Menghitung Reaksi Perletakan, Gaya Lintang, dan Momen pada Portal -Statis Tertentu-

Contoh Soal Menghitung Reaksi Perletakan Portal

Welcome back guys hihi, oh iya saya baru meng-update lagi yaa setelah libur bertahun-tahun lamanya wkwk. Seperti biasa tujuan saya bikin blog ini adalah untuk diri saya sendiri (oh iya jelas dong gue mentingin diri sendiri hahahaha *just kidding). So my point is saya bikin blog ini untuk bahan belajar saya juga, jika ada kesalahan dalam perhitungan saya, please give me a correction, agar saya bisa perbaiki, and we can discuss :)

Agar suatu sistem dalam keadaan statis atau tidak bergerak, maka harus ada beberapa perletakan pada sistem tersebut agar gaya-gaya luar dapat diimbangi oleh perletakan. Jenis Perletakan ada tiga yaitu perletakan jepit, perletakan sendi, dan perletakan rol. Dalam statika ada 3 syarat statis yang harus di penuhi yaitu :

ƩF_Y = 0

ƩF_X = 0

ƩM = 0

Materi kali ini yang dibahas adalah tentang menghitung reaksi perletakan, menghitung gaya lintang dan nilai momen pada portal.

SOAL 1

Jawaban :

Menghitung Reaksi Perletakan

ƩF_X = 0

H_A – P₂ = 0

H_A – 30 = 0

H_A = 30 kN

 

ƩM_A = 0

(–R_BV×5) – (P₂ × 6) + (P₁ × 4) + (q×3×1.5) = 0

(–R_BV×5) – (30 × 6) + (20 × 4) + (20×3×1.5) = 0

(–R_BV×5) – 10 = 0

R_BV = –10/5 = –2 kN

 

ƩM_B = 0

(R_AV×5) – (H_A×3) – (q×3×3.5) – (P₁×1) – (P₂×3)  = 0

(R_AV×5) – (30×3) – (20×3×3.5) – (20×1) – (30×3)  = 0

(R_AV×5) – 410 = 0

R_AV = 410/5 = 82 kN

 

CHECK

ƩF_Y = 0

R_AV + R_BV – P₁ – q×L = 0

82 – 2 – 20 – 20×3 = 0 (OK)

 

Menghitung Bidang Lintang

D_A = –H_A = –30 kN

D_C = R_AV = 82 kN

D_D = D_C – q×3 = 82 – 20×3 = 22 kN

D_E = D_D – P₁ = 22 – 20 = 2 kN

D_F = D_E + R_BV = 2 – 2 = 0 kN

 

Menghitung Bidang Momen

M_A = R_AV × 0 = 82 × 0 = 0 kN.m

M_C = –H_A × 6 = –30 × 6 = –180 kN.m

M_D = (82×3) – (30×6) – (20×3×1.5) = –24 kN.m

M_E = (82×4) – (30×6) – (20×3×2.5) = -2 kN.m

M_F = (82×5) – (30×6) – (20×3×3.5) – (20×1) = 0 kN.m

M_B = (82×5) – (30×3) – (20×3×3.5) – (20×1) – (30×3) = 0 kN.m

 

SOAL 2

Jawaban :

Menghitung Reaksi Perletakan

ƩF_X = 0

H_A + P₁ = 0

H_A + 25 = 0

H_A = –25 kN

 

ƩM_A = 0

(–R_BV×6) + (P₁×5) + (P₂×8) = 0

(–R_BV×6) + (25×5) + (35×8) = 0

(–R_BV×6) + 405 = 0

R_BV = 405/6 = 67.5 kN

 

ƩM_B = 0

(R_AV×6) + (P₁×5) + (P₂×2) + (H_A×0) = 0

(R_AV×6) + (25×5) + (35×2) + 0 = 0

(R_AV×6) + 195 = 0

R_AV = –195/6 = –32.5 kN

 

CHECK

ƩF_Y = 0

R_AV + R_BV – P₂ = 0

–32.5 + 67.5 – 35 = 0 (OK)

 

Menghitung Bidang Lintang

D_A = –H_A = –25 kN

D_C = R_AV = –32.5 kN

D_D = D_C + R_BV = –32.5 + 67.5 = 35 kN

D_E = –P₁ = –25 kN

D_F = D_D – P₂ = 35 – 35 = 0 kN

 

Menghitung Bidang Momen

M_A = R_AV × 0 = –32.5 × 0 = 0 kN.m

M_C = –H_A × 4 = 25 × 4 = 100 kN.m

M_D = (–32.5×6) + (25×4) = –95 kN.m

M_E = (–32.5×6) + (25×5) = –70 kN.m

M_F = (–32.5×8) + (25×4) + (67.5×2) + (25×1) = 0 kN.m

M_B = (–32.5×6) + (25×5) + (35×5)  = 0 kN.m

Contoh Soal Menghitung Reaksi Perletakan Balok Sederhana

Contoh Soal Menghitung Reaksi Perletakan Balok (Beam)

