1. OBLICZENIE REAKCJI:

ΣM_A = 0 $+ 4*2,5*\left( 3 + \frac{5}{4} \right) - 30 - \frac{15}{\sqrt{10}}*7 - V_{E}*6 + \frac{45}{\sqrt{10}}*3 = 0$

$$V_{E} = \frac{\frac{85}{2} - 30 - \frac{105}{\sqrt{10}} + \frac{135}{\sqrt{10}}}{6} = \frac{25 + 6\sqrt{10}}{12} \approx 3,6645$$

ΣY = 0 $+ V_{A} + V_{E} - 4*2,5 + \frac{15}{\sqrt{10}} = 0$

$$V_{A} = - \frac{25 + 6\sqrt{10}}{12} + 10 - \frac{15}{\sqrt{10}} = \frac{95 - 24\sqrt{10}}{12} \approx 1,5921$$

ΣX = 0 +$H_{A} - \frac{45}{\sqrt{10}} = 0$

$$H_{A} = \frac{45}{\sqrt{10}} = \frac{45\sqrt{10}}{10} = \frac{9\sqrt{10}}{2} \approx 14,2302$$

SPR.

ΣM_B = 0 $+ 4*2,5*\frac{5}{4} - 30 + V_{A}*3 + \frac{45}{\sqrt{10}}*3 - \frac{15}{\sqrt{10}}*4 - V_{E}*3 = 0$

$$\frac{50}{4} - 30 + \frac{95 - 24\sqrt{10}}{4} + \frac{135}{\sqrt{10}} - \frac{60}{\sqrt{10}} - \frac{25 + 6\sqrt{10}}{4} = 0$$

0 = 0

1. RÓWNANIA DO MOMENTÓW ZGINAJĄCYCH, SIŁ TNĄCYCH I NORMALNYCH:

x₁ ∈ <0, 3> M(x1) = +V_A * x

M(0) = 0 

$$M\left( 3 \right) = {+ V}_{A}*3 = \frac{95 - 24\sqrt{10}}{12}*3 = \frac{95 - 24\sqrt{10}}{4} \approx 4,7763$$

$$T\left( x1 \right) = {+ V}_{A} = \frac{95 - 24\sqrt{10}}{12} \approx 1,5921$$

$$N\left( x1 \right) = - H_{A} = - \frac{9\sqrt{10}}{2} \approx - 14,2302$$

x₂ ∈ <0, 3> M(x2) = 0

T(x2) = 0

N(x2) = 0

x₃ ∈ <0, 3> M(x3) = −30

T(x3) = 0

N(x3) = 0

x₄ ∈ <0, 2.5> $M\left( x4 \right) = + V_{A}*\left( 3 + x \right) - 30 - 4*\frac{x^{2}}{2}$

$$M\left( 0 \right) = \frac{95 - 24\sqrt{10}}{4} - 30 = \frac{- 25 - 24\sqrt{10}}{4} \approx - 25,2237$$

$$M\left( 2,5 \right) = \frac{95 - 24\sqrt{10}}{12}*5,5 - 30 - 4*\frac{{2,5}^{2}}{2} \approx - 33,7434$$

T(x4) = +V_A − 4x

$$T\left( 0 \right) = \frac{95 - 24\sqrt{10}}{12} \approx 1,5921$$

$$T\left( 2,5 \right) = \frac{95 - 24\sqrt{10}}{12} - 10 = \frac{- 25 - 24\sqrt{10}}{12} \approx - 8,4079$$

$$N\left( x4 \right) = - H_{A} = - \frac{9\sqrt{10}}{2} \approx - 14,2302$$

x₅ ∈ <0, 2.5> $M\left( x5 \right) = + V_{A}*\left( 5,5 + x \right) - 30 - 4*2,5*\left( \frac{5}{4} + x \right)$

$$M\left( 0 \right) = \frac{95 - 24\sqrt{10}}{12}*\frac{11}{2} - 30 - 10*\frac{5}{4} \approx 33,7434$$

$$M\left( 2,5 \right) = \frac{95 - 24\sqrt{10}}{12}*8 - 30 - 10*\left( \frac{5}{4} + 2,5 \right) = \frac{- 25 - 96\sqrt{10}}{6} \approx - 54,7631$$

$$T\left( x5 \right) = + V_{A} - 4*2,5 = \frac{95 - 24\sqrt{10}}{12} - 10 = \frac{- 25 - 24\sqrt{10}}{12} \approx - 8,4079$$

