1. REAKCJE PODPOROWE

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

$$R_{B} = q\frac{a}{4} = 2kN$$

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

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

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

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

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

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

1. SIŁY WEWNĘTRZNE

B – 4 0<x₁<4m

N (x₁) = -R_B = -2 kN

T (x₁) = H_B = -0,768 kN

M(x₁) = -H_B*x₁

M(x₁=0) =0

M(x₁=4 m) =3,072 kNm

4 – 3 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

3 – 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

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

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	45	38,66	30,96	21,80	11,31	0	-11,31	-21,80	-30,96	-38,66	-45
sinϕ	0,7071	0,6247	0,5144	0,3714	0,1961	0	-0,1961	-0,3714	-0,5144	-0,6247	-0,7071
cosϕ	0,7071	0,7809	0,8572	0,9285	0,9806	1	0,9806	0,9285	0,8572	0,7809	0,7071

1. REAKCJE PODPOROWE

$$\sum_{}^{}{M_{B} = 0\ \text{\ \ \ }R_{A}*l + q*f*\frac{f}{2} + \frac{1}{2}*\left( 2q - q \right)*f*\frac{f}{3} = 0}$$

$$R_{A} = - \frac{q*f^{2}*(\frac{1}{2} + \frac{1}{6})}{l} = - 0,833kN$$

$$\sum_{}^{}{y = 0\ \text{\ \ }R_{A} + R_{B} = 0}$$

R_B = −R_A = 0, 833kN

$$\sum_{}^{}{M_{C}^{P} = 0\ \ \text{\ \ }R_{B}*\frac{l}{2}} - H_{B}*f = 0$$

$$H_{B} = R_{B}*\frac{l}{2*f} = 1,667kN$$

$$\sum_{}^{}{x = 0\ \ \ \ q*f + \frac{1}{2}*\left( 2q - q \right)*f + H_{A} - H_{B} = 0}$$

$$H_{A} = H_{B} - \frac{3}{2}*q*f = - 5,833kN$$

1. RÓWNANIA SIŁ WEWNETRZNYCH

A-C 0<x<5m

Wyznaczenie ekstremum

C-B 5m<x<10m

Wyznaczenie ekstremum

1. WARTOŚCI SIŁ WEWNĘTRZNYCH

x	0	1	2	3	4	5	6	7	8	9	10
N	4,713	2,517	0,820	-0,436	-1,271	-1,667	-1,798	-1,857	-1,857	-1,822	-1,768
T	3,535	0,947	-0,479	-1,072	-1,104	-0,833	-0,490	-0,154	0,143	0,391	0,590
M	0	2,894	3,093	2,165	0,990	0	-0,659	-0,998	-0,998	-0,664	0