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Beam Deflection worksheet

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Solve each problem and show your work. Round answers to 2 decimal places where needed.

1.
Problem 1.
- Beam support: Cantilever (one end fixed)
- Load: Point load
- Span length (L): 10 ft
- Point load (P): 1,500 lbf
- Load position: 50% of span
- Material (Young's modulus E): Concrete: 30 GPa (typical)
- Cross-section: Enter I
- Second moment of area (I): 56 in⁴
- Show deflection in: Millimeters (mm)
- Show moment in: Kilonewton meters (kN·m)
Find: Maximum deflection
Answer
2.
Problem 2.
- Beam support: Simply supported (both ends)
- Load: Point load
- Span length (L): 6 ft
- Point load (P): 200 lbf
- Load position: 25% of span
- Material (Young's modulus E): Aluminum: 70 GPa (typical)
- Cross-section: Solid round
- Diameter (d): 2
- Section dimensions in: Inches (in)
- Show deflection in: Inches (in)
- Show moment in: Pound-force inches (lbf·in)
Find: Maximum deflection
Answer
3.
Problem 3.
- Beam support: Cantilever (one end fixed)
- Load: Uniform load along the span
- Span length (L): 6 ft
- Uniform load (w): 200 lbf/ft
- Material (Young's modulus E): Glass: 65 GPa (typical)
- Cross-section: Solid rectangle
- Width (b): 3
- Depth (h), in the direction of the load: 9.25
- Section dimensions in: Millimeters (mm)
- Show deflection in: Millimeters (mm)
- Show moment in: Pound-force inches (lbf·in)
Find: Maximum deflection
Answer
4.
Problem 4.
- Beam support: Cantilever (one end fixed)
- Load: Point load
- Span length (L): 4 ft
- Point load (P): 1,500 lbf
- Load position: 75% of span
- Material (Young's modulus E): Polystyrene: 3 GPa (typical)
- Cross-section: Solid rectangle
- Width (b): 1.5
- Depth (h), in the direction of the load: 5.5
- Section dimensions in: Inches (in)
- Show deflection in: Inches (in)
- Show moment in: Newton meters (N·m)
Find: Maximum deflection
Answer
5.
Problem 5.
- Beam support: Simply supported (both ends)
- Load: Uniform load along the span
- Span length (L): 6 ft
- Uniform load (w): 200 lbf/ft
- Material (Young's modulus E): Enter E
- Young's modulus (E): 40,000 ksi
- Cross-section: Solid rectangle
- Width (b): 5.5
- Depth (h), in the direction of the load: 11.25
- Section dimensions in: Centimeters (cm)
- Show deflection in: Millimeters (mm)
- Show moment in: Kilonewton meters (kN·m)
Find: Maximum deflection
Answer
6.
Problem 6.
- Beam support: Simply supported (both ends)
- Load: Point load
- Span length (L): 12 ft
- Point load (P): 200 lbf
- Load position: 75% of span
- Material (Young's modulus E): Enter E
- Young's modulus (E): 35,000 ksi
- Cross-section: Solid rectangle
- Width (b): 2
- Depth (h), in the direction of the load: 3.5
- Section dimensions in: Centimeters (cm)
- Show deflection in: Millimeters (mm)
- Show moment in: Pound-force inches (lbf·in)
Find: Maximum deflection
Answer
7.
Problem 7.
- Beam support: Simply supported (both ends)
- Load: Point load
- Span length (L): 16 ft
- Point load (P): 500 lbf
- Load position: 25% of span
- Material (Young's modulus E): Wood (along the grain): 13 GPa (typical)
- Cross-section: Solid round
- Diameter (d): 1.5
- Section dimensions in: Millimeters (mm)
- Show deflection in: Inches (in)
- Show moment in: Pound-force inches (lbf·in)
Find: Maximum deflection
Answer
8.
Problem 8.
- Beam support: Cantilever (one end fixed)
- Load: Point load
- Span length (L): 4 ft
- Point load (P): 200 lbf
- Load position: 50% of span
- Material (Young's modulus E): Aluminum: 70 GPa (typical)
- Cross-section: Solid round
- Diameter (d): 3
- Section dimensions in: Millimeters (mm)
- Show deflection in: Inches (in)
- Show moment in: Newton meters (N·m)
Find: Maximum deflection
Answer
9.
