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