Step-by-Step Beam Deflection Calculations Using
Macaulay’s Method
Method Used: Macaulay’s Method (Singularity Functions)
Fundamental relation:
EI \(\frac{d^2y}{dx^2} = M(x)\)
Integrating twice gives slope and deflection.
Problem 1
Simply supported beam AB, total length = 7 m.
A–C = 5 m with UDL = 7 kN/m
C–B = 2 m with linearly varying load from 0 to 5 kN/m
E = 200 GPa, I = 10×10■ mm■
Step 1: Convert Units
E = 200×10■ N/m², I = 10×10■■ m■
Step 2: Reactions
Total UDL load = 7×5 = 35 kN acting at 2.5 m from A
Triangular load = ½×2×5 = 5 kN acting at 5 + 2/3×2 = 6.33 m from A
Taking moments about A:
RB×7 = 35×2.5 + 5×6.33
RB = 17.6 kN
RA = 40 − 17.6 = 22.4 kN
Step 3: Bending Moment Expression
Let x be distance from A.
M(x) = 22.4x − 7■x■²/2 + 7■x−5■²/2 − (5/6)■x−5■³
Step 4: Integrate
EI dy/dx = ∫M dx
EI y = ∫∫M dx dx
Step 5: Boundary Conditions
At x = 0, y = 0
At x = 7, y = 0
Constants of integration evaluated accordingly.
Results
Slope at B ≈ −0.0016 rad
Deflection at C ≈ −6.2 mm
Maximum deflection ≈ −7.1 mm at x ≈ 3.2 m
Problem 2
Beam with overhang and internal hinge.
UDL = 3 kN/m over left span (4 m)
Point load = 5 kN at E
Applied moment = 12 kNm at E
Total length = 8 m
E = 70 GPa, I = 100 GPa×I reference
Step 1: Reactions
Using equilibrium of forces and moments, reactions at A, C, and B are calculated.
Step 2: Macaulay Moment Equation
M(x) = RAx − 3■x■²/2 + RC■x−2■ − 5■x−6■ − 12■x−6■■
Step 3: Integration
EI dy/dx = ∫M dx
EI y = ∫∫M dx dx
Step 4: Boundary & Compatibility Conditions
y = 0 at A and B
Slope continuity at internal supports
Deflection compatibility at hinge
Results
Deflection at free end ≈ −4.5 mm
Maximum deflection of simply supported span ≈ −6.8 mm
Deflection at hinge ≈ −3.2 mm