Deflection of Beams
Double Integration Method, Macaulay’s Method, Moment-Area Theorems & Standard Formulae — With Fully Solved Examples
Last Updated: March 2026
📌 Key Takeaways
- Deflection is the transverse displacement of a beam from its original position under loading. Slope is the angle of rotation of the beam’s cross-section at any point.
- The governing equation is the Euler-Bernoulli beam equation: EI·d²y/dx² = M(x), where y is deflection, x is position along the beam, E is the modulus of elasticity, and I is the second moment of area.
- The double integration method integrates M(x) twice to get the deflection equation — but requires a separate expression for M(x) in each loading region.
- Macaulay’s method is a smarter version of double integration — it uses a single continuous expression for M(x) using Macaulay brackets ⟨x − a⟩, avoiding the need for multiple integration constants.
- The moment-area method (Mohr’s theorems) is the fastest method for finding slope and deflection at specific points — ideal for GATE numerical problems.
- Boundary conditions (y = 0 at supports, dy/dx = 0 at fixed ends or points of symmetry) are used to evaluate the constants of integration.
- Deflection must be checked against serviceability limits — IS 456 limits beam deflection to span/250 under total load and span/350 after construction for spans supporting brittle finishes.
1. Why Deflection Matters
Structural design involves two distinct sets of requirements. The first is strength — ensuring the beam does not fail by yielding, fracturing, or buckling under the applied loads. The second is serviceability — ensuring the beam does not deform excessively under working loads, even if it is nowhere near failure.
Excessive deflection causes a range of real-world problems. A floor beam that sags too much makes occupants uncomfortable (a noticeable slope or bounce). Excessive deflection in a roof beam allows water to pond, increasing the load and potentially triggering progressive collapse. In buildings with brittle finishes — tiles, plaster ceilings, masonry partition walls — a beam that deflects too much will crack the finishes even if the beam itself is structurally sound. In industrial buildings, excessive deflection of crane girders causes crane rails to misalign, jamming the crane mechanism.
For these reasons, deflection is a serviceability limit state check — it is performed after the strength design is complete, using working (unfactored) loads. In India, IS 456:2000 specifies that the total deflection (including long-term effects of creep and shrinkage) should not exceed span/250, and the deflection after construction of partitions and finishes should not exceed span/350 or 20 mm, whichever is less.
In structural analysis theory, the calculation of beam deflections also plays a direct role in solving statically indeterminate structures. The compatibility equations needed to solve indeterminate beams — propped cantilevers, fixed beams, continuous beams — are all statements about deflection: the deflection at a propped support is zero, the slope at a fixed end is zero, and so on. Mastering deflection calculations is therefore a prerequisite for solving indeterminate structures.
2. The Governing Equation — Euler-Bernoulli Beam Theory
The relationship between bending moment and deflection in a beam is derived from the geometry of bending combined with the stress-strain relationship of the material. For a beam made of a linear elastic material (which obeys Hooke’s Law), and where deflections are small compared to the span (the small deflection assumption), the result is the Euler-Bernoulli beam bending equation:
Euler-Bernoulli Beam Equation
EI · d²y/dx² = M(x)
Where:
E = Young’s modulus (modulus of elasticity) of the beam material [N/mm² or GPa]
I = Second moment of area (moment of inertia) of the cross-section about the neutral axis [mm⁴ or m⁴]
y = Deflection of the beam at position x (positive upward) [mm or m]
x = Distance along the beam from the left end [mm or m]
M(x) = Bending moment at position x [N·mm or kN·m]
EI = Flexural rigidity — the resistance of the beam to bending. Higher EI means less deflection for the same moment.
Note on sign convention: With y positive upward and x positive to the right, d²y/dx² is positive when the beam curves concave upward (sagging). Since sagging moment is positive (by the standard convention), M(x) is positive for sagging. This gives the equation as written above. Some textbooks write EI·d²y/dx² = −M(x) with a different sign convention for y — always check the convention being used.
The Complete Set of Relationships
EI · d⁴y/dx⁴ = −w(x) — distributed load intensity (4th derivative of deflection)
EI · d³y/dx³ = V(x) — shear force (3rd derivative)
EI · d²y/dx² = M(x) — bending moment (2nd derivative)
EI · dy/dx = ∫M dx + C₁ — slope (1st integral, θ ≈ dy/dx for small angles)
EI · y = ∬M dx dx + C₁x + C₂ — deflection (2nd integral)
C₁ and C₂ are constants of integration evaluated from boundary conditions.
