Deflection of Beams

The moment-area method uses the area and the moment of the $M/EI$ diagram to find slopes and deflections of beams.

Basis

For small deflections the curvature is $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$. Consider two points A and B on the deflected beam and an elemental length $dx$ at a distance $x$ from B.

   tangent at A
   \
     \   elastic curve
 A ____\___________ B
      \_|_|__/
        dx         tangent at B
  <-- x --> (measured from B)

The tangents at the two ends of $dx$ meet at an angle $d\theta$:

$$d\theta = \frac{dx}{\rho} = \frac{M}{EI}\,dx$$

Theorem I (slope)

The change in slope between the tangents at two points A and B on the elastic curve equals the area of the $M/EI$ diagram between those points.

Proof: integrate $d\theta$ from A to B:

$$\theta_{AB} = \theta_B - \theta_A = \int_A^B \frac{M}{EI}\,dx = \text{Area}_{AB}\ \text{of the } \frac{M}{EI}\text{ diagram}$$

Theorem II (deviation)

The vertical deviation of point A from the tangent drawn at B equals the moment, about A, of the area of the $M/EI$ diagram between A and B.

Proof: the tangents at the ends of the element $dx$ intercept a small vertical distance on the vertical through A:

$$d\Delta = x\,d\theta = x\,\frac{M}{EI}\,dx$$

where $x$ is measured from A to the element. Integrating from A to B:

$$\Delta_{A/B} = \int_A^B x\,\frac{M}{EI}\,dx = \text{Area}_{AB}\times\bar{x}_A$$

where $\bar x_A$ is the distance from A to the centroid of the $M/EI$ area between A and B.

Notes

• A positive $M/EI$ area means a counter-clockwise (positive) change of slope from A to B and A lies above the tangent drawn at B.
• Theorems are valid for linear elastic beams with small slopes. Deviation is measured perpendicular to the original beam axis.

The conjugate beam method converts the problem of finding slope and deflection of a real beam into finding shear and moment in a fictitious beam (the conjugate beam), which is loaded with the $M/EI$ diagram of the real beam.

Theorems

• Theorem 1: The slope at a point in the real beam is numerically equal to the shear in the conjugate beam at the corresponding point.
• Theorem 2: The deflection at a point in the real beam is numerically equal to the bending moment in the conjugate beam at the corresponding point.

Proof

For a real beam, $\dfrac{d^2y}{dx^2} = \dfrac{M}{EI}$, with $\dfrac{d^2M}{dx^2} = w$, $\dfrac{dM}{dx} = V$ (load, shear, moment).

For the conjugate beam, loaded by $w' = \dfrac{M}{EI}$:

$$\frac{dV'}{dx} = w' = \frac{M}{EI},\qquad \frac{d^2M'}{dx^2} = w' = \frac{M}{EI}$$

Comparing with the real beam relations:

$$\frac{d\theta}{dx} = \frac{M}{EI}\ \Rightarrow\ \theta = \int\frac{M}{EI}dx = V'$$ $$\frac{d^2y}{dx^2} = \frac{M}{EI}\ \Rightarrow\ y = M'$$

The mathematical relations are the same, so $\theta = V'$ and $y = M'$ (provided the boundary conditions also match).

Supports of the conjugate beam

Procedure and example

1. Find $M$ of the real beam and draw $M/EI$.
2. Load the conjugate beam with $M/EI$ (sagging $M/EI$ acts downward).
3. Find the reactions, then shear (slope) and moment (deflection).

Example: simply supported beam, span $L$, central load $P$. $M_{max} = PL/4$.