Problem 9.
- Beam support: Cantilever (one end fixed)
- Load: Point load
- Span length (L): 16 ft
- Point load (P): 2,000 lbf
- Load position: 100% of span
- Material (Young's modulus E): Concrete: 30 GPa (typical)
- Cross-section: Solid round
- Diameter (d): 1
- Section dimensions in: Inches (in)
- Show deflection in: Millimeters (mm)
- Show moment in: Kilonewton meters (kN·m)
Find: Maximum deflection
Answer
10.
Problem 10.
- Beam support: Simply supported (both ends)
- Load: Point load
- Span length (L): 6 ft
- Point load (P): 2,000 lbf
- Load position: 100% of span
- Material (Young's modulus E): Enter E
- Young's modulus (E): 25,000 ksi
- Cross-section: Solid rectangle
- Width (b): 3
- Depth (h), in the direction of the load: 7.25
- Section dimensions in: Centimeters (cm)
- Show deflection in: Millimeters (mm)
- Show moment in: Newton meters (N·m)
Find: Maximum deflection
Answer

Answer key: Beam Deflection worksheet

1. Maximum deflection: 28.145 mm
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $10 ft \times 0.3048$ $= 3.048 m$
2. Use the typical stiffness (Young's modulus) for concrete, written in pascals
  $30 GPa$ $= 30,000,000,000 Pa$
3. Change I to m⁴, the standard unit the formula uses (1 in⁴ = 0.000000416231 m⁴)
  $56 in⁴ \times 0.000000416231$ $= 0.000023 m⁴$
4. Change the load to N, the standard unit the formula uses (1 lbf = 4.448221615261 N)
  $1,500 lbf \times 4.448221615261$ $= 6,672.332423 N$
5. Find the biggest bending moment, at the fixed end: load × its distance from that end
  $6,672.332423 \times 1.524$ $= 10,168.634612 N·m$
6. Find how far the free tip sags, which is the most anywhere along a cantilever: δ = P × a² × (3L − a) ÷ (6 × E × I)
  $\frac{6,672.332423 \times {1.524}^{2} \times (3 \times 3.048 - 1.524)}{6 \times 30,000,000,000 \times 0.000023}$ $= 0.0281453842 m$
7. Convert the sag from meters to mm so it's easy to read
  $\frac{0.0281453842 m}{0.001}$ $= 28.14538424 mm$
2. Maximum deflection: 0.1363 in
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $6 ft \times 0.3048$ $= 1.8288 m$
2. Use the typical stiffness (Young's modulus) for aluminum, written in pascals
  $70 GPa$ $= 70,000,000,000 Pa$
3. Work out how well the round bar's shape resists bending (its second moment of area): π × diameter⁴ ÷ 64
  $\frac{3.14159265 \times 2^{4}}{64}$ $= 0.785398 in⁴$
4. Change I to m⁴, the standard unit the formula uses (1 in⁴ = 0.000000416231 m⁴)
  $0.785398 in⁴ \times 0.000000416231$ $= 3.269074e-7 m⁴$
5. Change the load to N, the standard unit the formula uses (1 lbf = 4.448221615261 N)
  $200 lbf \times 4.448221615261$ $= 889.644323 N$
6. Find the biggest bending moment, right under the load: load × distance to each support ÷ span
  $\frac{889.644323 \times 0.4572 \times 1.3716}{1.8288}$ $= 305.059038 N·m$
7. Find the biggest sag, with the load a distance c from the nearer support: δ = P × c × (L² − c²)^1.5 ÷ (9√3 × E × I × L)
  $\frac{889.644323 \times 0.4572 \times {({1.8288}^{2} - {0.4572}^{2})}^{1.5}}{9 \times \sqrt{3} \times 70,000,000,000 \times 3.269074e-7 \times 1.8288}$ $= 0.0034616731 m$
8. Convert the sag from meters to in so it's easy to read
  $\frac{0.0034616731 m}{0.0254}$ $= 0.13628634 in$
