Płaskownik

B x h = 40x10mm; siła P, hipoteza T max; kg=200MPa;

$W = \frac{bh^{2}}{6}$; J$= \frac{bh^{3}}{12}$

E Fix=0 Rax=0; E Fiy=0 -P – P + Ray = 0 Ray= 2P

EMia=0; -P*L - $\frac{P*L}{2} + \text{Ma} = 0$; Ma=$\frac{3}{2}\text{PL}$

$\sigma_{\text{zred}} = \sqrt{\sigma_{\text{zg}}^{2} + 4T^{2}}$ ; $\sigma_{\text{ZG}} = \frac{\text{Mg}\ \max}{W} = \frac{6*\frac{3}{2}P*L}{bh^{2}}$ ;

$\sigma_{\text{ZG}} = \frac{\text{Mg}\ \max}{W} = \frac{6*\frac{3}{2}P*L}{bh^{2}}$ ; $\sigma_{\text{ZG}} = \frac{9*P*L}{4*1^{2}}$

$T = \frac{T*S}{b*J} = \frac{2P*b*\frac{h}{2}*\frac{h}{4}*12}{b*bh^{3}}$ ; $T = \frac{3P*bh^{2}}{b^{2}h^{3}} = \frac{3P}{b*h}$

$$\sigma_{\text{zred}} = \sqrt{\left( \frac{9*PL}{bh^{2}} \right)^{2} + 4\left( \frac{3P}{\text{bh}} \right)^{2} \leq \text{kg}}$$

Wyliczyć P … P <= jakaś wartość

= max dopuszczalna siła

Stalowa belka

b/h=3 obl. Wymiary i ugięcie kg=100MPa

$$\sigma = \frac{M}{W} \leq \text{kg}\text{\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ }W = \frac{bh^{2}}{6}$$

$$\frac{10000000}{100} \leq \frac{bh^{2}}{6}\ ;\ 6000 \leq 3h^{3}$$

$$\frac{600000}{3} \leq h^{3}\ ;h \geq \sqrt[3]{200000}\ ;h \geq 58,5\ \text{mm}$$

b = 3 * 58, 5 = 175, 5 mm

$\text{EJ} = \frac{d^{2}y}{dx^{2}} = - M\ ;\text{EJ}\frac{\text{dy}}{\text{dx}} = \ \int_{}^{}{- Mdx = \text{Mx} + C1}$

$$\text{EJ}\frac{\text{dy}}{\text{dx}} = \ \int_{}^{}{- M + C1\text{dx} = \frac{- Mx^{2}}{2} + C1x + D1}$$

Z warunków podparcia

$$y\left( l \right) = 0\ ;\left( \frac{\text{dy}}{\text{dx}} \right)l = 0$$

0 = −Ml + C1 = = > C1 = Ml

$$\text{EJy} = \frac{- Mx^{2}}{2} + \frac{2Ml^{2}}{2} + D1 = = > D1 = \frac{- Ml^{2}}{2}$$

$$\text{EJy} = \frac{- Mlx^{2}}{2} + Mlx - \frac{Ml^{2}}{2}$$

Strzałka ugięcia $y = - \frac{Ml^{2}}{2\text{EJ}}$

Średnica wału

P=300kW n=1200obr/min kr=100MPa Mg=100kNm

1) średnica Tmax<=kt ; kt=0,58 kn kt=58Mpa

$$T = \frac{\text{Ms}}{\text{Wo}}\ \ ;\text{Ms} = 9550*\frac{P}{n}\ ;\text{Ms} = 9550*\frac{300}{1200}$$

$$\text{Ms} = 2387,5\ \text{Nm}\ ;\text{Wo} = \frac{\pi D^{3}}{16}$$

$$\frac{16*Ms}{\pi D^{3}} \leq \text{kt}\ ;\ \sqrt[3]{\frac{16*Ms}{\text{πkt}}} \leq D$$

