ĆWICZENIE PROJEKTOWE NR 3

TEMAT:

DLA UKŁADU RAMOWEGO PODPARTEGO I OBCIĄŻONEGO JAK NA RYSUNKU

WYZNACZYĆ REAKCJE PODPÓR

I SPORZĄDZIĆ WYKRESY SIŁ WEWNĘTRZNYCH (N, T, M)

Magdalena Duszyńska

Gr II

Rok II

Semestr III

Rok Akad. 2010/2011

G=2 O=2 N=9 I=9

P = 25 kN M=25 kNm q=22 kN/m

∑MiB^L = 0 → V_A * 2 – P *4 – q * $\sqrt{8}$ * $\frac{\sqrt{8}}{2}$ - M = 0

V_A = 106,5 KN

∑Pix = 0 → - H_E + P – q * 2 + q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ = 0

H_E = 25 KN

∑Mic^P = 0 → V_E * 9 + H_E * 9 – M = 0

V_E = 22,222 KN

∑Piy = 0 → V_A - V_D + V_E – P – q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ = 0

V_D = 59,7 KN

∑Mic^L = 0 → - M_D -25 + V_A * 4 – q *2 * 1 – q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ * ($\frac{\sqrt{8}}{2}$ + 2) + q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ * ($\frac{\sqrt{8}}{2}$ + 2) = 0

M_D = 357,00 KNm

SPRAWDZENIE:

∑MiP = 0 → V_A * 1 + V_D * 3 - V_E * 12 + H_E * 12 – 2 * M – P * 3 + P * 3 + q * 2 * 2 - M_D = 0

0 = 0

X₁ ϵ (0;9)

T_(X) = 0 KN = T₍₀₎ = T₍₉₎

N_{(X )}= - V_A = - 106,5 KN = N₍₀₎= N_{(9 )}

M_(x) = 0 KNm = M₍₀₎ = M₍₉₎

X₂ ϵ (0;2)

T_(X) =- 25 KN = T₍₀₎ = T₍₂₎

N_{(X )}= - V_A = - 106,5 KN = N₍₀₎= N_{(2 )}

M_(x) = - P * x

M₍₀₎ = 0 KNm

M₍₂₎ = - 50 KNm

X₃ ϵ (0;$\ \sqrt{8}$ )

T_(X) = V_A * $\frac{2}{\sqrt{8}}$ - P * $\frac{2}{\sqrt{8}}$ - q * x

T₍₀₎ = 57,629 KN

T_{( 8 )} = - 4,596 KN

N_{(X )}= - V_A * $\frac{2}{\sqrt{8}}$ - P * $\frac{2}{\sqrt{8}}$

N₍₀₎= N_{( 8 )} = - 92,98 KN

M_(x) = V_A * x * $\frac{2}{\sqrt{8}}$ - P * ( 2 + x * $\frac{2}{\sqrt{8}}$ ) – M – q * x * $\frac{x}{2}$

M₍₀₎ = - 75 KNm

M_{( 8 )} = 0 KNm

X₄ ϵ (0; 2 )

T_(X) = V_A - q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$

T₍₀₎ = T_{(2) =} 62,5 KN

N_{(X )}= - P - q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$

N₍₀₎= N_{( 8 )} = - 69,00 KN

M_(x) = V_A * (2 + x) - P * 4 – M – q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ * (1 + x) - q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ * 1

M₍₀₎ = 0 KNm

M_{( 2)} = 125 KNm

X₅ ϵ (0; 2 )

T_(X) = P + q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$

T₍₀₎ = T_{(2) =} 69,00 KN

N_{(X )}= V_A - P - q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$

N₍₀₎= N_{( 8 )} = 37,5 KN

M_(x) = V_A * 4 - P *( 4 – x ) – M – q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ * (-1 + x) - q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ * 3

M₍₀₎ = 125 KNm

M_{( 2)} = 263 KNm

X₆ ϵ (0; 2 )

T_(X) = P + q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ - q * x

T₍₀₎ = 69 KN

T_{(2) =} 25 KN

N_{(X )}= V_A - P - q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$

N₍₀₎= N_{(2 )} = 37,5 KN

M_(x) = V_A * 4 - P *( 2 – x ) – M – q * $\frac{x}{2}$ * x - q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ * 3 + q * $\frac{2}{\sqrt{8}}$ * $\sqrt{8}$ * (1 + x)

M₍₀₎ = 263 KNm

M_{( 2)} = 357 KNm

X₇ ϵ (0; 9 )

T_(X) = 0 KN = T₍₀₎ = T₍₉₎

N_{(X )}= V_D = 59,7 KN = N₍₀₎= N_{(2 )}

M_(x) = M_D

M₍₀₎ = 357 KNm

M_{( 2)} = 357 KNm

X₈ ϵ (0;$\ 9\sqrt{2}$ )

T_(X) = - V_E * $\frac{9}{9\sqrt{2}}$ + H_E * $\frac{9}{9\sqrt{2}}$

T₍₀₎ = T_{(9 2 )} = 1,98 KN

N_{(X )}= - V_E * $\frac{9}{9\sqrt{2}}$ - H_E * $\frac{9}{9\sqrt{2}}$

N₍₀₎= N_{(9 2 )} = 33,75 KN

M_(x) = - V_E * $\frac{9}{9\sqrt{2}}$ * x - H_E * $\frac{9}{9\sqrt{2}}$ * x + 25

M₍₀₎ = 25 KNm

M_{( 9 2 )} = 0 KNm