1. Zadanie 1
1.1 Równanie linii wpływu reakcji
x ∈ <0 , 9>
ΣMB= MA(x) + 1(6-x) = 0
MA(x) = x-6
ΣY = -1 + RB (x) → RB (x) =1
x ∈ <9 , 15>
ΣML = RC (x) • 3 – 1(x-9)=0
RC (x)=$\ \frac{1}{3}$ (x-9)
ΣMC = -3 SL (x) • 3 +1(12-x)=0
SL (x)=$\text{\ \ }\frac{1}{3}$ (12-x)
ΣMB = -3 SL (x) + MA(x)=0
MA(x) = 12-x
ΣYDOL= RB (x) −SL (x) = 0
RB (x) =$\text{\ \ }\frac{1}{3}$ (12-x)
MA(x)= $\left\{ \begin{matrix} x - 6\ ;x\epsilon\ \left\langle 0,9 \right\rangle\ \\ 12 - x;x\epsilon\ \left\langle 9,15 \right\rangle \\ \frac{3}{2}\left( x - 17 \right);x\epsilon\ \left\langle 15,19 \right\rangle \\ \end{matrix} \right.\ $
RC(x)= $\left\{ \begin{matrix} 0\ ;\ x\epsilon\ \left\langle 0,9 \right\rangle\ \\ \frac{1}{3}(x - 9);\ \ x\epsilon\ \left\langle 9,15 \right\rangle \\ 17 - x\ ;\ \ \ x\epsilon\ \left\langle 15,19 \right\rangle \\ \end{matrix} \right.\ $
x ∈ <15, 19>
ΣMD = -2 SM (x) + 1(17-x) = 0
SM (x) $= \frac{1}{2}$ (17-x)
ΣML = 3RC (x) −6 SM (x)=0
RC (x) = 2 SM (x)
RC (x) = 17-x
ΣMC = SM (x) • (−3) − 3 SL (x) =0
SL (x) = − SM (x)
SL (x)$= - \ \frac{1}{2}$ (17-x)
ΣMB = -3 SL (x) + MA(x)=0
MA(x) = $\frac{3}{2}$ (x-17)
ΣYDOL= RB (x) −SL (x) = 0
RB (x) =$\text{\ \ } - \ \frac{1}{2}$ (17-x)
RB(x)= $\left\{ \begin{matrix} 1;\ x\epsilon\ \left\langle 0,9 \right\rangle\ \\ 4 - \frac{x}{3};\ \ x\epsilon\ \left\langle 9,15 \right\rangle \\ - \frac{1}{2}\ (17 - x)\ ;\ \ \ x\epsilon\ \left\langle 15,19 \right\rangle \\ \end{matrix} \right.\ $
SL(x)= $\left\{ \begin{matrix} 0;\ x\epsilon\ \left\langle 0,9 \right\rangle\ \\ 4 - \frac{x}{3};\ \ x\epsilon\ \left\langle 9,15 \right\rangle \\ - \frac{1}{2}\ (17 - x)\ ;\ \ \ x\epsilon\ \left\langle 15,19 \right\rangle \\ \end{matrix} \right.\ $
1.2 Równania linii wpływu sił przekrojowych w punkcie K
QK
x ∈ <0 , 3>
−QK(x) -1=0
QK(x)= -1
x ∈ <3 , 9>
QK(x) + RB(x) - 1= 0
QK(x) = 1 – 1 = 0
x ∈ <9, 15>
QK(x) + RB(x) - SL (x) =0
QK(x) =$\ 4 - \frac{x}{3}$ – 4 + $\frac{x}{3}$
QK(x) = 0
x ∈ <15 , 19>
QK(x) + RB(x) - SL (x) =0
QK(x) =$- \ \frac{1}{2}$ (17-x) $+ \ \frac{1}{2}$ (17-x)
QK(x) = 0
MK
x ∈ <0 , 3>
MK(x) + MA(x) +1(3-x)=0
MK(x) = 6 - x – 3 + x
MK(x) = 3
x ∈ <3 , 9>
−MK(x) + 3RB(x) - (x - 3) = 0
MK(x) = 3 – x +3
MK(x) = -x +6
x ∈ <9 , 15>
−MK(x) + 3RB(x) -6 SL (x)=0
MK(x) = 3$(4 - \frac{x}{3}$) - 6$(4 - \frac{x}{3}$)
MK(x) = -3$(4 - \frac{x}{3}$)
x ∈ <15 , 19>
−MK(x) + 3RB(x) -6 SL (x)=0
$M_{K}\left( x \right) = - \ \frac{3}{2}$ (17-x) + 3(17-x)
$M_{K}\left( x \right) = \frac{3}{2}$ (17-x)
1.3 Ekstremalne momenty względem punktu K dla zadanego obciążenia
MK(+)= 20 • 3 +70• 3 +50 • 2,5 = 395
MK(−)= 70•(−3) + 20• (-1,5) +50•(−1) = −290
2. Zadanie 2
2.1 Linie wpływu dla kratownicy
2.1 Reakcje
x ∈ <0 , 16>
ΣY = VA (x) - 1 =0
VA (x) = 1
ΣMB= 4HA - 1• x = 0
HA(x) = $\ \frac{x}{4}$
ΣMA=−4HA- x•1
HB(x) = $- \frac{x}{4}$
2.2 LwN1x ∈ <0 , 4>
ΣMKL = 4HA - 4VA + 1(x-4) + N1(x) = 0
4N1 (x)$= 4 - 4 \bullet \frac{x}{4}$ - 4 + x
N1 (x) = 0
x ∈ <8 , 16>
ΣMK P= 4N1 (x) + 1(x-4)=0
N1 (x) = -0,25(x-4)
2.3 LwN2
x ∈ <0 , 4>
ΣY = VA (x) - 1 - $\frac{\sqrt{2}}{2} \bullet \text{\ N}_{2}$ (x) = 0
N2 (x) = 0
x ∈ <8, 16>
ΣY = $\ \frac{\sqrt{2}}{2}\ $ • N2 (x) – 1 = 0
N2 (x) = $\sqrt{2}$