$$N\left( x5 \right) = - H_{A} = - \frac{9\sqrt{10}}{2} \approx - 14,2302$$

$x_{6} \in < 0,\sqrt{10} >$ M(x6) = −V_E * sin ∝ *x

M(0) = 0

$$M\left( \sqrt{10} \right) = - \frac{25 + 6\sqrt{10}}{12}*\frac{1}{\sqrt{10}}*\sqrt{10} = \frac{- 25 - 6\sqrt{10}}{12} \approx - 3,6645$$

$${T\left( x6 \right) = V_{E}*sin \propto = \frac{25 + 6\sqrt{10}}{12}*\frac{1}{\sqrt{10}} = \frac{12 + 5\sqrt{10}}{24} \approx 1,1588\backslash n}{N\left( x6 \right) = - V_{E}*cos \propto = - \frac{25 + 6\sqrt{10}}{12}*\frac{3}{\sqrt{10}} = - \frac{12 + 5\sqrt{10}}{8} \approx - 3,4764}$$

$x_{7} \in < 0,\sqrt{10} >$ $M\left( x7 \right) = {- V}_{E}*sin \propto - 15*x = - \frac{25 + 6\sqrt{10}}{12}*\frac{1}{\sqrt{10}}*\left( \sqrt{10} + x \right) - 15*x$

$$M\left( 0 \right) = \frac{- 25 - 6\sqrt{10}}{12} \approx - 3,6645$$

$$M\left( \sqrt{10} \right) = - \frac{25 + 6\sqrt{10}}{12\sqrt{10}}*2\sqrt{10} - 15\sqrt{10} = \frac{- 25 - 96\sqrt{10}}{6} \approx - 54,7631$$

$$T\left( x7 \right) = V_{E}*sin \propto + 15 = \frac{25 + 6\sqrt{10}}{12}*\frac{1}{\sqrt{10}} + 15 = \frac{372 + 5\sqrt{10}}{24} \approx 16,1588$$

$$N\left( x7 \right) = - V_{E}*cos \propto = - \frac{25 + 6\sqrt{10}}{12}*\frac{3}{\sqrt{10}} = - \frac{12 + 5\sqrt{10}}{8} \approx - 3,4764$$

OBLICZENIE EKSTREMUM W PRZEDZIALE X₄:

T(x4) = 0

+V_A − 4x = 0

$x = \frac{V_{A}}{4}$

$$x = \frac{95 - 24\sqrt{10}}{48} \approx 0,3980$$

$$M\left( x4 = \frac{95 - 24\sqrt{10}}{48} \right) = + V_{A}*\left( 3 + x \right) - 30 - 4*\frac{x^{2}}{2}$$

$M\left( x4 \right) = + V_{A}*\left( 3 + \frac{95 - 24\sqrt{10}}{48} \right) - 30 - 4*\frac{\frac{95 - 24\sqrt{10}}{48}^{2}}{2}$

M(x4) = −24, 9068 EKSTREMUM MINIMALNE

1. OBLICZENIE REAKCJI:

ΣM_A = 0 −x_D1 * 3 − V_E * 6 = 0

$$V_{E} = \frac{- 3}{6} = - \frac{1}{2}$$

ΣX = 0 H_A = 0

ΣY = 0 V_A + V_E + 1 = 0

$$V_{A} = - 1 + \frac{1}{2} = - \frac{1}{2}$$

1. RÓWNANIA DO MOMENTÓW ZGINAJĄCYCH, SIŁ TNĄCYCH I NORMALNYCH:

x₁ ∈ <0, 3> M(x1) = +V_A * x

M(0) = 0 

$$M\left( 3 \right) = {+ V}_{A}*3 = - \frac{1}{2}*3 = - \frac{3}{2}$$

$$T\left( x1 \right) = + V_{A} = - \frac{1}{2}$$

N(x1) = −H_A = 0

x₂ ∈ <0, 6> M(x2) = 0

T(x2) = 0

N(x2) = −1

x₃ ∈ <0, 5> M(x3) = +V_A * (3+x) + 1 * x

$$M\left( 0 \right) = - \frac{1}{2}*3 = - \frac{3}{2}\ $$

$$M\left( 5 \right) = - \frac{1}{2}*8 + 1*5 = - 4 + 5 = 1$$

$$T\left( x3 \right) = {+ V}_{A} + 1 = - \frac{1}{2} + 1 = \frac{1}{2}$$

N(x3) = 0

$x_{4} \in < 0,2\sqrt{10} >$ $M\left( x4 \right) = {- V}_{E}*sin\alpha*x = \frac{1}{2}*\frac{1}{\sqrt{10}}*x$

