Beam Deflection by Superposition

Assumptions

• Euler–Bernoulli beam theory: linear elastic, prismatic EI, small deflections, shear deformation neglected.
• Sign conventions: x from the left end; deflection positive UPWARD (downward gravity loads give negative deflections); slope is dy/dx in radians; applied moments positive counterclockwise.
• Simply supported: pin and roller at the two ends of the span; deflection is zero at both supports.

Solution steps

1. Flexural rigidity

   All handbook deflection formulas share the beam stiffness EI in the denominator.

   flexural rigidity EI = E × I

   EI = 10000 kN·m^2

2. Simply supported beam with a point load

   Handbook closed-form case; its deflection adds to the other loads by superposition.

   y = −Pbx(L² − b² − x²)/(6·L·EI) for x ≤ a (mirror form beyond the load)

   Contribution at x = 3 m: y = -4.5 mm

3. Maximum deflection

   For a single table case the extremum location follows from the closed form (where the slope is zero); the value below matches it.

   for a load right of center (a ≥ b): xmax = √((L² − b²)/3), on the longer segment

   δmax = -4.5 mm at x = 3 m (downward)

   = -4.5 mm

4. End slopes

   The support rotations bound the elastic curve; useful for checking serviceability and for slope-deflection methods.

   θA and θB from the slope function

   θA = -0.00225 rad; θB = 0.00225 rad

Results

Assumptions

• Euler–Bernoulli beam theory: linear elastic, prismatic EI, small deflections, shear deformation neglected.
• Sign conventions: x from the left end; deflection positive UPWARD (downward gravity loads give negative deflections); slope is dy/dx in radians; applied moments positive counterclockwise.
• Simply supported: pin and roller at the two ends of the span; deflection is zero at both supports.

Solution steps

1. Flexural rigidity

   All handbook deflection formulas share the beam stiffness EI in the denominator.

   flexural rigidity EI = E × I

   EI = 120800 kip·ft^2

2. Simply supported beam with a full-span uniform load

   Handbook closed-form case; its deflection adds to the other loads by superposition.

   y = −wx(L³ − 2Lx² + x³)/(24·EI); δmax = 5wL⁴/(384·EI) at midspan

   Contribution at x = 10 ft: y = -0.4138 in

3. Maximum deflection

   For a single table case the extremum location follows from the closed form (where the slope is zero); the value below matches it.

   slope = 0 at the extremum (or at a free end)

   δmax = -0.4138 in at x = 10 ft (downward)

   = -0.4138 in

4. End slopes

   The support rotations bound the elastic curve; useful for checking serviceability and for slope-deflection methods.

   θA and θB from the slope function

   θA = -0.005517 rad; θB = 0.005517 rad

Results

Assumptions

• Euler–Bernoulli beam theory: linear elastic, prismatic EI, small deflections, shear deformation neglected.
• Sign conventions: x from the left end; deflection positive UPWARD (downward gravity loads give negative deflections); slope is dy/dx in radians; applied moments positive counterclockwise.
• Cantilever fixed at the left end, free at the other.

Solution steps

1. Flexural rigidity

   All handbook deflection formulas share the beam stiffness EI in the denominator.

   flexural rigidity EI = E × I

   EI = 2800 kN·m^2

2. Cantilever with a point load

   Handbook closed-form case; its deflection adds to the other loads by superposition.

   y = −Px²(3a − x)/(6·EI) for x ≤ a; beyond the load the beam is straight; δtip = −Pa²(3L − a)/(6·EI)

   Contribution at x = 2 m: y = -4.762 mm

3. Maximum deflection

   For a single table case the extremum location follows from the closed form (where the slope is zero); the value below matches it.

   slope = 0 at the extremum (or at a free end)

   δmax = -4.762 mm at x = 2 m (downward)

   = -4.762 mm

4. Free-end deflection and slope

   The cantilever's tip carries the accumulated deflection and rotation.

   tip deflection and tip slope

   δtip = -4.762 mm; θtip = -0.003571 rad

Results