1. REAKCJE PODPOROWE

$$\sum_{}^{}{M_{A} = 0\ \ \ \text{\ \ \ }R_{B}*2a + P\cos{\varphi*\frac{3}{2}a - P\sin{\varphi*\frac{3}{2}a - q*\frac{a^{2}}{2}}}} = 0$$

$$R_{B} = q\frac{a}{2} = 2kN$$

$$\sum_{}^{}{M_{B} = 0\ \ \ \text{\ \ \ }R_{A}*2a - P\cos{\varphi*\frac{3}{2}a - P\sin{\varphi*\frac{1}{2}a + q*\frac{a^{2}}{2} = 0}}}$$

$$R_{A} = P\sin{\varphi - q*\frac{a}{2}} = 1,536kN$$

$$\sum_{}^{}{M_{2}^{L} = 0\ \ \ \text{\ \ \ }R_{A}*a - H_{A}*2a - q*a*\frac{3}{2}a = 0}$$

$$H_{A} = \frac{R_{A}}{2} - q*a*\frac{3}{2}a = - 5,232kN$$

$$\sum_{}^{}{M_{2}^{P} = 0\ \ \ \text{\ \ \ P}\sin{\varphi*\frac{a}{2} + P\cos{\varphi*\frac{a}{2} + H_{B}*2a} - R_{B}*a = 0}}$$

$$H_{B} = \frac{R_{B}}{2} - \frac{P}{2}\sin\varphi = - 0,768\text{kN}$$

1. SIŁY WEWNĘTRZNE

B – 3 0<x₁<4m

N (x₁) = -R_B = -1,536 kN

T (x₁) = -H_B = 0,768 kN

M(x₁) = -H_B*x₁

M(x₁=0) =0

M(x₁=4 m) =3,072 kNm

3 – 4 0<x₂<2$\sqrt{\mathbf{2}}$m

N (x₂) = -R_Bsinϕ-H_Bcosϕ = -0,871 kN

T (x₂) = R_Bcosϕ-H_Bsinϕ=1,957 kN

M(x₂) = -H_B­sinϕ(x₂+2$\sqrt{2}$) -H_B­cosϕ(2$\sqrt{2}$)+R_B­cosϕ(x₂+2$\sqrt{2}$) -R_B­sinϕ(2$\sqrt{2}$)

M(x₂) = x₂[(R_B-H_B)sinϕ]- 4$\sqrt{2}$H_Bcosϕ

M(x₂=0) = 3,072 kNm

M(x₂=2$\sqrt{2}$ m) = 8,608 kNm

4 – 2 2$\sqrt{\mathbf{2}}$m<x₂<4$\sqrt{\mathbf{2}}$m

N(x₂) = -R_Bsinϕ-H_Bcosϕ = -0,871 kN

T (x₂) = R_Bcosϕ-H_Bsinϕ-P= 3,043 kN

M(x₂) = -H_B­sinϕ(x₂+2$\sqrt{2}$) -H_B­cosϕ(2$\sqrt{2}$)+R_B­cosϕ(x₂+2$\sqrt{2}$) -R_B­sinϕ(2$\sqrt{2}$)-P(x₂-2$\sqrt{2}$)

M(x₂=2$\sqrt{2}$ m) = 8,608 kNm

M(x₂=4$\sqrt{2}$ m) = 0

A – 1 0<x₃<4m

N(x₃) = -R_A = -1,536 kN

T (x₃) = -H_A-q*x₃

T (x₃=0) = 5,232 kN

T (x₃=4 m) = -2,768 kN

M(x₃) = -H_A* x₃-q*$\frac{x_{3}^{2}}{2}$

Wyznaczenie ekstremum:

M(x₃)’= 5,232kN -2kN/m *x₃=0 → x₃ = 2,616 m

M(x₃=0 m) = 0 kNm

M(x₃=2,616 m) = 6,843 kNm

M(x₃=4 m) = 4,928 kNm

1 – 2 0<x₄<4$\sqrt{\mathbf{2}}$m

N(x₄) = -R_Acosϕ-H_Asinϕ-q*a*sinϕ = 3,043 kN

T(x₄) = R_Asinϕ-H_Acosϕ-q*a*cosϕ = 0,871 kN

M(x₄) = R_Asinϕ(x₄+2$\sqrt{2}$)-R_Acosϕ(2$\sqrt{2}$)-H_A(x₄+2$\sqrt{2}$)-H_Asinϕ(2$\sqrt{2}$)-q*a*cosϕ(x₄+$\sqrt{2}$)-q*a*sinϕ($\sqrt{2}$)

M(x₄=0) = 4,928 kNm

M(x₄=4$\sqrt{2}$m) = 0

1. RÓWNANIA OSI ŁUKU

$$y = \frac{4fx}{l^{2}} = x(1 - \frac{x}{10m})$$

1. PODZIAŁ ŁUKU NA RÓWNE CZĘŚCI ZE WZGLĘDU NA ZMIENNĄ PODSTAWOWĄ

x	0	1	2	3	4	5	6	7	8	9	10
y	0	0,9	1,6	2,1	2,4	2,5	2,4	2,1	1,6	0,9	0
ϕ
sinϕ
cosϕ