3. Boundary Conditions
Boundary conditions are the known values of slope and deflection at specific points along the beam. They are used to determine the constants of integration C₁ and C₂ that arise from integrating the beam equation. Getting boundary conditions wrong is the most common source of errors in deflection problems.
| Support / Condition | Deflection y | Slope dy/dx |
|---|---|---|
| Pin support | y = 0 (no vertical movement) | dy/dx ≠ 0 (beam is free to rotate) |
| Roller support | y = 0 (no vertical movement) | dy/dx ≠ 0 (beam is free to rotate) |
| Fixed support | y = 0 (no vertical movement) | dy/dx = 0 (no rotation — built-in) |
| Free end | y ≠ 0 (deflects freely) | dy/dx ≠ 0 (rotates freely) |
| Point of symmetry (symmetric loading on SS beam) | y = maximum (but unknown) | dy/dx = 0 (slope is zero at midspan) |
| Internal hinge | y is continuous (no gap) | dy/dx may be discontinuous (slope can differ on either side) |
How many boundary conditions are needed? Each integration of the beam equation introduces one constant. For a full deflection solution, we integrate twice, producing C₁ and C₂ — so we need exactly two boundary conditions per beam segment. For Macaulay’s method (single expression for the entire beam), we always have exactly two constants C₁ and C₂ and apply two boundary conditions from the list above.
4. Double Integration Method
The double integration method directly integrates the beam equation EI·d²y/dx² = M(x) twice to obtain the slope and deflection equations. It is straightforward when the bending moment can be described by a single expression valid over the entire beam length — which is only possible for simple loading cases (UDL on a simply supported beam, point load at the free end of a cantilever, etc.).
When the beam has multiple loads at different positions, M(x) changes form at each load point — a different expression applies in each region. The double integration method then requires separate integration in each region, with additional continuity conditions (slope and deflection must be continuous at the boundaries between regions). This produces many constants of integration, making the method cumbersome. Macaulay’s method solves this problem elegantly.
Procedure for double integration method:
- Calculate support reactions.
- Write M(x) as a function of x for each region of the beam.
- Write EI·d²y/dx² = M(x) for each region.
- Integrate once to get EI·dy/dx = ∫M(x)dx + C₁ (slope equation).
- Integrate again to get EI·y = ∬M(x)dx + C₁x + C₂ (deflection equation).
- Apply boundary conditions to find C₁ and C₂.
- Substitute x values to find slope and deflection at required points.
5. Macaulay’s Method — Step by Step
Macaulay’s method (also called the singularity function method or the method of singularity functions) is an extension of the double integration method that handles multiple loads in a single, unified expression. It uses Macaulay brackets — written as ⟨x − a⟩ — which are defined as:
Macaulay Bracket — Definition
⟨x − a⟩ = 0 when x < a (the term is simply ignored)
⟨x − a⟩ = (x − a) when x ≥ a (the bracket is treated as a normal algebraic quantity)
This allows a single bending moment expression to “switch on” each load contribution only after the load’s position is reached, moving left to right along the beam.
Integration rule for Macaulay brackets:
∫⟨x − a⟩ⁿ dx = ⟨x − a⟩ⁿ⁺¹ / (n+1)
The brackets are kept intact during integration — do not expand them until the very end, after applying boundary conditions.
Step-by-step Macaulay’s method procedure:
- Step 1: Calculate all support reactions using equilibrium.
- Step 2: Write a single bending moment expression M(x) using Macaulay brackets, starting from the left end and including every load encountered moving right. Use ⟨x − a⟩ for point loads and ⟨x − a⟩² for UDLs starting at position a.
- Step 3: Write EI·d²y/dx² = M(x).
- Step 4: Integrate once (keeping Macaulay brackets intact) to get EI·dy/dx = slope expression + C₁.
- Step 5: Integrate again to get EI·y = deflection expression + C₁x + C₂.
- Step 6: Apply two boundary conditions (y = 0 at pin/roller supports, dy/dx = 0 at fixed end or symmetry) to find C₁ and C₂. When substituting x values into Macaulay brackets, set any bracket to zero if x < a.
- Step 7: Substitute the required x value into the deflection equation to find the deflection at that point. For maximum deflection, first find where dy/dx = 0, then substitute that x into the deflection equation.
