Визначити опорні реакції балки, якщо $q=10$ кН/м, $Р=10$ кН $АС = 3$ м, $АВ = 5$ м і кут $ф=30°$ .

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Q. Визначити опорні реакції балки, якщо

q=10

кН/м,

Р=10

кН

АС = 3

м,

АВ = 5

м і кут

ф=30°

Calculate total distributed load: First, we need to calculate the total distributed load on the beam due to the uniform load $q$ . The total load will be $q$ multiplied by the length of the beam $AB$ .
Total distributed load $(Q) = q \times AB$
$Q = 10 \, \text{kN/m} \times 5 \, \text{m}$
Resolve point load components: Calculate the total distributed load. $Q = 50$ kN
Calculate vertical and horizontal components: Next, we need to resolve the point load $P$ into horizontal and vertical components because it is applied at an angle $\varphi$ . The vertical component ( $P_v$ ) will be $P \times \cos(\varphi)$ and the horizontal component ( $P_h$ ) will be $P \times \sin(\varphi)$ . $\newline$ $P_v = P \times \cos(\varphi)$ $\newline$ $P_h = P \times \sin(\varphi)$
Calculate reactions at supports: Calculate the vertical and horizontal components of the point load $P$ . $P_v = 10 \, \text{kN} \cdot \cos(30°)$ $P_h = 10 \, \text{kN} \cdot \sin(30°)$
Substitute values into equilibrium equation: Using the trigonometric values, $\cos(30°) = \sqrt{3}/2$ and $\sin(30°) = 1/2$ , calculate $P_v$ and $P_h$ . $\newline$ $P_v = 10 \, \text{kN} \times \sqrt{3}/2$ $\newline$ $P_h = 10 \, \text{kN} \times 1/2$
Consider moments around support A: Calculate the actual values for $P_v$ and $P_h$ . $\newline$ $P_v = 10 \, \text{kN} \times \sqrt{3}/2 \approx 8.66 \, \text{kN}$ $\newline$ $P_h = 10 \, \text{kN} \times 1/2 = 5 \, \text{kN}$
Simplify equation to find $R_b$ : Now, we will calculate the reactions at the supports. Let's denote the reaction at support A as $R_a$ and at support B as $R_b$ . Since the beam is in static equilibrium, the sum of vertical forces must be zero. $\newline$ $\sum F_y = 0 = R_a + R_b - Q - P_v$
Calculate reaction at support B: Substitute the known values into the equilibrium equation for vertical forces. $\newline$ $0 = R_a + R_b - 50\,\text{kN} - 8.66\,\text{kN}$
Use sum of vertical reactions for $Ra$ : Simplify the equation to find the sum of the reactions $Ra$ and $Rb$ . $\newline$ $Ra + Rb = 58.66 \, \text{kN}$
Calculate reaction at support A: To find the individual reactions, we also need to consider the moments around one of the supports. Let's take moments around support A (counterclockwise positive). $\newline$ $\sum M_A = 0 = R_b \cdot AB - Q \cdot (\frac{AB}{2}) - P_v \cdot AC$
Calculate reaction at support A: To find the individual reactions, we also need to consider the moments around one of the supports. Let's take moments around support A (counterclockwise positive). $\sum M_A = 0 = R_b \cdot AB - Q \cdot (AB/2) - P_v \cdot AC$ Substitute the known values into the moment equilibrium equation around point A. $0 = R_b \cdot 5 \, \text{m} - 50 \, \text{kN} \cdot (5 \, \text{m}/2) - 8.66 \, \text{kN} \cdot 3 \, \text{m}$
Calculate reaction at support A: To find the individual reactions, we also need to consider the moments around one of the supports. Let's take moments around support A (counterclockwise positive). $\newline$ $\sum M_A = 0 = R_b \cdot AB - Q \cdot (AB/2) - P_v \cdot AC$ Substitute the known values into the moment equilibrium equation around point A. $\newline$ $0 = R_b \cdot 5 \, \text{m} - 50 \, \text{kN} \cdot (5 \, \text{m}/2) - 8.66 \, \text{kN} \cdot 3 \, \text{m}$ Calculate the moment due to the distributed load and the point load. $\newline$ $0 = R_b \cdot 5 \, \text{m} - 125 \, \text{kN} \cdot \text{m} - 25.98 \, \text{kN} \cdot \text{m}$