3. Maximum deflection: 317,320.223 mm
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $6 ft \times 0.3048$ $= 1.8288 m$
2. Use the typical stiffness (Young's modulus) for glass, written in pascals
  $65 GPa$ $= 65,000,000,000 Pa$
3. Work out how well the rectangle's shape resists bending (its second moment of area): width × depth³ ÷ 12
  $\frac{3 \times {9.25}^{3}}{12}$ $= 197.863281 mm⁴$
4. Change I to m⁴, the standard unit the formula uses (1 mm⁴ = 0.000000000001 m⁴)
  $197.863281 mm⁴ \times 0.000000000001$ $= 1.978633e-10 m⁴$
5. Change the uniform load to N/m, the standard unit the formula uses (1 lbf/ft = 14.593902937206 N/m)
  $200 lbf/ft \times 14.593902937206$ $= 2,918.780587 N/m$
6. Find the biggest bending moment, at the fixed end: load per length × span² ÷ 2
  $\frac{2,918.780587 \times {1.8288}^{2}}{2}$ $= 4,880.944614 N·m$
7. Find how far the free tip sags, which is the most anywhere along a cantilever: δ = w × L⁴ ÷ (8 × E × I)
  $\frac{2,918.780587 \times {1.8288}^{4}}{8 \times 65,000,000,000 \times 1.978633e-10}$ $= 317.3202229976 m$
8. Convert the sag from meters to mm so it's easy to read
  $\frac{317.3202229976 m}{0.001}$ $= 317,320.22299764 mm$
4. Maximum deflection: 3.8669 in
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $4 ft \times 0.3048$ $= 1.2192 m$
2. Use the typical stiffness (Young's modulus) for polystyrene, written in pascals
  $3 GPa$ $= 3,000,000,000 Pa$
3. Work out how well the rectangle's shape resists bending (its second moment of area): width × depth³ ÷ 12
  $\frac{1.5 \times {5.5}^{3}}{12}$ $= 20.796875 in⁴$
4. Change I to m⁴, the standard unit the formula uses (1 in⁴ = 0.000000416231 m⁴)
  $20.796875 in⁴ \times 0.000000416231$ $= 0.000009 m⁴$
5. Change the load to N, the standard unit the formula uses (1 lbf = 4.448221615261 N)
  $1,500 lbf \times 4.448221615261$ $= 6,672.332423 N$
6. Find the biggest bending moment, at the fixed end: load × its distance from that end
  $6,672.332423 \times 0.9144$ $= 6,101.180767 N·m$
7. Find how far the free tip sags, which is the most anywhere along a cantilever: δ = P × a² × (3L − a) ÷ (6 × E × I)
  $\frac{6,672.332423 \times {0.9144}^{2} \times (3 \times 1.2192 - 0.9144)}{6 \times 3,000,000,000 \times 0.000009}$ $= 0.0982204974 m$
8. Convert the sag from meters to in so it's easy to read
  $\frac{0.0982204974 m}{0.0254}$ $= 3.86694872 in$
5. Maximum deflection: 0.236 mm
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $6 ft \times 0.3048$ $= 1.8288 m$
2. Change E to Pa, the standard unit the formula uses (1 ksi = 6,894,757.29316836 Pa)
  $40,000 ksi \times 6,894,757.29316836$ $= 275,790,291,726.73444 Pa$
3. Work out how well the rectangle's shape resists bending (its second moment of area): width × depth³ ÷ 12
  $\frac{5.5 \times {11.25}^{3}}{12}$ $= 652.587891 cm⁴$
4. Change I to m⁴, the standard unit the formula uses (1 cm⁴ = 0.00000001 m⁴)
  $652.587891 cm⁴ \times 0.00000001$ $= 0.000007 m⁴$
5. Change the uniform load to N/m, the standard unit the formula uses (1 lbf/ft = 14.593902937206 N/m)
  $200 lbf/ft \times 14.593902937206$ $= 2,918.780587 N/m$
6. Find the biggest bending moment, in the middle of the span: load per length × span² ÷ 8
