Civil's Guide

Useful Engineering Equations

Bending and deflection formulas

When designing beams (concrete, steel or timber), the bending and shear capacity is checked against applied bending theory. An applied force causes the element to bend and it is subjected to bending moments and ends react to shear loads.

The bending and shear profile of a beam/element depends on the type of beam support (i.e, pinned, fixed and free ends).

Elastic bending formula

\(\frac{M}{I}=\frac{\sigma}{y}=\frac{E}{R}\)

  • M is the applied moment
  • I is the section moment of inertia
  • \(\sigma\) is the fibre bending stress
  • y is the distance from the neutral axis to the fibre and R is the radius of curvature
  • Section modulus is Z=I/y
  • Applied bending stress can be simplified to \(\sigma\) = M/Z

Typical Beam design

An example of designing any beam can be simplified in the steps below:

  1. Determine the support conditions (fixed or pinned).
  2. Calculate the load applied to the beam.
  3. Calculate the bending moment and shear force in the beam based on typical diagrams below.
  4. Check applied bending moment and shear against required Eurocode or relevant codes (i.e, bending capacity, shear capacity)

Typical Bending and deflection formulas

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Single Point Load
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Single Point Load - Bending Moment

Moment:

\(M_{midspan} = \frac{PL}{4}\)

Deflection:

\(\delta = \frac{PL^3}{48EI}\)

Shear Force

Single Point Load Shear

Reaction:

\(R_A=R_B=\frac{P}{2}\)

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Single Point Load Anywhere load
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Single Point Load - Bending Moment v.1.1

Moment:

\(M_{C} = \frac{Pab}{L}\)

Deflection:

when a>b,

\(\delta_x = \frac{Pab(L+b)}{27EIL} \sqrt{3a(L+b)}\)

at \(x=\sqrt{\frac{a(L+b)}{3}}\) from A

Shear Force

Single Point Load Shear v1.1

Reaction:

\(R_A= \frac{Pb}{L}\)

\(R_B= \frac{Pa}{L}\)

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2-point load diagrams
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

2-point load diagrams - Bending moment Diagram

Moment:

\(M_{C} = Pa\)

\(M = \frac{PL}{3}\)

Deflection:

when a>b,

\(\delta_{midspan} = \frac{PL^3}{6EI} (\frac{3a}{4L}-(\frac{a}{L})^3)\)

\(\delta = \frac{23PL^3}{648EI}\)

Shear Force

2-point load diagrams - Shear Force Diagram

Reaction:

\(R_A = R_B = P\)

Thirdpoins: \(a=\frac{L}{3}\)

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UDL diagram
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

UDL diagram Bending Moment

Moment:

\(M_{midpan} = \frac{wl^2}{8}\)

Deflection:

\(\delta_{midspan} = \frac{5wl^4}{384EI} \)

Shear Force

UDL diagram - Shear Force

Reaction:

\(R_A = R_B = \frac{wL}{2}\)

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Continuos pin Beam UDL
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Continuos pin Beam - UDL - Bending moment

Moment:

\(M_{B} = \frac{wb^2}{2}\)

\(M_{C} = \frac{w(a+b)^2(a+b)^2}{8a^2}\)

maximum x = \(\frac{a}{b}(1-\frac{b^2}{a^2}\)

Deflection:

\(\delta_{c} = \frac{wL}{24EI}(x^4-2ax^3+\frac{2b^2}{a}x^3+a^3x-2ab^2x) \)

\(\delta_{freetip} = \frac{wb}{24EI}(3b^3 + 4ab^2-a^3) \)

Shear Force

Continuos pin Beam UDL Shear Force

Reaction:

\(R_A  = \frac{wL}{2a}(a^2-b^2)\)

\(R_B  = \frac{wL}{2a}(a^2+b^2)\)

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VDL Load
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

VDL Load Bending Moment

Moment:

\(M_{Midspan} = \frac{WL}{6}\)

Deflection:

\(\delta_{midspan} = \frac{WL^3}{60EI}\)

Shear Force

VDL Load - Shear Force

Reaction:

\(R_A  = R_B = \frac{W}{2}\)

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Partial UDL diagram
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Partial UDL diagram - Bending Moment

Moment:

\(M_{Max} = \frac{W}{b}(\frac{x_1^2-a^2}{2})\)