Calculate reaction at support A: To find the individual reactions, we also need to consider the moments around one of the supports. Let's take moments around support A (counterclockwise positive). $\newline$ $\sum M_A = 0 = R_b \cdot AB - Q \cdot (AB/2) - P_v \cdot AC$ Substitute the known values into the moment equilibrium equation around point A. $\newline$ $0 = R_b \cdot 5\,\text{m} - 50\,\text{kN} \cdot (5\,\text{m}/2) - 8.66\,\text{kN} \cdot 3\,\text{m}$ Calculate the moment due to the distributed load and the point load. $\newline$ $0 = R_b \cdot 5\,\text{m} - 125\,\text{kN}\cdot\text{m} - 25.98\,\text{kN}\cdot\text{m}$ Simplify the equation to solve for Rb. $\newline$ $R_b \cdot 5\,\text{m} = 150.98\,\text{kN}\cdot\text{m}$ $\newline$ $R_b = \frac{150.98\,\text{kN}\cdot\text{m}}{5\,\text{m}}$
Calculate reaction at support A: To find the individual reactions, we also need to consider the moments around one of the supports. Let's take moments around support A (counterclockwise positive). $\newline$ $\sum M_A = 0 = R_b \cdot AB - Q \cdot (AB/2) - P_v \cdot AC$ Substitute the known values into the moment equilibrium equation around point A. $\newline$ $0 = R_b \cdot 5 \, \text{m} - 50 \, \text{kN} \cdot (5 \, \text{m}/2) - 8.66 \, \text{kN} \cdot 3 \, \text{m}$ Calculate the moment due to the distributed load and the point load. $\newline$ $0 = R_b \cdot 5 \, \text{m} - 125 \, \text{kN} \cdot \text{m} - 25.98 \, \text{kN} \cdot \text{m}$ Simplify the equation to solve for $R_b$ . $\newline$ $R_b \cdot 5 \, \text{m} = 150.98 \, \text{kN} \cdot \text{m}$ $\newline$ $R_b = \frac{150.98 \, \text{kN} \cdot \text{m}}{5 \, \text{m}}$ Calculate the reaction at support B ( $R_b$ ). $\newline$ $R_b = 30.196 \, \text{kN}$
Calculate reaction at support A: To find the individual reactions, we also need to consider the moments around one of the supports. Let's take moments around support A (counterclockwise positive). $\newline$ $\sum M_A = 0 = R_b \cdot AB - Q \cdot (AB/2) - P_v \cdot AC$ Substitute the known values into the moment equilibrium equation around point A. $\newline$ $0 = R_b \cdot 5 \, \text{m} - 50 \, \text{kN} \cdot (5 \, \text{m}/2) - 8.66 \, \text{kN} \cdot 3 \, \text{m}$ Calculate the moment due to the distributed load and the point load. $\newline$ $0 = R_b \cdot 5 \, \text{m} - 125 \, \text{kN} \cdot \text{m} - 25.98 \, \text{kN} \cdot \text{m}$ Simplify the equation to solve for $R_b$ . $\newline$ $R_b \cdot 5 \, \text{m} = 150.98 \, \text{kN} \cdot \text{m}$ $\newline$ $R_b = \frac{150.98 \, \text{kN} \cdot \text{m}}{5 \, \text{m}}$ Calculate the reaction at support B ( $R_b$ ). $\newline$ $R_b = 30.196 \, \text{kN}$ Now, use the sum of the vertical reactions to find $R_a$ . $\newline$ $R_a + 30.196 \, \text{kN} = 58.66 \, \text{kN}$ $\newline$ $0 = R_b \cdot 5 \, \text{m} - 50 \, \text{kN} \cdot (5 \, \text{m}/2) - 8.66 \, \text{kN} \cdot 3 \, \text{m}$ $0$
Calculate reaction at support A: To find the individual reactions, we also need to consider the moments around one of the supports. Let's take moments around support A (counterclockwise positive). $\newline$ $\sum M_A = 0 = R_b \cdot AB - Q \cdot (AB/2) - P_v \cdot AC$ Substitute the known values into the moment equilibrium equation around point A. $\newline$ $0 = R_b \cdot 5 \, \text{m} - 50 \, \text{kN} \cdot (5 \, \text{m}/2) - 8.66 \, \text{kN} \cdot 3 \, \text{m}$ Calculate the moment due to the distributed load and the point load. $\newline$ $0 = R_b \cdot 5 \, \text{m} - 125 \, \text{kN} \cdot \text{m} - 25.98 \, \text{kN} \cdot \text{m}$ Simplify the equation to solve for $R_b$ . $\newline$ $R_b \cdot 5 \, \text{m} = 150.98 \, \text{kN} \cdot \text{m}$ $\newline$ $R_b = \frac{150.98 \, \text{kN} \cdot \text{m}}{5 \, \text{m}}$ Calculate the reaction at support B ( $R_b$ ). $\newline$ $R_b = 30.196 \, \text{kN}$ Now, use the sum of the vertical reactions to find $R_a$ . $\newline$ $R_a + 30.196 \, \text{kN} = 58.66 \, \text{kN}$ $\newline$ $0 = R_b \cdot 5 \, \text{m} - 50 \, \text{kN} \cdot (5 \, \text{m}/2) - 8.66 \, \text{kN} \cdot 3 \, \text{m}$ $0$ Calculate the reaction at support A ( $R_a$ ). $\newline$ $0 = R_b \cdot 5 \, \text{m} - 50 \, \text{kN} \cdot (5 \, \text{m}/2) - 8.66 \, \text{kN} \cdot 3 \, \text{m}$ $2$