Welcome to my blog guys hehe, oh iya sebelum masuk materi saya perkenalan dulu deh, tujuan saya bikin blog ini adalah untuk diri saya sendiri (oh iya jelas dong gue mentingin diri sendiri hahahaha *just kidding). So my point is saya bikin blog ini untuk bahan belajar saya juga, jika ada kesalahan dalam perhitungan saya, please give me a correction, agar saya bisa perbaiki, and we can discuss :)

Agar suatu sistem (dalam hal ini balok) dalam keadaan statis atau tidak bergerak, maka harus ada beberapa perletakan pada sistem tersebut agar gaya-gaya luar dapat diimbangi oleh perletakan. Jenis Perletakan ada tiga yaitu perletakan jepit, perletakan sendi, dan perletakan rol. Dalam statika ada 3 syarat statis yang harus di penuhi yaitu :

ƩF_Y = 0

ƩF_X = 0

ƩM = 0

Materi kali ini yang dibahas adalah tentang menghitung reaksi perletakan, menghitung gaya lintang dan nilai momen dari suatu struktur balok (beam).

SOAL 1 (BEBAN TEPUSAT)

Jawaban :

Menghitung Reaksi Perletakan

ƩM_A = 0

(–R_BV×22) + (30 × 15) + (10 × 9) + (20×4) = 0

(–R_BV×22) + 620 = 0

R_BV = 620/22 = 28.18 kN

 

ƩM_B = 0

(R_AV×22) – (20×18) – (10×13) – (30×7) = 0

(R_AV×22) – 700 = 0

R_AV = 700/22 = 31.82 kN

 

CHECK

ƩF_Y = 0

R_AV + R_BV – P₁ – P₂ – P₃ = 0

31.82 + 20.18 – 20 – 10 – 30 = 0 (OK)

 

Menghitung Bidang Lintang

D_A = R_AV = 31.82 kN

D_C = R_AV – P₁ = 31.82 – 20 = 11.82 kN

D_D = D_C – P₂ = 11.82 – 10 = 1.82 kN

D_E = D_D – P₃ = 1.82 – 30 = –28.18 kN

D_B = D_E + R_BV = –28.18 + 28.18 = 0 kN

 

Menghitung Bidang Momen

M_A = R_AV × L = 31.82 × 0 = 0 kN.m

M_C = (31.82×4) – (20×0) = 127.28 kN.m

M_D = (31.82×9) – (20×5) – (10×0) = 186.38 kN.m

M_E = (31.82×15) – (20×11) – (10×6) – (30×0) = 197.3 kN.m

M_B = (31.82×22) – (20×18) – (10×13) – (30×7) + (R_BV×0) = 0 kN.m

SOAL 2 (BEBAN MERATA)

Jawaban :

Menghitung Reaksi Perletakan

ƩM_A = 0

(–R_BV×6) + (20×6×3) = 0

(–R_BV×6) + 360 = 0

R_BV = 360/6 = 60 kN

 

ƩM_B = 0

(R_AV×6) – (20×6×3) = 0

(R_AV×6) – 360 = 0

R_AV = 360/6 = 60 kN

 

CHECK

ƩF_Y = 0

R_AV + R_BV – (q×L) = 0

60 + 60 – (20×6) = 0 (OK)

 

Menghitung Bidang Lintang

D_A = R_AV = 60 kN

D_A1 = R_AV – (20×1) = 60 – 20 = 40 kN

D_A2 = R_AV – (20×2) = 60 – 40 = 20 kN

D_A3 = R_AV – (20×3) = 60 – 60 = 0 kN

D_A4 = R_AV – (20×4) = 60 – 80 = –20 kN

D_A5 = R_AV – (20×5) = 60 – 100 = –40 kN

D_A6 = R_AV – (20×6) = 60 – 120 = –60 kN

 

Menghitung Bidang Momen

M_A = R_AV × L = 60 × 0 = 0 kN.m

L_MAX = R_AV/q = 60/20 = 3 m

M_MAX = (R_AV×L_MAX) – (1/2×q×L_MAX²) = (60×3) – (1/2×20×3²) = 90 kN.m

M_B = (60×6) – (20×6×3) + (R_BV×0) = 0 kN.m

SOAL 3 (BEBAN TERPUSAT DAN BEBAN MERATA)

Jawaban :

Menghitung Reaksi Perletakan

ƩM_A = 0

(–R_BV×10) + (10×8) + (5×6×3)  = 0

(–R_BV×10) + 170 = 0

R_BV = 170/10 = 17 kN

 