  $\frac{2,918.780587 \times {1.8288}^{2}}{8}$ $= 1,220.236153 N·m$
7. Find the biggest sag, in the middle of the span: δ = 5 × w × L⁴ ÷ (384 × E × I)
  $\frac{5 \times 2,918.780587 \times {1.8288}^{4}}{384 \times 275,790,291,726.73444 \times 0.000007}$ $= 0.0002362039 m$
8. Convert the sag from meters to mm so it's easy to read
  $\frac{0.0002362039 m}{0.001}$ $= 0.23620392 mm$
6. Maximum deflection: 36.75 mm
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $12 ft \times 0.3048$ $= 3.6576 m$
2. Change E to Pa, the standard unit the formula uses (1 ksi = 6,894,757.29316836 Pa)
  $35,000 ksi \times 6,894,757.29316836$ $= 241,316,505,260.8926 Pa$
3. Work out how well the rectangle's shape resists bending (its second moment of area): width × depth³ ÷ 12
  $\frac{2 \times {3.5}^{3}}{12}$ $= 7.145833 cm⁴$
4. Change I to m⁴, the standard unit the formula uses (1 cm⁴ = 0.00000001 m⁴)
  $7.145833 cm⁴ \times 0.00000001$ $= 7.145833e-8 m⁴$
5. Change the load to N, the standard unit the formula uses (1 lbf = 4.448221615261 N)
  $200 lbf \times 4.448221615261$ $= 889.644323 N$
6. Find the biggest bending moment, right under the load: load × distance to each support ÷ span
  $\frac{889.644323 \times 2.7432 \times 0.9144}{3.6576}$ $= 610.118077 N·m$
7. Find the biggest sag, with the load a distance c from the nearer support: δ = P × c × (L² − c²)^1.5 ÷ (9√3 × E × I × L)
  $\frac{889.644323 \times 0.9144 \times {({3.6576}^{2} - {0.9144}^{2})}^{1.5}}{9 \times \sqrt{3} \times 241,316,505,260.8926 \times 7.145833e-8 \times 3.6576}$ $= 0.0367501329 m$
8. Convert the sag from meters to mm so it's easy to read
  $\frac{0.0367501329 m}{0.001}$ $= 36.75013295 mm$
7. Maximum deflection: 45,766,008.7254 in
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $16 ft \times 0.3048$ $= 4.8768 m$
2. Use the typical stiffness (Young's modulus) for wood (along the grain), written in pascals
  $13 GPa$ $= 13,000,000,000 Pa$
3. Work out how well the round bar's shape resists bending (its second moment of area): π × diameter⁴ ÷ 64
  $\frac{3.14159265 \times {1.5}^{4}}{64}$ $= 0.248505 mm⁴$
4. Change I to m⁴, the standard unit the formula uses (1 mm⁴ = 0.000000000001 m⁴)
  $0.248505 mm⁴ \times 0.000000000001$ $= 2.485049e-13 m⁴$
5. Change the load to N, the standard unit the formula uses (1 lbf = 4.448221615261 N)
  $500 lbf \times 4.448221615261$ $= 2,224.110808 N$
6. Find the biggest bending moment, right under the load: load × distance to each support ÷ span
  $\frac{2,224.110808 \times 1.2192 \times 3.6576}{4.8768}$ $= 2,033.726922 N·m$
7. Find the biggest sag, with the load a distance c from the nearer support: δ = P × c × (L² − c²)^1.5 ÷ (9√3 × E × I × L)
  $\frac{2,224.110808 \times 1.2192 \times {({4.8768}^{2} - {1.2192}^{2})}^{1.5}}{9 \times \sqrt{3} \times 13,000,000,000 \times 2.485049e-13 \times 4.8768}$ $= 1,162,456.6216249052 m$
8. Convert the sag from meters to in so it's easy to read
  $\frac{1,162,456.6216249052 m}{0.0254}$ $= 45,766,008.72538997 in$
8. Maximum deflection: 23,756.5497 in
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $4 ft \times 0.3048$ $= 1.2192 m$
2. Use the typical stiffness (Young's modulus) for aluminum, written in pascals
  $70 GPa$ $= 70,000,000,000 Pa$
