rbh. 3,7990 +(2)(5)(2,5)
= 24,8393 (5) – 22,6965(3,7990) +25
= 62,9725 tn
Untuk titik D, x = 2m dari A, h = 2m , L = 25,3590 m, x = 2 m
Y = 4h ( x )(L−x )
L2=4(2) ( x )(25,3590−x )
25,35902
= 8
643,0789(25,3590x−x2)
Untuk x = 2m , y = 8
643,0789(25,3590(2)−(2)2) = 0,5812 m
{dydx }= 24643,0789
(25,3590−2 x)
{dydx }= 24643,0789
(25,3590−2 (2 ) )=0,2657
tan θ = 0,26570 → θ = 14,8797o sin θ = 0,2567cos θ = 0,9665
TITIK D (2 ; 0,5812)Vx = Rav – q.x = 15,1607 – 2(2) = 11,1607 t (↓)Hx = Rah = 28,6965 t (←)
GAYA LINTANG (SFx)SFx = Vcos θ – H sin θ = (11,1607(0,9665))-(28,6965(0,2567)) = 3,4204 t ≈ 3t
Gaya Normal (NFx)NFx= -(Vsin θ + Hcos θ) = -((11,1607(0,2567)) +((28,6965)(0,9665)) = -30,6001 tMD = RAv.2 – Rah. 0,5812 – (2)(2)(1) = 15,1607(2) – 28,6965(0,5812) – 4= 19,1128 tm
UNTUK TITIK A (0,0) DARI A
X = 0 → y = 0 {dydx }x=0→8
643,0789(25,3590−0 )=0,3155
Tan θ = 0,3155 → θ = 17,5105 sin θ = 0,3009cos θ = 0,9537
TITIK A (0,0)Vx = Rav = 15,1607 t (↓)Hx = Rah = 28,6965 t (←)
Gaya Lintang (SFx)SFx = Vcos θ - Hsin θ (15,1607)(0,9537) – (28,6965)(0,3009) = 5,8240 t ≈ 6 t
Gaya Normal (NFx)NFx = -(Vsin θ + H cos θ) = -((15,1607) (0,3009) + (28,6965)(0,9537) = -31,9297 tM dititik A → Ma = 0
Untuk titik S (12,6795;6) dari B
{dydx }x=12,6795= 24
643,0789(25,3590−(2 ) (12,6795 ) )=0
Tan θ = 0 → sin θ = 0, cos θ = 1S = (12,6795;6)Vx = Rbv – q.x = 24,8393 –(2)(12,6795) = -0,5197 t (↓)
Hx = Rbh + q.x = 22,q.y = 22.,6965 +(1)(6) = 28,6965 t (→)
Gaya Lintang (SFx)SFx = Vcos θ – Hsin θ = (-0,5197.1) – (28,6965.0) =-0,5197 t Gaya Normal (NFx) NFx = Vsin θ + Hcos θ = (-0,5197.0) – (28,6965.1) = 28,6965 t
Untuk titik B (0,0) dari B
{dydx }x=0= 24
643,0789(25,3590−(2 ) (0 ) )=0,9464
Tan θ = 0,9464 → θ = 43,4226Sin θ = 0,6874Cos θ = 0,7263
Titik B (0,0)Vx = Rbv = 24,8393 t (↓)HX = Rbh = 22,6965 t (→)Gaya Lintang (SFx)SFx = Vcos θ – Hsin θ = (24,8393)(0,7263) – (22,6965)(0,6874) =2,4392 t Gaya Normal (NFx) NFx = Vsin θ + Hcos θ = (-0,5197)(0,6874) – (28,6965)(0,7263) = 28,6965 t
S O A L 5
S
q = 2 kN/m
C10 m
P1 = 5kN
P2 = 4 kN
Persamaan dasar parabola yang digunakan adalah : 4h ( x )(L−x )
L2
Dimana :
Y = tinggi titik yang ditinjau dari tumpuan
H = tinggi puncak parabola dari tumpuan
X = jarak mendatar dari tumpuan terdekat
L = jarak mendatar dua tumpuan
SOLUSI :
Y : 4h ( x ) (L−x )
L2 = 4 (10 ) (10 ) (40−10 )
402 = 400 (30 )1600
=12001600
=7.5m
∑MA = 0(-Rbv.40)-(4.7,5) + (5.30) + (q.20.10) = 0-40 Rbv – 30 +150 + 400 = 0
-40 Rbv = -520Rbv = 13 kNm
∑MB = 0(-Rav.40) – (q.20.30) – (P1.10) – (P2.7,5) = 0(-Rav.40) – 1200 – 50 – 30 = 0
40 Rav = 1280Rav = 32 kN
∑Ms kanan = 0
A B20 m 10 m 10 m
(-Rbv.20) + (Rbh.10) + (4.2,5) + (5.10) = 0(-20.13) + (10 Rbh) + 10 + 50 = 0-260 + 10 Rbh + 60 = 0
Rbh = 20 kN
∑Ms kiri = 0(-Rav.20) - (Rah.10) - (q.20.10) = 0(-20.32) - (10 Rah) – 400 = 0-10 Rah = 0
Rah = 24 kN
∑V = 0Rav + Rbv = 40 +532 + 13 = 45 ………..ok!
∑H = 0Rah - Rbh = 424 – 20 = 4 …………..ok!
S O A L 6
5 kN 5 kN
1 kN
4 m
56
F8F6
F7
F9
Pertanyaan : Hitung gaya gaya batang yang terjadi pada struktur rangka batang tegrambar di bawah ini menggunakan metode kesetimbanan titik :
m = 2j – r
9 = 2(6) – 3
9 = 9….(statis tentu)
1. REAKSI PERLETAKAN
∑MA = 0(-Rbv.3) + (5.3) + (-1.8) +(-1.4) = 0-3Rbv – 15 – 8 – 4 = 0
Rbv = 1
∑MB = 0(-Rav.3) + (-5.3) + (-1.8) +(-1.4) = 0
Rav = 9
∑V = 0Rav + Rbv = 05 + 5 = 0…………………………...ok!