Conjugate beam: simply supported, triangular load with peak $\dfrac{PL}{4EI}$. Total load $= \tfrac12\cdot L\cdot\dfrac{PL}{4EI} = \dfrac{PL^2}{8EI}$, so each reaction is $\dfrac{PL^2}{16EI}$.

$$\theta_A = V'_A = \frac{PL^2}{16EI}$$

At mid-span, the conjugate moment:

$$y_C = M'_C = \frac{PL^2}{16EI}\times\frac{L}{2} - \frac{PL^2}{16EI}\times\frac{L}{6}\ \ \Rightarrow\ \ y_C = \frac{PL^3}{32EI} - \frac{PL^3}{96EI} = \frac{PL^3}{48EI}$$

which is the known mid-span deflection.

The moment-area method finds slopes and deflections of beams from the area and the moment of the $M/EI$ diagram.

Theorem I

The change of slope between two points A and B of the elastic curve equals the area of the $M/EI$ diagram between those points:

$$\theta_B - \theta_A = \int_A^B \frac{M}{EI}\,dx$$

Theorem II

The vertical deviation of a point A from the tangent drawn at another point B equals the moment of the $M/EI$ area between A and B about the vertical through A:

$$\Delta_{A/B} = \int_A^B x\,\frac{M}{EI}\,dx = \text{Area}\times\bar{x}_A$$

Here $\bar x_A$ is the horizontal distance from A to the centroid of the area.

        tangent at B
  A ______________________ B (fixed: tangent horizontal)
     \                    |
      \_____ elastic curve|
   A'<--- deviation ---->

Simple example

A cantilever of length $L$ with a point load $P$ at the free end B, fixed end A. Tangent at A is horizontal.

$M = -Px$ (x from B), the $M/EI$ diagram is a triangle with base $L$, peak $\dfrac{PL}{EI}$ at A.

$$\text{Area} = \frac12\times L\times\frac{PL}{EI} = \frac{PL^2}{2EI}$$

Slope at B (Theorem I, since $\theta_A = 0$):

$$\theta_B = \frac{PL^2}{2EI}$$

Deflection at B (Theorem II): the centroid of the triangle is at $\dfrac{2L}{3}$ from B:

$$\Delta_B = \frac{PL^2}{2EI}\times\frac{2L}{3} = \frac{PL^3}{3EI}$$

These are the standard cantilever results.

Given data

Simply supported beam, span $L = 14$ m (support 0 at the left, support 1 at the right). Load $P_2 = 240$ kN at point 2 (3 m from support 0); $P_3 = 160$ kN at point 3 (9.5 m from support 0, 4.5 m from support 1). $EI = 3.36\times10^{11}$ kNmm$^2 = 3.36\times10^{5}$ kNm$^2$.

Step 1: Reactions and bending moments

$$R_1 = \frac{240\times3 + 160\times9.5}{14} = 160\ \text{kN},\qquad R_0 = 400 - 160 = 240\ \text{kN}$$ $$M_2 = 240\times3 = 720\ \text{kNm},\qquad M_3 = 160\times4.5 = 720\ \text{kNm}$$

Step 2: Conjugate beam (simply supported, loaded with $M/EI$)

The $M/EI$ diagram is a trapezoid with peak $720/EI$ between points 2 and 3.

       720/EI ____________
            /|            |\
          /  |            |  \
  0 ----/----2------------3-----\---- 1
        3 m       6.5 m       4.5 m

Loads on the conjugate beam (in units of $1/EI$):

• Triangle 0-2: $\tfrac12\times3\times720 = 1080$, centroid 2.0 m from 0.
• Rectangle 2-3: $6.5\times720 = 4680$, centroid at 6.25 m from 0.
• Triangle 3-1: $\tfrac12\times4.5\times720 = 1620$, centroid 11 m from 0.

Step 3: Conjugate reactions = slopes at supports

$$\theta_0 = R'_0 = \frac{1080\times12 + 4680\times7.75 + 1620\times3}{14} = \frac{54090}{14} = \frac{3863.57}{EI}$$ $$\theta_1 = R'_1 = 1080 + 4680 + 1620 - 3863.57 = \frac{3516.43}{EI}$$

Step 4: Slopes at points 2 and 3 (conjugate shear)

$$\theta_2 = 3863.57 - 1080 = \frac{2783.57}{EI}$$ $$\theta_3 = 3863.57 - 1080 - 4680 = -\frac{1896.43}{EI}$$

(The sign changes between 2 and 3, so the maximum deflection lies between the loads.)

