Wyznaczenie reakcji H_A , V_A , V_B , H_B

1.

Σ M_A=0

-F∙1 + M + Q∙7,5 + Q_x∙8 + Q_y∙12 - V_B∙10 + H_B∙1 = 0

-10∙1 + 30 + 100∙7,5 + 80∙8 + 80∙12 - V_B∙10 + H_B∙1 = 0

-10 + 30 + 750 + 640 + 960 - V_B∙10 + H_B∙1 = 0

- V_B∙10 + H_B∙1 + 2370 = 0

H_B = -2370 + 10V_B

Σ M^II_c = 0

Q∙3,5 + Q_x∙2 + Q_y∙8 - V_B∙6 - H_B∙5 = 0

100∙3,5 + 80∙2 + 80∙8 - V_B∙6 - H_B∙5 = 0

350 + 160 + 640 - V_B∙6 - H_B∙5 = 0

- V_B∙6 - H_B∙5 + 1150 = 0

6V_B = 1150 - 5H_B

6V_B = 1150 - 5(-2370 + 10V_B)

6V_B = 1150 + 11850 - 50V_B

56V_B = 13000

V_B = 232,1428 kN

H_B = -2370 + 10(232,1428)

H_B = -48,572 kN

2.

Σ M_B=0

-H_A + V_A∙10 + M - Q∙2,5 + Q_x∙7 + Q_y∙2 = 0

-H_A + V_A∙10 + 30 - 100∙2,5 + 80∙7 + 80∙2 = 0

-H_A + V_A∙10 + 30 - 250 + 560 + 160 = 0

H_A = 500 + 10V_A

Σ M^I_c = 0

V_A∙4 - H_A∙6 + F∙5 + M = 0

V_A∙4 - H_A∙6 + 10∙5 + 30 = 0

V_A∙4 - H_A∙6 + 80 = 0

4V_A = -80 + 6H_A

4V_A = -80 + 6(500 + 10V_A)

4V_A = -80 + 3000 + 60V_A

-56 V_A = 2920

V_A = -52,1428 kN

H_A = 500 + 10(-52,1428)

H_A = 500 - 521,428

H_A = -21,428 kN

Sprawdzenie:

ΣF_x=0

H_A - F + H_B + Q_x = 0

-21,428 - 10 - 48,572 + 80 = 0

0 = 0

ΣF_y=0

V_A - Q_y + V_B - Q = 0

-52,1428 - 100 + 232,1428 - 80 = 0

0 = 0

Obliczenie reakcji V_C i H_C w przegubie

ΣF^II_y=0

V_C - Q - Q_y + V_B = 0

V_C - 100 - 80 + 232,1428 = 0

V_C = -52,1428 kN

ΣF^II_x=0

H_C + H_B + 80 = 0

H_C - 48,572 + 80 = 0

H_C = 31,428 kN

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