Moment area: girder with 120 kN and 80 kN

Given data

Simply supported girder, span 14 m. 120 kN at 3 m from the left support A; 80 kN at 4.5 m from the right support B (i.e. 9.5 m from A). $I = 16\times10^8$ mm$^4 = 1.6\times10^{-3}$ m$^4$, $E = 210$ kN/mm$^2 = 2.1\times10^8$ kN/m$^2$, so

$$EI = 2.1\times10^8\times1.6\times10^{-3} = 3.36\times10^5\ \text{kNm}^2$$

Part 1: Moment-area method

Step 1: Reactions

$R_{A} = 120.00$ kN, $R_{B} = 80.00$ kN (positive = upward; check: $\sum F_y = 200.00$ kN equals the total load 200.00 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 $x = 14$ m from the tangent at A (Theorem II, moment of the $M/EI$ area about that support):

$$t = (540.00)(12.000) + (2340.00)(7.750) + (810.00)(3.000) = \frac{27045.00}{EI}$$

That support does not deflect, so $\theta_A = -t/14$:

$$\theta_A = \frac{-1931.786}{EI}\ \text{(clockwise)}$$

At the 120 kN load ($x = 3$ m):

$$\theta = \theta_A + \sum A = -1931.786 + [(540.00)] = \frac{-1391.786}{EI}\ \text{(clockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -5795.357 + [(540.00)(1.000)] = \frac{-5255.357}{EI}\ \text{(downward)}$$

At the 80 kN load ($x = 9.5$ m):

$$\theta = \theta_A + \sum A = -1931.786 + [(540.00) + (2340.00)] = \frac{948.214}{EI}\ \text{(anticlockwise)}$$ $$y = \theta_A\,x + t_{P/A} = -18351.964 + [(540.00)(7.500) + (2340.00)(3.250)] = \frac{-6696.964}{EI}\ \text{(downward)}$$

At the point of maximum deflection ($x = 6.866$ m):

$$\theta = \theta_A + \sum A = -1931.786 + [(540.00) + (1391.76)] \approx 0\ \text{(horizontal tangent)}$$ $$y = \theta_A\,x + t_{P/A} = -13263.641 + [(540.00)(4.866) + (1391.76)(1.933)] = \frac{-7945.729}{EI}\ \text{(downward)}$$

Maximum deflection

Between the loads the bending moment is constant, $M = 360$ kNm, so the slope changes by $360/EI$ per metre. The slope is $-1391.79/EI$ under the 120 kN load ($x = 3$ m), so it is zero at

$$-1391.79 + 360\,(x - 3) = 0 \;\Rightarrow\; x = 6.866\ \text{m from A}$$

where the deflection is $y_{max} = \dfrac{7945.7}{EI}$ downward (same formula as above).

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

Part 2: Verification by the conjugate beam method

Step 1: Reactions

$R_{A} = 120.00$ kN, $R_{B} = 80.00$ kN (positive = upward; check: $\sum F_y = 200.00$ kN equals the total load 200.00 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 conjugate of a simply supported beam is a simply supported beam of the same span. 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 $x = 14$ m (conjugate moment is zero there):

$$R'_A\times14 = (540.00)(12.000) + (2340.00)(7.750) + (810.00)(3.000) = 27045.00$$ $$R'_A = \frac{1931.786}{EI}\ \text{(slope at A, clockwise)}$$

At the 120 kN load ($x = 3$ m):

$$V' = R'_A - \sum A = 1931.786 - [(540.00)] = \frac{1391.786}{EI}\ \text{(slope, clockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 5795.357 - [(540.00)(1.000)] = \frac{5255.357}{EI}\ \text{(deflection, downward)}$$

At the 80 kN load ($x = 9.5$ m):

$$V' = R'_A - \sum A = 1931.786 - [(540.00) + (2340.00)] = \frac{-948.214}{EI}\ \text{(slope, anticlockwise)}$$ $$M' = R'_A\,x - \sum A(x-\bar x) = 18351.964 - [(540.00)(7.500) + (2340.00)(3.250)] = \frac{6696.964}{EI}\ \text{(deflection, downward)}$$

The conjugate beam gives the same slopes and deflections as the moment-area method, so the results are verified.

Answer: slope at A $= 0.00575$ rad (clockwise); under the 120 kN load: slope $0.00414$ rad (clockwise), deflection $15.64$ mm downward; under the 80 kN load: slope $0.00282$ rad (anticlockwise), deflection $19.93$ mm downward. Maximum deflection $= 23.65$ mm downward at $6.866$ m from the left support.