Step 5: Deflections (conjugate moment)

$$y_2 = 3863.57\times3 - 1080\times1 = \frac{10510.7}{EI}$$ $$y_3 = 3863.57\times9.5 - 1080\times7.5 - 4680\times3.25 = \frac{13393.9}{EI}$$

Step 6: Numerical values with $EI = 3.36\times10^5$ kNm$^2$

Answer: slopes at supports: $0.0115$ rad and $0.01047$ rad; slopes under loads: $0.00828$ rad (point 2) and $0.00564$ rad (point 3, opposite sense); deflections: $31.28$ mm under the 240 kN load and $39.86$ mm under the 160 kN load.

Given data

Beam ABC: A hinge, B roller, $AB = 6$ m ($2EI$) with UDL 15 kN/m, overhang $BC = 2$ m ($EI$) with 3 kN at C.

   15 kN/m (2EI)               3 kN
 A =============== B ----------v C
 ^        6 m      ^   2 m (EI)

Step 1: Reactions

$R_{A} = 44.00$ kN, $R_{B} = 49.00$ kN (positive = upward; check: $\sum F_y = 93.00$ kN equals the total load 93.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Conjugate beam

The real beam has a hinge at A, a roller at the second support and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

Step 4: Conjugate reaction at A (slope at A)

Taking moments of the conjugate beam about the support at $x = 6$ m (conjugate moment is zero there):

$$R'_A\times6 = (126.00)(3.071) = 387.00$$ $$R'_A = \frac{64.500}{EI}\ \text{(slope at A, clockwise)}$$

At C (free end) ($x = 8$ m):

$$V' = R'_A - \sum A = 64.500 - [(126.00) + (-6.00)] = \frac{-55.500}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 516.000 - [(126.00)(5.071) + (-6.00)(1.333)] = \frac{-115.000}{EI}\ \text{(deflection, upward)}$$

Answer: slope at C $= \dfrac{55.5}{EI}$ (anticlockwise) and deflection at C $= \dfrac{115}{EI}$ upward (the hogging moment from the UDL span lifts the overhang more than the 3 kN load pushes it down).

Given data

Beam ABC: A hinge at 0, B roller at 8 m. UDL 2 kN/m over AB; 10 kN at D (4 m from A); overhang $BC = 4$ m with 5 kN at 2 m from B (at 10 m from A). C is the free end at 12 m. $EI$ constant.

        10 kN   2 kN/m          5 kN
 A ------v============ B ------v------- C
 ^  4 m  D     4 m     ^  2 m     2 m

Step 1: Reactions

$R_{A} = 11.75$ kN, $R_{B} = 19.25$ kN (positive = upward; check: $\sum F_y = 31.00$ kN equals the total load 31.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Conjugate beam

The real beam has a hinge at A, a roller at the second support and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

Step 4: Conjugate reaction at A (slope at A)

Taking moments of the conjugate beam about the support at $x = 8$ m (conjugate moment is zero there):

$$R'_A\times8 = (72.67)(5.431) + (52.67)(3.038) = 554.67$$ $$R'_A = \frac{69.333}{EI}\ \text{(slope at A, clockwise)}$$

At D (10 kN) ($x = 4$ m):

$$V' = R'_A - \sum A = 69.333 - [(72.67)] = \frac{-3.333}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 277.333 - [(72.67)(1.431)] = \frac{173.333}{EI}\ \text{(deflection, downward)}$$

At C (free end) ($x = 12$ m):

$$V' = R'_A - \sum A = 69.333 - [(72.67) + (52.67) + (-10.00)] = \frac{-46.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 832.000 - [(72.67)(9.431) + (52.67)(7.038) + (-10.00)(3.333)] = \frac{-190.667}{EI}\ \text{(deflection, upward)}$$

Answer: slope at C $= \dfrac{46}{EI}$ (anticlockwise); deflection at C $= \dfrac{190.67}{EI}$ upward (kNm$^2$/EI and kNm$^3$/EI). (The deflection under the 10 kN load at D is $173.33/EI$ downward.)

