Calculate the reactions at the supports of a beam
1. 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. ΣFx = 0: HA = 0 ΣMA = 0: The sum of the moments about the pin support at the point A: - P1*1 - q1*2*(2 + 2/2) + RB*5 = 0 ΣMB = 0: The sum of the moments about the roller support at the point B: - RA*5 + P1*4 + q1*2*(3 - 2/2) = 0 2. Calculate reaction of roller support at the point B: RB = ( P1*1 + q1*2*(2 + 2/2)) / 5 = ( 10*1 + 5*2*(2 + 2/2)) / 5 = 8.00 (kN) 3. Calculate reaction of pin support at the point A: RA = ( P1*4 + q1*2*(3 - 2/2)) / 5 = ( 10*4 + 5*2*(3 - 2/2)) / 5 = 12.00 (kN) 4. Solve this system of equations: HA = 0 (kN) 5. The sum of the forces about the Oy axis is zero: ΣFy = 0: RA - P1 - q1*2 + RB = 12.00*1 - 10 - 5*2 + 8.00*1 = 0
Draw diagrams for the beam
Consider first span of the beam 0 ≤ x1 < 1
Determine the equations for the shear force (Q): Q(x1) = + RA The values of Q at the edges of the span: Q1(0) = + 12 = 12 (kN) Q1(1) = + 12 = 12 (kN) Determine the equations for the bending moment (M): M(x1) = + RA*(x1) The values of M at the edges of the span: M1(0) = + 12*(0) = 0 (kN*m) M1(1) = + 12*(1) = 12 (kN*m)
Consider second span of the beam 1 ≤ x2 < 2
Determine the equations for the shear force (Q): Q(x2) = + RA - P1 The values of Q at the edges of the span: Q2(1) = + 12 - 10 = 2 (kN) Q2(2) = + 12 - 10 = 2 (kN) Determine the equations for the bending moment (M): M(x2) = + RA*(x2) - P1*(x2 - 1) The values of M at the edges of the span: M2(1) = + 12*(1) - 10*(1 - 1) = 12 (kN*m) M2(2) = + 12*(2) - 10*(2 - 1) = 14 (kN*m)
Consider third span of the beam 2 ≤ x3 < 4
Determine the equations for the shear force (Q): Q(x3) = + RA - P1 - q1*(x3 - 2) The values of Q at the edges of the span: Q3(2) = + 12 - 10 - 5*(2 - 2) = 2 (kN) Q3(4) = + 12 - 10 - 5*(4 - 2) = -8 (kN) The value of Q on this span that crosses the horizontal axis. Intersection point: x = 0.40 Determine the equations for the bending moment (M): M(x3) = + RA*(x3) - P1*(x3 - 1) - q1*(x3 - 2)2/2 The values of M at the edges of the span: M3(2) = + 12*(2) - 10*(2 - 1) - 5*(2 - 2)2/2 = 14 (kN*m) M3(4) = + 12*(4) - 10*(4 - 1) - 5*(4 - 2)2/2 = 8 (kN*m) Local extremum at the point x = 0.40: M3(2.40) = + 12*(2.40) - 10*(2.40 - 1) - 5*(2.40 - 2)2/2 = 14.40 (kN*m)
Consider fourth span of the beam 4 ≤ x4 < 5
Determine the equations for the shear force (Q): Q(x4) = + RA - P1 - q1*(4 - 2) The values of Q at the edges of the span: Q4(4) = + 12 - 10 - 5*(4 - 2) = -8 (kN) Q4(5) = + 12 - 10 - 5*(4 - 2) = -8 (kN) Determine the equations for the bending moment (M): M(x4) = + RA*(x4) - P1*(x4 - 1) - q1*(4 - 2)*[(x4 - 4) + (4 - 2)/2] The values of M at the edges of the span: M4(4) = + 12*(4) - 10*(4 - 1) - 5*2*(0 + 1) = 8 (kN*m) M4(5) = + 12*(5) - 10*(5 - 1) - 5*2*(1 + 1) = 0 (kN*m)
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1.) For the beam below draw the BMD and SFD diagram using the "method of sections!!...