#### Calculate the reactions at the supports of a beam

A beam is in equilibrium when it is stationary relative to an inertial reference frame. The following conditions are satisfied when a beam, acted upon by a system of forces and moments, is in equilibrium:

1. The fixed support is located at point A (on the left). A fixed support will resist translation in all directions and rotation (moment) - **H**_{A}, R_{A}, M_{A}.

2. The sum of the forces and moment about any point is zero: **ΣF**_{x} = 0, ΣF_{y} = 0, ΣM_{A} = 0.

**ΣF**_{x} = 0: H_{A} = 0

**ΣF**_{y} = 0: R_{A} - q_{1}*1.8 - P_{1} = 0;

**ΣM**_{A} = 0: M_{A} - q_{1}*1.8*(1.8/2) + M_{1} - 3*P_{1} = 0;

3. Solve this system of equations:

H_{A} = 0 (kN)

R_{A} = q_{1}*1.8 + P_{1} = 2*1.8 + 7 = 10.60 (kN)

M_{A} = q_{1}*1.8*(1.8/2) - M_{1} + 3*P_{1} = 2*1.8*(1.8/2) - 19 + 3*7 = 5.24 (kN*m)

4. Verification equation of equilibrium about the point B (on the right):

- 3*R_{A} + M_{A} + q_{1}*1.8*(1.2+1.8/2) + M_{1} + 0*P_{1} = - 3*10.60 + 5.24 + 2*1.8*(1.2+1.8/2) + 19.00 + 0*7 = 0

#### Draw diagrams for the beam

**First span of the beam: 0 ≤ x**_{1} < 1.8Determine the equations for the shear force (Q):

Q(x_{1}) = + R_{A} - q_{1}*(x_{1} - 0)

Q_{1}(0) = + 10.60 - 2*(0 - 0) = 10.60 (kN)

Determine the equations for the bending moment (M):

M(x_{1}) = + R_{A}*(x_{1}) - M_{A} - q_{1}*(x_{1})^{2}/2

M_{1}(0) = + 10.60*(0) - 5.24 - 2*(0 - 0)^{2}/2 = -5.24 (kN*m)

**Second span of the beam: 1.8 ≤ x**_{2} < 2.4Determine the equations for the shear force (Q):

Q(x_{2}) = + R_{A} - q_{1}*(1.8 - 0)

Q_{2}(1.80) = + 10.60 - 2*(1.8 - 0) = 7 (kN)

Q_{2}(2.40) = + 10.60 - 2*(1.8 - 0) = 7 (kN)

Determine the equations for the bending moment (M):

M(x_{2}) = + R_{A}*(x_{2}) - M_{A} - q_{1}*(1.8 - 0)***[**(x_{2} - 1.80) + (1.80 - 0)/2**]**

M_{2}(1.80) = + 10.60*(1.80) - 5.24 - 2*1.8*(0 + 0.90) = 10.60 (kN*m)

M_{2}(2.40) = + 10.60*(2.40) - 5.24 - 2*1.8*(0.60 + 0.90) = 14.80 (kN*m)

**Third span of the beam: 2.4 ≤ x**_{3} < 3Determine the equations for the shear force (Q):

Q(x_{3}) = + R_{A} - q_{1}*(1.8 - 0)

Q_{3}(2.40) = + 10.60 - 2*(1.8 - 0) = 7 (kN)

Q_{3}(3) = + 10.60 - 2*(1.8 - 0) = 7 (kN)

Determine the equations for the bending moment (M):

M(x_{3}) = + R_{A}*(x_{3}) - M_{A} - q_{1}*(1.8 - 0)***[**(x_{3} - 1.80) + (1.80 - 0)/2**]** - M_{1}

M_{3}(2.40) = + 10.60*(2.40) - 5.24 - 2*1.8*(0.60 + 0.90) - 19 = -4.20 (kN*m)

M_{3}(3) = + 10.60*(3) - 5.24 - 2*1.8*(1.20 + 0.90) - 19 = 0 (kN*m)

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