In the conjugate beam method, a real beam is replaced by a conjugate beam so that slope = shear and deflection = bending moment in the conjugate beam. For this to hold, the support conditions of the conjugate beam must reproduce the boundary conditions of the real beam.

 Real:      fixed -----------  free       simple --- interior --- simple
 Conjugate: free  -----------  fixed      simple --- hinge ------ simple

The conjugate beam must be statically stable and determinate, which these rules guarantee when the real beam is determinate.

Both methods use the $M/EI$ diagram of the beam to find slope and deflection, but they apply it in different ways.

Example 1: moment-area (cantilever)

Cantilever with end load $P$: $\Delta_B = \dfrac{PL^2}{2EI}\times\dfrac{2L}{3} = \dfrac{PL^3}{3EI}$ (tangent horizontal at the fixed end).

Example 2: conjugate beam (simply supported beam, central load)

The conjugate beam is simply supported with a triangular load of peak $\dfrac{PL}{4EI}$. The conjugate reaction (the slope at the support) is $\dfrac{PL^2}{16EI}$ and the conjugate moment at mid-span is $\dfrac{PL^3}{48EI}$ (the deflection).

Both methods give the same results.

Given data

Simply supported beam AB, $L = 12$ m, 60 kN at C (4 m from A). D is 6 m from B (6 m from A). $EI$ constant.

       60 kN
 A ------v--------D-------- B
 ^  4 m  C   2 m     6 m    ^

Reactions and bending moments

$$R_A = \frac{60\times8}{12} = 40\ \text{kN},\qquad R_B = 20\ \text{kN}$$ $$M_C = 40\times4 = 160\ \text{kNm},\qquad M_D = 20\times6 = 120\ \text{kNm}$$

The $M/EI$ diagram is a triangle A-C-B with peak $160/EI$ at C.

Slope at A

Deviation of B from the tangent at A (Theorem II): area of the triangle $= \tfrac12\times12\times160 = 960$; its centroid is $(0 + 4 + 12)/3 = 5.333$ m from A, i.e. 6.667 m from B.

$$\Delta_{B/A} = \frac{960\times6.667}{EI} = \frac{6400}{EI}$$ $$\theta_A = \frac{\Delta_{B/A}}{L} = \frac{6400}{12\,EI} = \frac{533.33}{EI}\ \text{(clockwise)}$$

Slope at B (Theorem I)

$$\theta_B = \theta_A - \text{Area}_{AB} = \frac{533.33 - 960}{EI} = -\frac{426.67}{EI}$$

so $\theta_B = 426.67/EI$ (anticlockwise).

Deflection at D

Deviation of D from the tangent at A (moment of the $M/EI$ area between A and D about D):

• Triangle A to C: area $= \tfrac12\times4\times160 = 320$, centroid 3.333 m from D: $320\times3.333 = 1066.67$
• Rectangle C to D: $2\times120 = 240$, centroid 1 m from D: $240$
• Triangle (extra height 40 at C, zero at D): $\tfrac12\times2\times40 = 40$, centroid 1.333 m from D: $53.33$

$$\Delta_{D/A} = \frac{1066.67 + 240 + 53.33}{EI} = \frac{1360}{EI}$$

Tangent offset at D: $\theta_A\times6 = \dfrac{533.33\times6}{EI} = \dfrac{3200}{EI}$.

$$y_D = \frac{3200 - 1360}{EI} = \frac{1840}{EI}\ \text{(downward)}$$

Answer: $\theta_A = 533.3/EI$ (clockwise), $\theta_B = 426.7/EI$ (anticlockwise), deflection at D $= 1840/EI$ downward (kNm$^3$/EI, i.e. in metres with $EI$ in kNm$^2$).