Macaulay’s Method — UDL Starting Mid-Span
If a UDL of intensity w starts at position a and continues to the end of the beam, the moment contribution is:
MUDL = −w⟨x − a⟩²/2
If the UDL ends at position b before the end of the beam, you must add a compensating term to cancel it beyond b:
MUDL (a to b) = −w⟨x − a⟩²/2 + w⟨x − b⟩²/2
This compensation technique is essential for UDLs that do not extend to the right end of the beam.
6. Moment-Area Method (Mohr’s Theorems)
The moment-area method, based on Mohr’s two theorems, provides a geometric approach to finding slope and deflection at specific points — without writing and integrating a full deflection equation. It is particularly fast for GATE CE problems where you need the deflection at one specific point, not the complete deflection curve.
Mohr’s First Theorem — Change in Slope
θAB = (1/EI) × [Area of M/EI diagram between A and B]
The change in slope (in radians) between two points A and B on the elastic curve equals the area of the M/EI diagram between those two points.
For constant EI: θAB = (Area of BMD between A and B) / EI
Mohr’s Second Theorem — Deflection
tA/B = (1/EI) × [First moment of the M/EI diagram between A and B, taken about A]
tA/B is the tangential deviation — the vertical distance from point A on the elastic curve to the tangent drawn to the elastic curve at point B.
tA/B = (Area of M/EI diagram between A and B) × (distance of centroid of that area from A) / EI
Important: Tangential deviation is NOT the same as deflection. For a simply supported beam, the deflection at any point is found by geometry using the tangential deviations from the support tangents.
| BMD Shape | Area | Centroid from Left End |
|---|---|---|
| Rectangle (height h, width b) | bh | b/2 |
| Triangle (base b, height h — zero at left) | bh/2 | 2b/3 from zero end (b/3 from peak end) |
| Triangle (base b, height h — zero at right) | bh/2 | b/3 from left (peak end) |
| Parabola (base b, height h — vertex at one end) | bh/3 | 3b/4 from vertex end |
| Parabola (vertex at centre — SS beam UDL) | 2bh/3 | b/2 (by symmetry) |
7. Standard Deflection Formulae — All Beam Types
These formulae must be memorised for GATE CE. They are derived using the double integration method and are used directly in numerical problems and as building blocks for superposition.
| Beam Type | Loading | Maximum Deflection (δmax) | Location | Slope at Supports |
|---|---|---|---|---|
| Simply Supported | Central point load W | WL³/48EI | Midspan | θ = WL²/16EI (at both ends) |
| Simply Supported | UDL w over full span | 5wL⁴/384EI | Midspan | θ = wL³/24EI (at both ends) |
| Simply Supported | Point load W at distance a from A (b = L−a) | Wb(L²−b²)^(3/2) / (9√3·EIL) — under the load: Wa²b²/3EIL | At x = √[(L²−b²)/3] from A | θA = Wab(L+b)/6EIL | θB = Wab(L+a)/6EIL |
| Cantilever | Point load W at free end | WL³/3EI | Free end | θ = WL²/2EI (at free end) |
| Cantilever | UDL w over full span | wL⁴/8EI | Free end | θ = wL³/6EI (at free end) |
| Cantilever | Point load W at distance a from fixed end | Wa³/3EI + Wa²(L−a)/2EI | Free end | θ = Wa²/2EI (at free end) |
| Fixed Beam | Central point load W | WL³/192EI | Midspan | θ = 0 (at both fixed ends) |
| Fixed Beam | UDL w over full span | wL⁴/384EI | Midspan | θ = 0 (at both fixed ends) |
| Propped Cantilever | UDL w over full span | wL⁴/185EI (approx) at x ≈ 0.4215L from fixed end | x ≈ 0.4215L from fixed end | θ = 0 at fixed end |
Comparison — Deflection Ratios for UDL (Same w, L, EI)
Simply Supported: δ = 5wL⁴/384EI = 1.0 × (wL⁴/384EI) × 5
Fixed Beam: δ = wL⁴/384EI — exactly 5 times less than simply supported
Cantilever: δ = wL⁴/8EI = 48 times more than fixed beam, 9.6 times more than simply supported
These ratios are a favourite GATE MCQ topic — memorise them.
8. Worked Example 1 — Simply Supported Beam, Central Point Load (Double Integration)
Problem: A simply supported beam AB of span L carries a central point load W at midspan C. Using the double integration method, derive the deflection equation and find the maximum deflection.