2) naprężenie zredukowane

$$\text{σzred} = \sqrt{\sigma^{2} + 3T^{2}} \leq \text{kr}\ ;\ \sigma = \frac{\text{Mg}}{W}\ ;\text{Mg} = 100\text{kNm}$$

$W = \frac{\pi D^{3}}{32}$ ; $\text{σzred} = \sqrt{\left( \frac{32*Mg}{\pi D^{3}} \right)^{2} + 3*\left( \frac{16*Ms}{\pi D^{3}} \right)^{2}} \leq \text{kr}\ $

Belka dwuteowa – wytrzymałość i ugięcie

Q=10kN/m ; kg=150MPa ; l=2m

$$0 \leq x \leq l;\text{Mg} = q*x*\frac{x}{2}$$

$$\text{Mgmax} - g*l*\frac{l}{2} = 10000*2 = 20000Nm = 20000000Nmm$$

$$W = \frac{\left( BH^{3} - bh^{3} \right)}{6H} = = >$$

$$W = \frac{80*120^{3} - 60*100^{3}}{6*120} = 108667mm^{3}$$

$$\frac{\text{Mgmax}}{W} \leq \text{kg}\ ;\frac{20000000}{108667} \leq 150\ ;184 \leq 150$$

Warunek niespełniony

$$\text{EJ}\frac{d^{2}y}{dx^{2}} = - q*\frac{x^{2}}{2}\ ;\text{EJ}\frac{\text{dy}}{\text{dx}} = \int_{}^{}{- q\frac{x^{2}}{2}\text{dx} = - q\frac{x^{3}}{6} + C1}$$

$$\text{EJy} = \int_{}^{}{- q\frac{x^{3}}{6} + C1 = - q\frac{x^{3}}{24} + C1x + D}$$

Z warunków podparcia $Y\left( l \right) = 0\ ;\left( \frac{\text{dy}}{\text{dx}} \right)l = 0$

$$0 = - q\frac{l^{3}}{6} + C1 = > C1 = q\frac{l^{3}}{6}\ ;\text{EJy} = - q\frac{x^{4}}{24} + q\frac{l^{3}}{6}x + D1$$

$$0 = - q\frac{l^{4}}{24} + q\frac{l^{3}}{6} + D1 = > D1 = \frac{ql^{4}}{8}$$

$$\text{EJy} = \frac{{- qx}^{4}}{24} + \frac{\text{ql}^{3}}{6}x + \frac{ql^{4}}{8}\ ;x = 0\ ;f = \frac{ql^{4}}{8\text{EJ}}$$

Belka utw. 1 końcem

B=1,5m ; P=15kN=1500N; h/b=3

$\frac{h}{b} = 3 = > b = \frac{h}{3}$ ; a=1m b=1,5m

$$q = 15\frac{\text{kN}}{m} = 15000\frac{N}{m}\ ;J = 1,45*10^{- 5}\ m^{4} = 14500000\ m^{4}$$

$$\text{Mg}\max{= P*a + q*b\left( a + \frac{b}{2} \right) = 15000*1 + 15000*}$$

*1, 5 * (1+0,75) = 54375Nm = 54375000Nmm

$J = \frac{bh^{3}}{12}$ ; $12*1,45*10^{- 5} = \frac{h^{4}}{3}\ ;h = \sqrt[4]{3*12*1,45*10^{- 5}}$

$$h = 0,15m = 150\text{mm}\ ;\ \sigma_{g = \frac{\text{Mgmax}}{J}*\frac{3}{8}h = \frac{54375000}{12500000}*\frac{3}{8}*150 = 211\text{MPa}}$$