Given data

Beam ABC: A hinge at the left end, D at 4 m from A (24 kN down), B roller support at 6 m from A, overhang BC = 4 m (stiffness $2EI$) with UDL 10 kN/m. Portion AB has $EI$. Required: slope and deflection at the free end C.

        24 kN                    10 kN/m
 A ------v------- B ======================= C
 ^   4 m   D 2 m  ^        4 m (2EI)
   (EI)

Step 1: Reactions and bending moments

Moments about B: $R_A\times6 = 24\times2 - 40\times2$, so $R_A = -5.333$ kN (downward); $R_B = 24 + 40 + 5.333 = 69.33$ kN.

$$M_D = -5.333\times4 = -21.33\ \text{kNm},\qquad M_B = -5.333\times6 - 24\times2 = -80\ \text{kNm (hogging)}$$

(Check from the overhang: $M_B = -10\times4^2/2 = -80$ kNm.)

Step 2: $M/EI$ diagram (hogging, so conjugate loads act upward)

Step 3: Conjugate beam

Real hinge A stays a hinge; the interior support B becomes an internal hinge; the free end C becomes a fixed end.

Moment about the hinge B of the part AB (moment of the conjugate beam at B = 0):

$$R'_A\times6 = 42.67(6-2.667) + 42.67(6-5) + 58.67(6-5.333) = 142.2 + 42.67 + 39.11 = 224$$ $$R'_A = \theta_A = \frac{37.33}{EI}\ \text{(anticlockwise)}$$

Step 4: Slope at C (shear at fixed end C)

Total upward load on AB $= 42.67 + 42.67 + 58.67 = 144$; shear at B (continuous through the internal hinge):

$$\theta_B = 37.33 - 144 = -\frac{106.67}{EI}$$

Shear at C: $\theta_C = -106.67 - 53.33 = -\dfrac{160}{EI}$

Step 5: Deflection at C (moment at fixed end C)

$$y_C = \theta_B\times4 + (-53.33)(3) = -426.67 - 160 = -\frac{586.67}{EI}$$

The negative sign means downward.

Answer: slope at C $= 160/EI$ (clockwise); deflection at C $= 586.67/EI$ downward (kNm$^2$ and kNm$^3$ per $EI$ respectively; with $EI$ in kNm$^2$ the answers are in rad and m).

Given data

Simply supported beam ABCD, span $AD = 10$ m: A hinge, D roller, $AB = 3$ m (200 kN at B), $BC = 5$ m (300 kN at C), $CD = 2$ m. $EI$ constant. $x$ is measured from A.

       200 kN         300 kN
 A ------v--------------v----- D
 ^  3 m  B     5 m      C 2 m  ^

Step 1: Reactions

$R_{A} = 200.00$ kN, $R_{D} = 300.00$ kN (positive = upward; check: $\sum F_y = 500.00$ kN equals the total load 500.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Tangent at A

Deviation of the support at $x = 10$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (900.00)(8.000) + (3000.00)(4.500) + (600.00)(1.333) = \frac{21500.00}{EI}$$

That support does not deflect, so $\theta_A = -t/10$:

$$\theta_A = \frac{-2150.000}{EI}\ \text{(clockwise)}$$

At B ($x = 3$ m):

$$\theta = \theta_A + \sum A = -2150.000 + [(900.00)] = \frac{-1250.000}{EI}\ \text{(clockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -6450.000 + [(900.00)(1.000)] = \frac{-5550.000}{EI}\ \text{(downward)}$$

