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Calculate reactions, shear forces, bending moments, and deflections for simply supported beams with various load configurations including point loads, distributed loads, and trapezoidal loading.
Example: Steel W8x24 = 82.8 in⁴
Steel: 29,000,000 psi | Aluminum: 10,100,000 psi
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A simply supported beam has pinned and roller supports at both ends, allowing rotation but no vertical displacement. Reactions are calculated from equilibrium equations.
| Load Case | RA | RB | Max Moment | Max Deflection |
|---|---|---|---|---|
| Center Point Load P | P/2 | P/2 | PL/4 | PL³/(48EI) |
| Off-Center Load P at a | Pb/L | Pa/L | Pab/L | Pb(L²-b²)/(9√3 EI) |
| Two Point Loads | (P₁b₁+P₂b₂)/L | (P₁a₁+P₂a₂)/L | Max at loads | Varies |
| UDL w over length | wL/2 | wL/2 | wL²/8 | 5wL⁴/(384EI) |
| Partial UDL | w(end-start)/L | w(end-start)/L | At centroid | Varies |
| Trapezoidal Load | (W·c)/L | (W·(L-c))/L | Variable | Numerical |
Where L = span, a = distance from left support, b = distance from right support, w = load per unit length, W = total load
The principle of superposition allows us to solve complex loading by breaking it into simple load cases and combining results.
Break complex load into multiple simple cases (point loads, UDL, etc.)
Use formulas for each simple load case separately
Add reactions, shears, and moments algebraically at each point
Determine maximum positive/negative values and locations
Given: 20 ft span, 2000 lb load at center
Solution:
Given: 20 ft span, 1500 lb load at 8 ft from left
Solution:
Given: 20 ft span, 200 lbs/ft across entire length
Solution:
A simply supported beam is supported at two points (ends) with a pin support on one side and a roller support on the other. It can freely rotate at supports but cannot translate vertically, making it statically determinate.
For standard steel sections (W, C, I beams), refer to steel manuals or online tables. For custom sections, use I = ∫y²dA. Online calculators can compute I for rectangles, circles, and composite shapes quickly.
Center loading (point load at midspan) creates equal reactions and maximum moment at center. Off-center loading creates unequal reactions and shifts the maximum moment location toward the heavier side.
Use superposition when you have complex loads combining point loads, distributed loads, and trapezoidal loads. Break into simple cases, solve each, and add the results algebraically.
Shear sign convention: positive shear acts downward on left face. Moving from left to right, reactions create positive shear; downward loads create negative shear. Shear changes sign where concentrated loads occur.
Building codes typically specify: L/240 for roofs, L/360 for floors, L/480 for heavily used floors or industrial. Check local codes. Excessive deflection causes cracked drywall and sticking doors.
No, this is specifically for simply supported beams with pin and roller supports. Cantilever beams (fixed at one end) have different boundary conditions and require different formulas.
These calculations follow standard engineering theory for idealized beams with linear elastic material behavior. Real beams may have residual stresses, imperfections, and material nonlinearity. Always verify with professional analysis for critical applications.
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