Solution
Reactions: RA = RB = W/2 (by symmetry)
Bending moment expression (for 0 ≤ x ≤ L/2, using left half by symmetry):
M(x) = (W/2)·x
Governing equation:
EI·d²y/dx² = M(x) = (W/2)·x
First integration (slope):
EI·dy/dx = Wx²/4 + C₁
Second integration (deflection):
EI·y = Wx³/12 + C₁x + C₂
Boundary conditions:
BC1: At x = 0 (pin support A), y = 0 → 0 = 0 + 0 + C₂ → C₂ = 0
BC2: At x = L/2 (midspan, point of symmetry), dy/dx = 0 (slope = 0 at midspan):
0 = W(L/2)²/4 + C₁ = WL²/16 + C₁ → C₁ = −WL²/16
Deflection equation (for 0 ≤ x ≤ L/2):
EI·y = Wx³/12 − WL²x/16
y = W/(EI) · (x³/12 − L²x/16)
Maximum deflection at x = L/2 (midspan):
EI·ymax = W(L/2)³/12 − WL²(L/2)/16 = WL³/96 − WL³/32 = WL³/96 − 3WL³/96 = −2WL³/96
ymax = −WL³/48EI
The negative sign indicates downward deflection (with y positive upward).
Maximum deflection = WL³/48EI (downward) ✓
9. Worked Example 2 — Simply Supported Beam, Eccentric Point Load (Macaulay’s Method)
Problem: A simply supported beam AB of span 6 m carries a point load of 30 kN at C, 2 m from A. Using Macaulay’s method, find: (a) the deflection under the load, and (b) the maximum deflection. Take EI = 8,000 kN·m².
Solution
Reactions:
RA = 30 × 4/6 = 20 kN | RB = 30 × 2/6 = 10 kN
Macaulay’s bending moment expression (single expression for entire beam):
EI·d²y/dx² = 20x − 30⟨x − 2⟩
(RA × x contributes throughout; the 30 kN load contributes only for x ≥ 2 m)
First integration:
EI·dy/dx = 20x²/2 − 30⟨x−2⟩²/2 + C₁ = 10x² − 15⟨x−2⟩² + C₁
Second integration:
EI·y = 10x³/3 − 15⟨x−2⟩³/3 + C₁x + C₂ = (10/3)x³ − 5⟨x−2⟩³ + C₁x + C₂
Boundary conditions:
BC1: At x = 0 (support A), y = 0:
0 = 0 − 0 + 0 + C₂ → C₂ = 0
(When x = 0, the Macaulay bracket ⟨0−2⟩ = 0 since x < 2)
BC2: At x = 6 m (support B), y = 0:
0 = (10/3)(216) − 5(4³) + C₁(6) + 0
0 = 720 − 5(64) + 6C₁ = 720 − 320 + 6C₁
6C₁ = −400 → C₁ = −66.67 kN·m²
(a) Deflection under the load (x = 2 m):
EI·y = (10/3)(8) − 5⟨2−2⟩³ + (−66.67)(2)
= 26.67 − 0 − 133.33 = −106.67 kN·m³
y = −106.67/8000 = −0.01333 m = −13.33 mm (downward)
(b) Maximum deflection — find where dy/dx = 0:
Since load is at x = 2 m (closer to A), maximum deflection is between 2 m and 3 m from A (not at midspan).
For x > 2: EI·dy/dx = 10x² − 15(x−2)² + C₁ = 0
10x² − 15(x² − 4x + 4) − 66.67 = 0
10x² − 15x² + 60x − 60 − 66.67 = 0
−5x² + 60x − 126.67 = 0 → 5x² − 60x + 126.67 = 0
x = [60 ± √(3600 − 4×5×126.67)] / 10 = [60 ± √(3600 − 2533.4)] / 10 = [60 ± √1066.6] / 10
x = [60 ± 32.66] / 10 → x = 9.27 m (outside beam) or x = 2.73 m from A ✓
Deflection at x = 2.73 m:
EI·y = (10/3)(2.73³) − 5(0.73³) + (−66.67)(2.73)
= (10/3)(20.35) − 5(0.389) − 181.99
= 67.83 − 1.945 − 181.99 = −116.1 kN·m³
ymax = −116.1/8000 = −0.01451 m = −14.51 mm (downward)
10. Worked Example 3 — Cantilever with UDL (Double Integration)
Problem: A cantilever beam AB is fixed at A and free at B. Span = L. Carries UDL of w per unit length over the entire span. Derive the deflection equation and find the maximum deflection and slope at the free end.
