REAKCJE PODPOROWE
$$\sum_{}^{}{M_{A} = 0\ \ \ \text{\ \ \ }R_{B}*2a + P\cos{\varphi*\frac{3}{2}a - P\sin{\varphi*\frac{3}{2}a - q*\frac{a^{2}}{2}}}} = 0$$
$$R_{B} = q\frac{a}{2} = 2kN$$
$$\sum_{}^{}{M_{B} = 0\ \ \ \text{\ \ \ }R_{A}*2a - P\cos{\varphi*\frac{3}{2}a - P\sin{\varphi*\frac{1}{2}a + q*\frac{a^{2}}{2} = 0}}}$$
$$R_{A} = P\sin{\varphi - q*\frac{a}{2}} = 1,536kN$$
$$\sum_{}^{}{M_{2}^{L} = 0\ \ \ \text{\ \ \ }R_{A}*a - H_{A}*2a - q*a*\frac{3}{2}a = 0}$$
$$H_{A} = \frac{R_{A}}{2} - q*a*\frac{3}{2}a = - 5,232kN$$
$$\sum_{}^{}{M_{2}^{P} = 0\ \ \ \text{\ \ \ P}\sin{\varphi*\frac{a}{2} + P\cos{\varphi*\frac{a}{2} + H_{B}*2a} - R_{B}*a = 0}}$$
$$H_{B} = \frac{R_{B}}{2} - \frac{P}{2}\sin\varphi = - 0,768\text{kN}$$
SIŁY WEWNĘTRZNE
B – 3 0<x1<4m
N (x1) = -RB = -1,536 kN
T (x1) = -HB = 0,768 kN
M(x1) = -HB*x1
M(x1=0) =0
M(x1=4 m) =3,072 kNm
3 – 4 0<x2<2$\sqrt{\mathbf{2}}$m
N (x2) = -RBsinϕ-HBcosϕ = -0,871 kN
T (x2) = RBcosϕ-HBsinϕ=1,957 kN
M(x2) = -HBsinϕ(x2+2$\sqrt{2}$) -HBcosϕ(2$\sqrt{2}$)+RBcosϕ(x2+2$\sqrt{2}$) -RBsinϕ(2$\sqrt{2}$)
M(x2) = x2[(RB-HB)sinϕ]- 4$\sqrt{2}$HBcosϕ
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) = -RBsinϕ-HBcosϕ = -0,871 kN
T (x2) = RBcosϕ-HBsinϕ-P= 3,043 kN
M(x2) = -HBsinϕ(x2+2$\sqrt{2}$) -HBcosϕ(2$\sqrt{2}$)+RBcosϕ(x2+2$\sqrt{2}$) -RBsinϕ(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) = -RA = -1,536 kN
T (x3) = -HA-q*x3
T (x3=0) = 5,232 kN
T (x3=4 m) = -2,768 kN
M(x3) = -HA* 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) = -RAcosϕ-HAsinϕ-q*a*sinϕ = 3,043 kN
T(x4) = RAsinϕ-HAcosϕ-q*a*cosϕ = 0,871 kN
M(x4) = RAsinϕ(x4+2$\sqrt{2}$)-RAcosϕ(2$\sqrt{2}$)-HA(x4+2$\sqrt{2}$)-HAsinϕ(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
RÓWNANIA OSI ŁUKU
$$y = \frac{4fx}{l^{2}} = x(1 - \frac{x}{10m})$$
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 |