1.1. Wyznaczenie wartości reakcji w belce

Belka 1

1. S P_y = 0

R₁ + R_c – W₁ + P = 0

R₁ + R_c – 8 + 8 = 0

R₁ = - R_c

2. S P_x = 0

3. S M₁ = 0

W₁*2 – R_c*4 – P*8 = 0

8*2 – R_c*4 – 8*8 = 0

16 - R_c*4 -64 = 0

4*R_c = -48

R_c = -12 => R₁ = 12

Belka 2

1. S P_y = 0

R_A + R_B – W₂ – R₁ = 0

R_A + R_B – 24 -12 = 0

R_A + R_B = 36 => R_A = 36 – R_B = 16

2. S P_x = 0

H_A = 0

3. S M_A = 0

6*R_B + 12 – R₁*9 – W₂*1 = 0

6*R_B + 12 – 108 – 24 = 0

6*R_B = 120

R_B = 20

1.2. Wyznaczenie wartości sił tnących i momentów zginających

Belka 1

1. 0 < x < 4

q(x) = x₁/2

T(x) = -x/2*x + 12 = x²/2

T^L(0) = 12

T^P(4) = 4

M(x) = 12*x – x³/4

M^L(0) = 0

M^P(4) = 32

2. 0 < x₁ < 4

T(x₁) = -8

M(x₁) = 8*x₁

M^L(0) = 0

M^P(0) = 32

Belka 2

1. 0 < x < 3

q(x) = x²/6

T(x) = 12 + x²/6

T^L(0) = 12

T^P(3) = 13,5

M(x) = -12*x – x2/6*x/3 = -12x – x³/18

M^L(0) = 0

M^P(3) = -37,5

2. 3 < x₁ < 6

T(x₁) = -8 + x₁²/6

T^L(3) = -6,5

T^P(6) = -2

M(x₁) = -12*x₁ + 20(x₁ – 3) – x₁³/18

M^L(3) = -37,5

M^P(6) = -24

3. 6 < x₂ <9

T(x₂) = -8 + x₂²/6

T^L(6) = -2

T^P(9) = 5,5

M(x2) = 8*x2 – x₂³/18 – 48

M^L(6) = -12

M^P(9) = -16,5

4. 9 < x₃ < 12

T(x₃) = -24 + x₃²/6

T^L(9) = -10,5

T^P(12) = 0

M(x₃) = 24*x₃ – 192 – x₃³/18

M^L(9) = -16,5

M^P(12) = 0

1.3. Sprawdzenie linii wpływu

R_B = 24*1/6 + 12 * 5/6 + 8*3/4 = 20

2.1. Obliczenie wartości reakcji w ramie

Rama 1

S P_x = 0

H_A – ql = 0 => H_A = ql

S P_y = 0

R_A + R_B = 3/2ql => R_A = 3/2ql – 5/4ql = 1/4ql

S M_A = 0

ql² + 9/8ql² – 1/4ql² – 3/2R_B = 0

15/8ql – 3/2R_B = 0

R_B = 5/4ql

Rama 2

S M₁^L = 0

ql² -2ql² – H – 1/8ql² +1/8ql² = 0

H = -ql

2.2. Wyznaczenie wartości sił tnących, normalnych i momentów zginających

A – C 0 < y < 2l

T(y) = ql

N(y) = -1/4ql

M(y) = ql*y – 2ql*(y -1)

M^L(0) = 2ql²

M^P(2l) = 0

C – 1 0 < x < ½

T(x) = 1/4ql – qx

T^L(0) = 1/4ql

T^P(1/2) = -1/4ql

T(1/4) = 0

N(x) = - ql

M(x) = 1/4ql*x – 1/2qx²

M^L(0) = 0

M^P(1/2) = 0

1 – D 0 < x₁ < 1

T(x₁) = -5/4ql + qx1

T^L(0) = -5/4ql

T^P(1) = -1/4ql

N(x₁) = -ql

M(x₁) = 5/4ql²x₁ – 3/4ql² – 1/2qx₁²

M^L(0) = - 3/4ql²

M^P(1) = 0

E – D 0 < y₁ < 2l

T(y₁) = ql

N(y₁) = -5/4ql

M(y₁) = ql*(y₁ – 1) + 1/4ql²

M^L(0) = -3/4ql²

M^P(2_l) = 5/4ql²