3. Work out how well the round bar's shape resists bending (its second moment of area): π × diameter⁴ ÷ 64
  $\frac{3.14159265 \times 3^{4}}{64}$ $= 3.976078 mm⁴$
4. Change I to m⁴, the standard unit the formula uses (1 mm⁴ = 0.000000000001 m⁴)
  $3.976078 mm⁴ \times 0.000000000001$ $= 3.976078e-12 m⁴$
5. Change the load to N, the standard unit the formula uses (1 lbf = 4.448221615261 N)
  $200 lbf \times 4.448221615261$ $= 889.644323 N$
6. Find the biggest bending moment, at the fixed end: load × its distance from that end
  $889.644323 \times 0.6096$ $= 542.327179 N·m$
7. Find how far the free tip sags, which is the most anywhere along a cantilever: δ = P × a² × (3L − a) ÷ (6 × E × I)
  $\frac{889.644323 \times {0.6096}^{2} \times (3 \times 1.2192 - 0.6096)}{6 \times 70,000,000,000 \times 3.976078e-12}$ $= 603.4163633754 m$
8. Convert the sag from meters to in so it's easy to read
  $\frac{603.4163633754 m}{0.0254}$ $= 23,756.54973919 in$
9. Maximum deflection: 561,143.547 mm
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $16 ft \times 0.3048$ $= 4.8768 m$
2. Use the typical stiffness (Young's modulus) for concrete, written in pascals
  $30 GPa$ $= 30,000,000,000 Pa$
3. Work out how well the round bar's shape resists bending (its second moment of area): π × diameter⁴ ÷ 64
  $\frac{3.14159265 \times 1^{4}}{64}$ $= 0.049087 in⁴$
4. Change I to m⁴, the standard unit the formula uses (1 in⁴ = 0.000000416231 m⁴)
  $0.049087 in⁴ \times 0.000000416231$ $= 2.043171e-8 m⁴$
5. Change the load to N, the standard unit the formula uses (1 lbf = 4.448221615261 N)
  $2,000 lbf \times 4.448221615261$ $= 8,896.443231 N$
6. Find the biggest bending moment, at the fixed end: load × its distance from that end
  $8,896.443231 \times 4.8768$ $= 43,386.174347 N·m$
7. Find how far the free tip sags, which is the most anywhere along a cantilever: δ = P × a² × (3L − a) ÷ (6 × E × I)
  $\frac{8,896.443231 \times {4.8768}^{2} \times (3 \times 4.8768 - 4.8768)}{6 \times 30,000,000,000 \times 2.043171e-8}$ $= 561.1435471692 m$
8. Convert the sag from meters to mm so it's easy to read
  $\frac{561.1435471692 m}{0.001}$ $= 561,143.54716915 mm$
10. Maximum deflection: 0 mm
1. Change the span to m, the standard unit the formula uses (1 ft = 0.3048 m)
  $6 ft \times 0.3048$ $= 1.8288 m$
2. Change E to Pa, the standard unit the formula uses (1 ksi = 6,894,757.29316836 Pa)
  $25,000 ksi \times 6,894,757.29316836$ $= 172,368,932,329.209 Pa$
3. Work out how well the rectangle's shape resists bending (its second moment of area): width × depth³ ÷ 12
  $\frac{3 \times {7.25}^{3}}{12}$ $= 95.269531 cm⁴$
4. Change I to m⁴, the standard unit the formula uses (1 cm⁴ = 0.00000001 m⁴)
  $95.269531 cm⁴ \times 0.00000001$ $= 9.526953e-7 m⁴$
5. Change the load to N, the standard unit the formula uses (1 lbf = 4.448221615261 N)
  $2,000 lbf \times 4.448221615261$ $= 8,896.443231 N$
6. Find the biggest bending moment, right under the load: load × distance to each support ÷ span
  $\frac{8,896.443231 \times 1.8288 \times 0}{1.8288}$ $= 0 N·m$
7. The load sits directly on a support, so the beam doesn't bend
  $0 m$
8. Convert the sag from meters to mm so it's easy to read
  $\frac{0 m}{0.001}$ $= 0 mm$