At C ($x = 8$ m):

$$\theta = \theta_A + \sum A = -2150.000 + [(900.00) + (3000.00)] = \frac{1750.000}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -17200.000 + [(900.00)(6.000) + (3000.00)(2.500)] = \frac{-4300.000}{EI}\ \text{(downward)}$$

At the point of maximum deflection ($x = 5.08333$ m):

$$\theta = \theta_A + \sum A = -2150.000 + [(900.00) + (1250.00)] = \frac{0.000}{EI}\ \text{(zero)}$$ $$y = \theta_A\,x + t_{P/A} = -10929.167 + [(900.00)(3.083) + (1250.00)(1.042)] = \frac{-6852.083}{EI}\ \text{(downward)}$$

Maximum deflection

The deflection is maximum where the slope is zero. The slope at B is $-1250/EI$ (clockwise) and at C it is $+1750/EI$, so the zero lies between B and C. In BC, $M/EI = 600/EI$ is constant, so the slope changes by $600/EI$ per metre:

$$-1250 + 600\,(x - 3) = 0 \;\Rightarrow\; x = 3 + 2.083 = 5.083\ \text{m from A}$$

The deflection there (Step 3 tangent offset, from the same formula) is

$$y_{max} = \theta_A x + t_{P/A} = \frac{-6852.08}{EI}$$

Answer: deflection at B $= 5550/EI$ downward; at C $= 4300/EI$ downward (kNm$^3$/EI). Maximum deflection $= 6852.1/EI$ downward at $5.083$ m from A (and 4.917 m from D).

Given data

Beam ABCD: A hinge at 0, B at 4 m (50 kN down), C roller at 8 m, D free end at 11 m. $AB$ and $BC$ have $EI$; $BC$ carries 20 kN/m; overhang $CD = 3$ m has $2EI$ with 30 kN at D. Required: deflection at B and slope at D.

      50 kN    20 kN/m                30 kN
 A ----v---------------------- C ---------v D
 ^ 4 m B        4 m (EI)       ^   3 m (2EI)

Step 1: Reactions

$R_{A} = 33.75$ kN, $R_{C} = 126.25$ kN (positive = upward; check: $\sum F_y = 160.00$ kN equals the total load 160.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Conjugate beam

The real beam has a hinge at A, a roller at the second support C and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

Step 4: Conjugate reaction at A (slope at A)

Taking moments of the conjugate beam about the support at $x = 8$ m (conjugate moment is zero there):

$$R'_A\times8 = (270.00)(5.333) + (196.67)(3.525) = 2133.33$$ $$R'_A = \frac{266.667}{EI}\ \text{(slope at A, clockwise)}$$

At B ($x = 4$ m), deflection needed:

$$V' = R'_A - \sum A = 266.667 - [(270.00)] = \frac{-3.333}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 1066.667 - [(270.00)(1.333)] = \frac{706.667}{EI}\ \text{(deflection, downward)}$$

At D ($x = 11$ m), slope needed:

$$V' = R'_A - \sum A = 266.667 - [(270.00) + (196.67) + (-67.50)] = \frac{-132.500}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 2933.333 - [(270.00)(8.333) + (196.67)(6.525) + (-67.50)(2.000)] = \frac{-465.000}{EI}\ \text{(deflection, upward)}$$

Answer: deflection at B $= \dfrac{706.67}{EI}$ downward; slope at D $= \dfrac{132.5}{EI}$ (anticlockwise, i.e. the free end tilts upward). (Values in kNm$^3$/EI and kNm$^2$/EI; with $EI$ in kNm$^2$ they are metres and radians.)

Given data

Beam ABCD: A hinge at 0, B at 6 m (300 kN), C roller at 12 m, D free end at 14 m with 150 kN. $AC$ ($= 12$ m) has $2EI$; overhang $CD = 2$ m has $EI$.