Pręt – obl. naprężenie i wydłużenie

A=6cm E=2*10^5MPa; dz=0,015m=15mm

D2=0,025m=25mm; N=0,5T=4905N

σ₁ = ε₁E ; $\sigma_{1} = \frac{N}{S1}\ ;\ \varepsilon_{1} = \frac{l1}{l1}\ ;\ l1 = \frac{N*l1}{\text{ES}1}$

$$\sigma_{2} = \varepsilon_{2}E\ ;\ \sigma_{2} = \frac{N}{S2}\ ;\ \varepsilon_{2} = \frac{l2}{l2}\ ;\ l2 = \frac{N*l2}{\text{ES}2}$$

$$lc = l1 + l2 = \frac{N*l1}{E*S1} + \frac{N*l2}{E*S2} = \frac{4905*6}{2*10^{5}*490,6} =$$

$$= \frac{29430}{353,2*10^{5}} + \frac{29430}{981,2*10^{5}} = 0,001\text{cm}$$

Pręt – 2 momenty

A=100 cm; $\frac{M1}{M2} = 1,5\ ;d = 0,04m;G = 0,85*10^{5}\text{MPa}$

Re = 250MPa ; n = 1, 5 ;  φ dop = 0, 2 = 0, 003rad

$$k = \frac{\text{Re}}{n} = \frac{250}{1,5} = 166\text{MPa};w_{0} = \frac{\pi d^{3}}{16};J_{0} = \frac{\pi d^{4}}{32}$$

0 ≤ x1 ≤ a ; Msx1 = M1 = 1, 5M2

a ≤ x2 ≤ 2a; Msx2 = M1 − M2 = 1, 5M2 − M2 = 0, 5M2

Msmax = 1, 5M2 ; $\tau = \frac{\text{Msmax}}{W0} \leq k;\text{Msmax} \leq k*w_{0}$

$$M_{s\ \max} \leq 166*\frac{\pi*40^{3}}{16};\ M_{s\ \max} \leq 2084,96\text{Nm} = M1$$

$= > M2 = \frac{2084,96}{1,5} = 1389,9\text{Nm}$;

$$\varphi_{\max} = \frac{M_{\max}*l}{G*J_{0}} \leq \varphi_{\text{dop}\ ;\ }M_{\max} \leq \frac{\varphi_{dop*G*J_{0}\ }}{l}$$

$$M_{\max} \leq \frac{0,003*0,85*10^{5}*\pi*40^{4}}{32*1000};$$

M_max ≤ 64056Nm = M1 = >M2 = 42704Nm

Obc. Dop. To M1=2084Nm i M2=1389Nm

Z I koła Gr A - wymiary belki o przekroju prostokątnym i

obl jej ugięcie

A=10cm, b=24cm, c=8cm, L=150cm, P=400kN

$$\text{σzred} = \ \sqrt{\left( \sigma_{R} + \sigma_{\text{ZG}} \right)^{2} + 3\tau^{2}}\ ;\ \text{σR} = \frac{N}{A}\ ;\ $$

$$\text{σZG} = \frac{\text{Mg}}{J_{0}}*r\ ;\ \tau = \frac{T*S}{a*J0} = \frac{P*S}{b*J0}\ \text{gdzie}$$

$J_{0} = \frac{b^{3}*a}{12}$ ; $r = c + \frac{b}{4}\ ;s = \left( \frac{b}{2} - c \right)a*r;$

$$\text{σzred} \leq \text{kdop} = = > \frac{\text{podstawic}}{\text{wyliczyc}}i\ \text{spr}\ \text{warunek}$$

Gr B belka o przekroju kołowym

D=10cm, L=1000mm, M=10kNm, kdop=120N/mm^2

$$\text{σzred} = \ \sqrt{\left( \sigma_{R} + \sigma_{\text{ZG}} \right)^{2} + 4\tau^{2}} \leq \text{kdop}$$

$$\sigma_{R} = \frac{4P}{\pi d^{2}}\ ;\ \sigma_{\text{ZG}} = \frac{\text{Mg}}{W}\ \text{gdzie}\ \text{Mg} = 2P*L\ ;W = \frac{\text{πd}^{3}}{32}$$

$$\sigma_{\text{ZG}} = \frac{2PL*32}{\text{πd}^{3}}\ ;\ \tau = \frac{\text{Ms}}{W_{0}} = \frac{M*16}{\text{πd}^{3}}\text{\ \ }\text{bo}\ w_{0} = \frac{\pi d^{3}}{16}$$

Podstawić I wyliczyć