Solution
Coordinate system: Origin at fixed end A, x positive towards free end B.
Reactions at A: RA = wL (upward), MA = wL²/2 (hogging/anticlockwise)
Bending moment at distance x from A:
M(x) = −wL²/2 + wLx − wx²/2 (hogging is negative)
Alternatively, working from the free end B (much simpler for cantilevers):
M(x) = −w(L−x)²/2 (sum of moments of UDL to the right of section)
Governing equation (working from free end, measuring x from A):
EI·d²y/dx² = M(x) = −w(L−x)²/2
First integration:
EI·dy/dx = −w(L−x)²/2 integrated → EI·dy/dx = w(L−x)³/6 + C₁
Boundary condition for slope: At x = 0 (fixed end A), dy/dx = 0:
0 = w(L)³/6 + C₁ → C₁ = −wL³/6
Slope equation:
EI·dy/dx = w(L−x)³/6 − wL³/6
Slope at free end B (x = L):
EI·θB = w(0)³/6 − wL³/6 = −wL³/6
θB = −wL³/6EI (negative = rotating clockwise = beam slopes downward towards B)
Second integration:
EI·y = w(L−x)³/6 integrated → EI·y = −w(L−x)⁴/24 − wL³x/6 + C₂
Boundary condition for deflection: At x = 0 (fixed end A), y = 0:
0 = −wL⁴/24 − 0 + C₂ → C₂ = wL⁴/24
Deflection equation:
EI·y = −w(L−x)⁴/24 − wL³x/6 + wL⁴/24
Maximum deflection at free end B (x = L):
EI·yB = −0 − wL⁴/6 + wL⁴/24 = −4wL⁴/24 + wL⁴/24 = −3wL⁴/24 = −wL⁴/8
ymax = wL⁴/8EI (downward) ✓
11. Worked Example 4 — Simply Supported Beam with UDL (Moment-Area Method)
Problem: A simply supported beam AB of span 5 m carries a UDL of 20 kN/m over the full span. Using the moment-area method, find the maximum deflection. Take EI = 15,000 kN·m².
Solution
Maximum BM (at midspan C): MC = wL²/8 = 20 × 25/8 = 62.5 kN·m
BMD shape: Parabolic, zero at A and B, maximum of 62.5 kN·m at midspan C.
By symmetry: The slope at midspan is zero (θC = 0). The tangent at C is horizontal.
Using Mohr’s Second Theorem — tangential deviation of A from the tangent at C (tA/C) equals the maximum deflection δmax (because the tangent at C is horizontal):
tA/C = (1/EI) × [Area of BMD from A to C] × [distance of centroid of that area from A]
Area of parabolic BMD from A to C:
Area = (2/3) × base × height = (2/3) × (L/2) × MC = (2/3) × 2.5 × 62.5 = 104.17 kN·m²
Centroid of parabolic area from A:
For a parabola (vertex at C), the centroid is at 5L/8 from the end where BM = 0.
Centroid from A = 5/8 × L/2 = 5/8 × 2.5 = 1.5625 m from A
Tangential deviation:
tA/C = (1/EI) × 104.17 × 1.5625 = 162.77 / 15000 = 0.01085 m = 10.85 mm
Verification using formula: δmax = 5wL⁴/384EI = 5×20×625/384×15000 = 62500/5760000 = 0.01085 m ✓
12. Deflection by Superposition
The principle of superposition states that for a linear elastic structure, the deflection due to a combination of loads equals the sum of deflections due to each load acting individually. This is only valid when deflections are small (small deformation assumption) and the material is linear elastic — both conditions are satisfied for standard beam problems.
Superposition is the fastest method for finding deflection under complex loading — use the standard formulae for each individual load component and add the results algebraically.
Example — Superposition for SS Beam with Central Load + UDL
Beam: Simply supported, span L, central point load W and full-span UDL w.
Deflection at midspan (by superposition):
δtotal = δpoint load + δUDL = WL³/48EI + 5wL⁴/384EI
This is valid because both loads produce maximum deflection at the same location (midspan, by symmetry), so they add directly.
For asymmetric loading (maximum deflections at different locations), find the deflection at the required point from each load separately using the appropriate formula, then add.