∑H = 0
1 kN
1 kN
4 m
12
34
F2F3
F1
F4
3 m
Rav Rbv
Rah Rbh
F5
M= jumlah batang
J = jumlah titik buhul
R = jumlah reaksi perletakan
Rah = 1+1+1Rah = 3…………………………...ok!
II. GAYA DI TITIK BUHUL
BUHUL 1
BUHUL 2
∑ V = 0
Rbv + F4 + F3y = 0
1 + F4 + 3,3345
= 0
F4 + 3,66 = 0 → F4 = - 3,66
∑ H = 0F1 + F3x + 1 = 0-3 + F3x + 1 = 0F3 = 3,333
BUHUL 3
αRah
Rav
F1
F2
4 m
3 m
∑V = 0
Rav + F2 = 0
9 + F2 = 0
F2 = -9 kN
∑H = 0
Rah + F1 = 0
3 + F1 = 0
F1 = -3 kN
F4F3yF3
F3x
F1
F3x
F3y
3 m
4 m5 mα
F3x = F3 sin α = F335
F3y = F3 cos α = F345
F6
F3x = F3 cos α = F335
F3y = F3 sin α = F345
∑ V = 0
F2 + F3y – F6 = 0
-9 + 3,33 45
= 0 → F6 = - 6,33
∑ H = 0
F3x + F5 = 0 → F5 = - 1,99
BUHUL 5
o ∑ V = 0
F6 + F7y + 5 = 0
-6,33 + F7 .45
+ 5= 0 → F7 = 1,66
o ∑ H = 0F7x + F9 = 0 → F9 = - 0,99
BUHUL 6
F2 F3y
F3
F3x
F5
F9
F7x
F7F7yF6
5 kN
F7x = F7 cos α = F7 .35
F7y = F7 sin α = F7.45
o ∑ V = 0
F8 + 5 = 0
F8 = -5
BUHUL 4 (kontrol)
>>>>>>>>>>>>>>>>>>>>>>>>>>>>>>> SELESAI<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<<
5 kN
F9
F8
1 kN
F8
F7yF7
F7x
F5
F4
F7x = F7 cos α = 1,66 .35 =0,99
F7y = F7 sin α = F7.45
= 1,33
o ∑ V = 0
F6 + F7y - F4 = 0
-5 +1,33 – (-3,36) = 0 → ok
o ∑ H = 0
PORTAL TIGA SENDI
Struktur portal tiga sendi dengan ukuran dan pembebanan seperti Gambar di bawah, hitunglah :
1. Reaksi Perletakan2. Persamaan Gaya-Gaya Dalam dan Diagram Benda Bebas (FBD)3. Diagram Gaya-Gaya Dalam (Momen, Geser, dan Normal)
S DC
q1 = 5 kN/mP = 5 kN
SOLUSI
DIAGRAM BENDA LEPAS (DBL)
S DC
B
R1 = 10 kN
R3 = 12 kN
P = 5 kN
2 m
2 m 2 m 4 m
R2 = 20 kN
A
B
q2 = 3 kN/m4 m
2 m
Rah
Rav
Rah
Rav
q2 = 3 kN/m
2 m
MENENTUKAN BESARNYA REAKSI PERLETAKAN
∑MA = 0
-RBV.6 + RBH.2 – 12.4 – 20.4 + 10.1 - 5.2 = 0
-RBV.6 + RBH.2 = -32 ………………………………………………………………………………………..(1)
∑MB = 0
-RAV.6 + RAH.2 –5.8 – 10.5 – 20.2 – 12.2 = 0
-RAV.6 - RAH.2 = 154 ………………………………………………………………………………………..(2)
∑Ms kanan = 0
-RBV.4 + RBH.4 + 12. 2 +20.2 = 0
-RBV.4 - RBH.4 = -64………………………………………………………………………………………..(3)
∑Ms kiri = 0
RAV.2 - RAH.6 – 5.4 – 10.1 = 0
RAV.2 - RAH.6 = 30…………………………………………………………………………………………..(4)
A
2 m
2 m 1 m 2 m1 m 2 m
Rah
Rav
Rbh
Rbv
ELIMINASI PERSAMAAN (1) DAN (3)
-RBV.6 + RBH.2 = -32 ×2 -RBV.12 + RBH.4 = -64
-RBV.4 - RBH.4 = -64 ×1 -RBV.4 - RBH.4 = -64 +
-16 RBV = -128
RBV = 8 kN
Substitusi nilai RBV ke persamaan 1
-RBV.6 + RBH.2 = -32 → -6(8) + RBH.2 = -32 → RBH = 8 kN
Eliminasi Pers. 2 dan Pers 4
-RAV.6 - RAH.2 = -154 ×1 -RAV.6 - RAH.2 = -154
RAV.2 - RAH.6 = 30 ×3 RAV.6 - RAH.18 = 90 -
16 RAH = 154
RAH = 4 kN
Substitusi nilai RAH ke Persamaan 2
-RAV.6 - RAH.2 = -154 → -6(RAV) – 2(4) = 154 → RAV = 27 kN
DIAGRAM BENDA BEBAS
S DC
q2 = 3 kN/m
q1 = 5 kN/m
P = 5 kN
E
5
10C
4
27
24 C
34
4
22
R1 = 30 kN
A
B
Rah = 4
Rav = 27
Rbh = 8
Rbv = 8