When \(x_1=a+\frac{R_Ab}{W}\)

Deflection:

\(\delta_{max} = \frac{W}{384EI}(8L^3-4Lb^2+b^3)\)

Shear Force

Partial UDL diagram -Shear Diagram

Reaction:

\(R_A  = \frac{W}{L} (\frac{b}{2}+c)\)

\(R_B  = \frac{W}{L} (\frac{b}{2}+a)\)

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Partial VDL
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Partial VDL - Bending Moment

Moment:

\(M_{x} = \frac{Wx(L^2-x^2)}{3L^2}\)

maximum at x=0.5774L

Deflection:

\(\delta_{midspan} = \frac{5WL^3}{384EI}\)

\(\delta_{max} = \frac{0.01304WL^3}{EI}\)

when x = 0.5193L

Shear Force

Partial VDL - Shear Moment

Reaction:

\(R_A  = \frac{W}{3}\)

\(R_B  = \frac{2W}{3}\)

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Fixed Connection
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Fixed Connection - Bending Moment

Moment:

\(M_{A} = \frac{-Pab^2}{L^2}\)

\(M_{B} = \frac{-Pba^2}{L^2}\)

\(M_{C} = \frac{2Pa^2b^2}{L^3}\)

Deflection:

\(\delta_{max} = \frac{2Pa^3b^2}{3EI(L+2a)^2}\)

when \(x=\frac{L^2}{(3L-2a)}\)

\(\delta_{c} = \frac{Pa^3b^2}{3EIL}\)

Shear Force

Fixed Connection - Shear Force

Reaction:

\(R_A  = \frac{Pb^2(L+2a)}{L^3}\)

\(R_B  = \frac{Pb^2(L+2b)}{L^3}\)

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Fix - Fix - load
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Fix - Fix - Bending Moment

Moment:

\(M_{A} = M_{B} = \frac{-wL^2}{12}\)

\(M_{C} = \frac{wL^2}{24}\)

Deflection:

\(\delta_{midspan} = \frac{wL^4}{384EI}\)

Shear Force

Fix - Fix - Shear Force

Reaction:

\(R_A  = R_B \frac{wL}{2}\)

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Fix - Free load
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Fix - Free Bending

Moment:

\(M_{A} = \frac{-wL^2}{2}\)

Deflection:

\(\delta_{B} = \frac{wL^4}{8EI}\)

Shear Force

Fix Fix Shear Force 1

Reaction:

\(R_A  = wL\)

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Fix - Pin Point Load-
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Fix - Pin Point Load-

Moment:

\(M_{A} = \frac{-Pb(L^2-b^2)}{2L^2}\)

\(M_{C} = \frac{Pb}{2}(2-\frac{3b}{L}+\frac{b^3}{L^3}\)

Deflection:

\(\delta_{c} = \frac{Pa^3b^2}{12EIL^3}(4L-a)\)

Shear Force

Fix Pin Point Load Shear

Reaction:

\(R_A  = P-R_B\)

\(R_B  = \frac{Pa^2(2L+b)}{2L^3}\)

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Fix Pin UDL
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Fix - Pin UDL - Bending

Moment:

\(M_{A} = \frac{wL^2}{8}\)

\(M_{sagging} = \frac{9wL^2}{128}\) at 0.62L from A

Deflection:

\(\delta_{max} = \frac{wL^4}{185EI}\) at 0.58L from A

 

Shear Force

Fix - Pin UDL - Shear

Reaction:

\(R_A  = \frac{5wL}{8}\)

\(R_B  = \frac{3wL}{8}\)

 

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Fix-pin continuous Load
  • P is Force in kN
  • L is total length in mm
  • E is young’s modulus 
  • I is the second moment of area (\(mm^2\))
  • W is total load (UDL x length)
  • w is UDL (force per unit length, kN/m)

Maximum moment and deflection

Fix - Pin UDL - Bending

Moment:

\(M_{A} = \frac{Pb}{2}\)

\(M_{B} = -Pb = -2M_A\)

Deflection:

\(\delta_{c} = \frac{Pab^2}{4EI}(a+\frac{4b}{3})\)

Shear Force

Fix - Pin UDL - Shear

Reaction:

\(R_A  = \frac{-3Pb}{2a}\)

\(R_B  = \frac{P}{a}(a+\frac{3b}{2}\)