      300 kN                          150 kN
 A ------v------------------ C --------v D
 ^   6 m B      6 m   (2EI)  ^  2 m (EI)

Step 1: Reactions

$R_{A} = 125.00$ kN, $R_{C} = 325.00$ kN (positive = upward; check: $\sum F_y = 450.00$ kN equals the total load 450.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Conjugate beam

The real beam has a hinge at A, a roller at the second support and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

Step 4: Conjugate reaction at A (slope at A)

Taking moments of the conjugate beam about the support at $x = 12$ m (conjugate moment is zero there):

$$R'_A\times12 = (1125.00)(8.000) + (675.00)(5.333) = 12600.00$$ $$R'_A = \frac{1050.000}{EI}\ \text{(slope at A, clockwise)}$$

At B (300 kN) ($x = 6$ m):

$$V' = R'_A - \sum A = 1050.000 - [(1125.00)] = \frac{-75.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 6300.000 - [(1125.00)(2.000)] = \frac{4050.000}{EI}\ \text{(deflection, downward)}$$

At D (free end) ($x = 14$ m):

$$V' = R'_A - \sum A = 1050.000 - [(1125.00) + (675.00) + (-300.00)] = \frac{-450.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 14700.000 - [(1125.00)(10.000) + (675.00)(7.333) + (-300.00)(1.333)] = \frac{-1100.000}{EI}\ \text{(deflection, upward)}$$

Answer: at the free end D the deflection is $\dfrac{1100}{EI}$ upward (the 150 kN load is small compared with the span load, so the overhang rises), and the rotation is $\dfrac{450}{EI}$ anticlockwise (kNm$^3$/EI and kNm$^2$/EI).

Given data

Overhanging beam ABC: A hinge, B roller, $AB = 4$ m ($2EI$, UDL 10 kN/m), overhang $BC = 2$ m ($EI$) with 100 kN at C.

   10 kN/m (2EI)                100 kN
 A =============== B ------------v C
 ^       4 m       ^    2 m (EI)

Step 1: Reactions

$R_{A} = -30.00$ kN, $R_{B} = 170.00$ kN (positive = upward; check: $\sum F_y = 140.00$ kN equals the total load 140.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Tangent at A

Deviation of the support at $x = 4$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (-173.33)(1.231) = \frac{-213.33}{EI}$$

That support does not deflect, so $\theta_A = -t/4$:

$$\theta_A = \frac{53.333}{EI}\ \text{(anticlockwise)}$$

At C (free end) ($x = 6$ m):

$$\theta = \theta_A + \sum A = 53.333 + [(-173.33) + (-200.00)] = \frac{-320.000}{EI}\ \text{(clockwise)}$$ $$y = \theta_A\,x + t_{P/A} = 320.000 + [(-173.33)(3.231) + (-200.00)(1.333)] = \frac{-506.667}{EI}\ \text{(downward)}$$

Answer: vertical deflection at the free end C $= \dfrac{506.67}{EI}$ (kNm$^3$/EI), downward.

Given data

Beam ABC: A hinge, 30 kN at 3 m from A, B roller 3 m beyond the load (at 6 m), overhang $BC = 2$ m with 20 kN at C. $EI$ constant.

Step 1: Reactions

$R_{A} = 8.33$ kN, $R_{B} = 41.67$ kN (positive = upward; check: $\sum F_y = 50.00$ kN equals the total load 50.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Conjugate beam

The real beam has a hinge at A, a roller at the second support and a free end at the overhang tip, so the conjugate beam has a hinge at A, an internal hinge at the support and a fixed end at the tip. It carries the $M/EI$ diagram of Step 2 as load (sagging $M/EI$ acts downward, hogging acts upward). Conjugate shear $V'$ = slope (positive = clockwise) and conjugate moment $M'$ = deflection (positive = downward).