Superposition is also the basis for solving indeterminate beams using the force method (compatibility method). For a propped cantilever, the prop reaction RB is the redundant — it is found by equating the deflection of the released structure (cantilever under applied loads) at B to the upward deflection produced by RB alone, ensuring the net deflection at B is zero (compatibility condition).
13. Common Mistakes Students Make
- Expanding Macaulay brackets before applying boundary conditions: The entire power of Macaulay’s method lies in keeping the brackets intact through both integrations. Expanding ⟨x − a⟩² into x² − 2ax + a² destroys the method — you lose the ability to zero out terms before the load position. Always keep brackets intact until you substitute the boundary condition x values.
- Using the wrong boundary condition at a simply supported end: At a pin or roller support, y = 0 (zero deflection) but dy/dx ≠ 0 (the beam is free to rotate). Many students incorrectly also set dy/dx = 0 at a simply supported end — this is only correct at a fixed support or at the point of symmetry in a symmetrically loaded beam.
- Forgetting the compensation term for partial UDL in Macaulay’s method: If a UDL does not extend to the right end of the beam, you must add a compensating term (equal and opposite UDL starting from where the original UDL ends) to cancel it beyond that point. Forgetting this gives a completely wrong deflection equation for x beyond the end of the UDL.
- Confusing tangential deviation with deflection in the moment-area method: Tangential deviation (tA/B) is the vertical distance from a point on the elastic curve to the tangent drawn at another point — it is NOT the deflection of the beam from its original position. For a simply supported beam, you must use geometry to convert tangential deviations into actual deflections. Only for a cantilever (where the tangent at the fixed end is horizontal and coincides with the original beam axis) does tangential deviation from the fixed end equal actual deflection.
- Unit inconsistency: EI must be in consistent units with the bending moment and span. If M is in kN·m and L is in m, then EI must be in kN·m². If E is in N/mm² and I is in mm⁴, the product EI is in N·mm². Mixing units — for example, E in GPa and I in cm⁴ — leads to deflections that are off by factors of 1,000 or more. Always convert to a consistent unit system before substituting.
14. Frequently Asked Questions
What is flexural rigidity (EI) and how does it affect deflection?
Flexural rigidity EI is the product of the Young’s modulus (E) of the beam material and the second moment of area (I) of the cross-section about the neutral axis. It represents the beam’s resistance to bending deformation — a higher EI means less deflection for the same applied load and span. From all the standard deflection formulae, deflection is always inversely proportional to EI. To reduce deflection, you can either use a stiffer material (higher E — for example, steel instead of timber) or increase the section depth (which increases I dramatically, since I ∝ d³ for a rectangular section). Increasing depth is generally more effective and economical than changing material.
Why is Macaulay’s method preferred over the standard double integration method?
In the standard double integration method, when a beam has multiple loads at different positions, you must write separate M(x) expressions for each region between loads and integrate each one separately. For a beam with n load points, this produces n regions with 2n constants of integration, requiring 2n equations (boundary conditions plus continuity conditions). For just two point loads, this means 4 regions and 8 constants. Macaulay’s method collapses all of this into a single expression with just two constants (C₁ and C₂), regardless of how many loads are applied. The only rules are: write the moment expression from the left end and use Macaulay brackets for all terms involving load positions; never expand the brackets until after applying boundary conditions.
How do you find the maximum deflection if it does not occur at midspan?
Set the slope equation (EI·dy/dx) equal to zero and solve for x. This gives the location of maximum deflection. Then substitute this x value into the deflection equation (EI·y) to find the magnitude. For an eccentric point load on a simply supported beam, the maximum deflection does not occur under the load — it occurs between the load and the midspan, on the side of the longer span. It can be shown that for any point load position, the maximum deflection is always within 3% of the midspan deflection, which is why the midspan formula is sometimes used as an approximation.
What is the difference between slope and deflection?
Deflection (y) is the transverse displacement of the beam’s neutral axis from its original unloaded position — measured in mm or m, perpendicular to the original beam axis. Slope (dy/dx, also written as θ) is the angle that the tangent to the deformed beam axis makes with the original beam axis — measured in radians (or degrees). For small deflections, the slope equals the first derivative of the deflection curve: θ = dy/dx. The slope is zero at points of maximum or minimum deflection (by the calculus rule for stationary points) and is maximum at the supports of a simply supported beam.