ƩM_B = 0

(R_AV×10) – (5×6×(3+4)) – (10×2) = 0

(R_AV×10) – 230 = 0

R_AV = 230/10 = 23 kN

 

CHECK

ƩF_Y = 0

R_AV + R_BV – (q×L) – P = 0

23 + 17 – (5×6) – 10 = 0 (OK)

 

Menghitung Bidang Lintang

D_A = R_AV = 23 kN

D_A1 = R_AV – (5×1) = 23 – 5 = 18 kN

D_A2 = R_AV – (5×2) = 23 – 10 = 13 kN

D_A3 = R_AV – (5×3) = 23 – 15 = 8 kN

D_A4 = R_AV – (5×4) = 23 – 20 = 3 kN

D_A5 = R_AV – (5×5) = 23 – 25 = –2 kN

D_A6 = R_AV – (5×6) = 23 – 30 = –7 kN

D_C = R_AV – (5×6) = 23 – 30 = –7 kN

D_D = D_C – P = –7 – 10 = –17 kN

D_B = D_D + R_BV = –17 + 17 = 0 kN

 

Menghitung Bidang Momen

M_A = R_AV × L = 23 × 0 = 0 kN.m

L_MAX = R_AV/q = 23/5 = 4.6 m

M_MAX = (R_AV×L_MAX) – (1/2×q×L_MAX²) = (23×4.6) – (1/2×5×4.6²) = 52.9 kN.m

M_C = (23×6) – (5×6×3) = 48 kN.m

M_D = (23×8) – (5×6×(3+2)) = 34 kN.m

M_B = (23×10) – (5×6×(3+4)) – (10×2) + (R_BV×0) = 0 kN.m

SOAL 4 (BEBAN TERPUSAT DAN BEBAN MERATA)

Jawaban :

Menghitung Reaksi Perletakan

ƩM_A = 0

(20×4×(2+14)) – R_BV×14 + (20×12) + (20×10) + (40×8×4)  = 0

– R_BV×14 – 3000 = 0

R_BV = 3000/14 = 214.29 kN

 

ƩM_B = 0

(R_AV×14) – (40×8×(4+6)) – (20×4) – (20×2) + (20×4×2) = 0

(R_AV×14) – 3610 = 0

R_AV = 3160/14 = 225.71 kN

 

CHECK

ƩF_Y = 0

R_AV + R_BV – (q₁×L₁) – P₁ – P₂ – (q₂×L₂)  = 0

214.29 + 225.71 – (40×8) – 20 – 20 – (20×4) = 0 (OK)

 

Menghitung Bidang Lintang

D_A = R_AV = 225.71 kN

D_A1 = 225.71 – (40×1) = 225.71 – 40 = 185.71 kN

D_A2 = 225.71 – (40×2) = 225.71 – 80 = 145.71 kN

D_A3 = 225.71 – (40×3) = 225.71 – 120 = 105.71 kN

D_A4 = 225.71 – (40×4) = 225.71 – 160 = 65.71 kN

D_A5 = 225.71 – (40×5) = 225.71 – 200 = 25.71 kN

D_A6 = 225.71 – (40×6) = 225.71 – 240 = –14.29 kN

D_A7 = 225.71 – (40×7) = 225.71 – 280 = –54.29 kN

D_A8 = 225.71 – (40×8) = 225.71 – 320 = –94.29 kN

D_C = R_AV – (40×8) = 225.71 – 320 = –94.29 kN

D_D = D_C – P₁ = –94.29 – 20 = –114.29 kN

D_E = D_D – P₂ = –114.29 – 20 = –134.29 kN

D_B = D_E + R_BV = –134.29 + 214.29 = 80 kN

D_F = D_B – (q₂×L₂) = 80 – (20×4) = 0 kN

 

Menghitung Bidang Momen

M_A = R_AV × L = 225.71 × 0 = 0 kN.m

L_MAX = R_AV/q₁ = 225.71/40 = 5.64 m

M_MAX = (R_AV×L_MAX) – (1/2×q₁×L_MAX²) = (225.71×5.64) – (1/2×40×5.64²) = 636.81 kN.m

M_C = (225.71×8) – (40×8×4) = 525.68 kN.m

M_D = (225.71×10) – (40×8×(4+2)) – (20×0) = 337.1 kN.m

M_E = (225.71×12) – (40×8×(4+4)) – (20×2) – (20×0) = 108.52 kN.m

M_B = (225.71×14) – (40×8×(4+6)) – (20×4) – (20×2) + (R_BV×0) = –160 kN.m

M_F = (225.71×18) – (40×8×(4+10)) – (20×8) – (20×6) + (R_BV×4) – (20×4×2) = 0 kN.m