M(0) = 0

$$M\left( 2\sqrt{10} \right) = \frac{1}{2\sqrt{10}}*2\sqrt{10} = 1$$

$$T\left( x4 \right) = V_{E}*sin\alpha = - \frac{1}{2}*\frac{1}{\sqrt{10}} = \frac{- \sqrt{10}}{20} \approx - 0,1581$$

$$N\left( x4 \right) = {- V}_{E}*cos\alpha = + \frac{1}{2}*\frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{20} \approx 0,4743$$

1. OBLICZENIE REAKCJI:

ΣM_A = 0 −x_D2 * 6 − V_E * 6 = 0

$$V_{E} = \frac{- 6}{6} = - 1$$

ΣX = 0 H_A + 1 = 0

H_A = −1

ΣY = 0 V_A + V_E = 0

V_A = 1

1. RÓWNANIA DO MOMENTÓW ZGINAJĄCYCH, SIŁ TNĄCYCH I NORMALNYCH:

x₁ ∈ <0, 3> M(x1) = +V_A * x

M(0) = 0 

M(3) = +V_A * 3 = 1 * 3 = 3

T(x1) = +V_A = 1

N(x1) = −H_A = 1

x₂ ∈ <0, 6> M(x2) = −1 * x

M(0) = 0 

M(6) = −6

T(x2) = −1

N(x2) = 0

x₃ ∈ <0, 5> M(x3) = +V_A * (3+x) − 1 * 6

M(0) = +3 − 6 = −3 

M(5) = +8 − 6 = 2

T(x3) = +V_A = 1

N(x3) = −H_A − 1 = +1 − 1 = 0

$x_{4} \in < 0,2\sqrt{10} >$ $M\left( x4 \right) = {- V}_{E}*sin\alpha*x = 1*\frac{1}{\sqrt{10}}*x$

M(0) = 0

$$M\left( 2\sqrt{10} \right) = \frac{1}{\sqrt{10}}*2\sqrt{10} = 2$$

$$T\left( x4 \right) = V_{E}*sin\alpha = - 1*\frac{1}{\sqrt{10}} = \frac{- \sqrt{10}}{10} \approx - 0,3162$$

$$N\left( x4 \right) = - V_{E}*cos\alpha = + 1*\frac{3}{\sqrt{10}} = \frac{3\sqrt{10}}{10} \approx 0,9487$$

1. OBLICZENIE REAKCJI:

ΣM_A = 0 +x_D3 − V_E * 6 = 0

$$V_{E} = \frac{1}{6}$$

ΣX = 0 H_A = 0

ΣY = 0 V_A + V_E = 0

$$V_{A} = - \frac{1}{6}$$

1. RÓWNANIA DO MOMENTÓW ZGINAJĄCYCH, SIŁ TNĄCYCH I NORMALNYCH:

x₁ ∈ <0, 3> M(x1) = +V_A * x

M(0) = 0 

$$M\left( 3 \right) = {+ V}_{A}*3 = - \frac{1}{6}*3 = - \frac{1}{2}$$

$$T\left( x1 \right) = + V_{A} = - \frac{1}{6}$$

N(x1) = −H_A = 0

x₂ ∈ <0, 6> M(x2) = +1

T(x2) = 0

N(x2) = 0

x₃ ∈ <0, 5> M(x3) = +V_A * (3+x) + 1

$$M\left( 0 \right) = - \frac{1}{6}*3 + 1 = \frac{1}{2}$$

$$M\left( 5 \right) = - \frac{1}{6}*8 + 1 = - \frac{8}{6} + 1 = - \frac{4}{3} + 1 = - \frac{1}{3}$$

$$T\left( x3 \right) = + V_{A} = - \frac{1}{6}$$

N(x3) = −H_A = 0

$x_{4} \in < 0,2\sqrt{10} >$ $M\left( x4 \right) = {- V}_{E}*sin\alpha*x = - \frac{1}{6}*\frac{1}{\sqrt{10}}*x$

M(0) = 0

$$M\left( 2\sqrt{10} \right) = - \frac{1}{6\sqrt{10}}*2\sqrt{10} = - \frac{1}{3}$$

$$T\left( x4 \right) = V_{E}*sin\alpha = \frac{1}{6}*\frac{1}{\sqrt{10}} = \frac{\sqrt{10}}{60} \approx 0,0527$$

$$N\left( x4 \right) = - V_{E}*cos\alpha = - \frac{1}{6}*\frac{3}{\sqrt{10}} = - \frac{\sqrt{10}}{20} \approx - 0,1581$$