Step 4: Conjugate reaction at A (slope at A)

Taking moments of the conjugate beam about the support at $x = 6$ m (conjugate moment is zero there):

$$R'_A\times6 = (37.50)(4.000) + (-22.50)(-0.667) = 165.00$$ $$R'_A = \frac{27.500}{EI}\ \text{(slope at A, clockwise)}$$

At the 30 kN load ($x = 3$ m):

$$V' = R'_A - \sum A = 27.500 - [(37.50)] = \frac{-10.000}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 82.500 - [(37.50)(1.000)] = \frac{45.000}{EI}\ \text{(deflection, downward)}$$

At C (free end) ($x = 8$ m):

$$V' = R'_A - \sum A = 27.500 - [(37.50) + (-22.50) + (-40.00)] = \frac{52.500}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 220.000 - [(37.50)(6.000) + (-22.50)(1.333) + (-40.00)(1.333)] = \frac{78.333}{EI}\ \text{(deflection, downward)}$$

Answer: slope at A $= \dfrac{27.5}{EI}$ (clockwise); deflection at C $= \dfrac{78.33}{EI}$ downward (kNm$^2$/EI and kNm$^3$/EI).

Given data

Simply supported beam AB, $L = 12$ m. $AC = CD = DB = 4$ m. 50 kN at D (8 m from A). $EI$ constant.

                       50 kN
 A ------- C ------- D v ----- B
 ^   4 m     4 m         4 m   ^

Step 1: Reactions

$R_{A} = 16.67$ kN, $R_{B} = 33.33$ kN (positive = upward; check: $\sum F_y = 50.00$ kN equals the total load 50.00 kN).

Step 2: Bending moment diagram (sagging +), as $M/EI$ pieces

(Stiffness is in terms of $EI$; "Area" is $\int M/EI\,dx$ in units of $1/EI$; $\bar x$ is the centroid distance from A.)

Step 3: Tangent at A

Deviation of the support at $x = 12$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (533.33)(6.667) + (266.67)(2.667) = \frac{4266.67}{EI}$$

That support does not deflect, so $\theta_A = -t/12$:

$$\theta_A = \frac{-355.556}{EI}\ \text{(clockwise)}$$

At C ($x = 4$ m):

$$\theta = \theta_A + \sum A = -355.556 + [(133.33)] = \frac{-222.222}{EI}\ \text{(clockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -1422.222 + [(133.33)(1.333)] = \frac{-1244.444}{EI}\ \text{(downward)}$$

At D ($x = 8$ m):

$$\theta = \theta_A + \sum A = -355.556 + [(533.33)] = \frac{177.778}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -2844.444 + [(533.33)(2.667)] = \frac{-1422.222}{EI}\ \text{(downward)}$$

At the point of maximum deflection ($x = 6.532$ m):

$$\theta = \theta_A + \sum A = -355.556 + [(355.56)] = \frac{0.003}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -2322.489 + [(355.56)(2.177)] = \frac{-1548.319}{EI}\ \text{(downward)}$$

Slope at the supports

$\theta_A = 355.56/EI$ (clockwise) as found above. At B ($x = 12$ m): $\theta_B = \theta_A + \text{total area} = -355.56 + 800 = 444.44$, so $\theta_B = \dfrac{444.44}{EI}$ (anticlockwise). (Total area of the $M/EI$ triangle $= \tfrac12\times12\times133.33 = 800$.)

Maximum deflection

The slope is zero in the portion AD where $M = 16.667x$. Slope: $\theta(x) = -355.56 + 8.333x^2$. Setting it to zero:

$$x = \sqrt{\frac{355.56}{8.333}} = 6.532\ \text{m from A}$$

(this lies in AD, so the assumption is valid.) Deflection there from the same formula: $y_{max} = \dfrac{1548.3}{EI}$ downward.

Answer: $\theta_A = 355.56/EI$ (clockwise), $\theta_B = 444.44/EI$ (anticlockwise); deflection at C $= 1244.4/EI$, at D $= 1422.2/EI$ (both downward); maximum deflection $= 1548.3/EI$ downward at $6.532$ m from A.