How to derive fixed end moment of a beam fixed at both ends with a point force not in the center

You are on the right track. The way I'd solve it, is the following.

I would break your problem in two subproblems right next to the application of the force F (at point K which is at a distance $x=a^+$).

In that case, there compliance constraints that should be met. I.e. the point K is common at both problem 1 and problem 2, so:

• The shear forces at $x=a^+$ for problem 1 (i.e. $Q_1(a^+)$ or $Q_{K;1}$) and problem 2 (i.e. $Q_2(a^+)$ or or $Q_{K;2}$) should be equal. (Eventually this will not be used but its for the sake of completeness).

$$Q_{K;1}= Q_{K;2}$$

• The bending moments at $x=a^+$ for problem 1 (i.e. $M_1(a^+)$ or $M_{K;1}$) and problem 2 (i.e. $M_2(a^+)$ or or $M_{K;2}$) should be equal.

$$M_1(K;2) = M_{K;2}$$

• The angle at $x=a^+$ for problem 1 (i.e. $dot{y}_1(a^+)$) and problem 2 (i.e. $dot{y}_2(a^+)$) should be equal.

$$dot{y}_1(a^+) = dot{y}_2(a^+)$$

• The displacement at $x=a^+$ for problem 1 (i.e. $y_1(a^+)$) and problem 2 (i.e. $y_2(a^+)$) should be equal.

$$y_1(a^+) = y_2(a^+)$$

So right now from the original problem you had 4 unknowns ($M_A, R_A, M_B, R_B$) and two equations (equilibrium $sum F_x, sum M$). By breaking the problem up you are introducing two more unknowns (the internal reactions at point K, i.e. $Q_K, M_K$) and you have 4 more additional constraints. Therefore this is a system of 6 equation with 6 unknowns.

Solving the system

I am not keen in performing the mathematics ( especially for problem 2 there is a tricky part depending on how you obtain the coordinate system), so I will perform only problem 1, for you.

Problem 1

To revise, problem 1 will look like :

Therefore you have:

• $Q_1(x)= begin{cases} R_A & 0<x<a -F+R_A & a<x<a+ end{cases}$

• $M_1(x)= begin{cases} -M_A +R_A x & 0<x<a -M_A +R_A x - Fcdot (x-a) & a<x<a+ end{cases}$

• $dot{y}_(K_1) $ this will be calculated from $$EI y_1'' = M_1(x)$$ by integrating $$int_0^{a^+} EI y_1'' dx = int_0^{a^+}M_1(x)dx$$

$$ EI int_0^{a^+} y_1''dx = int_0^{a^+}M_1(x)dx$$

because $a^+$ is infinitesimally close to a and , $int_0^{a} y_1''dx= y_1'(a)= y'_{K;1} $ we can write

$$ EI y'_{K;1} = int_0^{a}M_1(x)dx$$ $$ EI y'_{K;1} = int_0^{a}( -M_A +R_A x)dx$$ $$ EI y'_{K;1} = left[-M_A x +frac{R_A}{2} x^2right]_0^a$$ Finally: $$y'_{K;1} = frac{1}{EI}left(-M_A a+frac{R_A}{2} a^2right)$$

• For the deflection of the beam at point K using a similar approach $$int_0^{a^+} left(int_0^{a^+} EI y_1'' dx right)dx= int_0^{a^+} left(int_0^{a^+}M_1(x)dxright)dx$$

$$ y_{K;1}= frac{1}{EI}int_0^{a} left( -M_A x +frac{R_A}{2} x^2 right)dx$$ $$ y_{K;1}= frac{1}{EI} left[ -frac{M_A}{2} x^2 +frac{R_A}{6} x^3 right]_0^a$$ $$ y_{K;1}= frac{1}{EI} left( -frac{M_A}{2} a^2 +frac{R_A}{6} a^3 right)$$

Problem 2

The way we broke up the problem (at $a^+$) simplifies the equation for the second par. The tricky part about problem two is that you integrating from a to L.

• $Q_2(x)= -R_B $ for $ a<xle L$

• $M_2(x)= -M_B+ R_Bcdot(L-x) $ for $ a<xle L$

• $dot{y}_(K_2) $ this will be calculated from $$EI y_2'' = M_2(x)$$ then $$int_{a^+}^L EI y_2'' dx = int{a^+}^L M_2(x)dx$$

Here, because $y_2'(x=a+ )=y'_{K;2}$ and $y_2'(x=a)=0$ you obtain $$EI (0 - y'_{K;2}) = int_{a}^L -M_B+ R_Bcdot(L-x) dx$$

$$ y'_{K;2} = -frac{1}{EI}int_{a}^L -M_B+ R_Bcdot(L-x) dx$$ $$ y'_{K;2} = -frac{1}{EI}left[ -M_B x+ R_Bcdot(L x-frac{x^2}{2}) right]_{a}^L$$ $$ y'_{K;2} = -frac{1}{EI}left(left( -M_B L+ R_Bcdot(L^2-frac{L^2}{2}) right) - left( -M_B a+ R_Bcdot(L a-frac{a^2}{2}) right)right)$$

There is another step with the displacement, which I am not keen to go through (I leave it as an exercise to the interested reader).

System of equation from constraints

Going back to the constraints you should get for

• The shear forces at $x=a^+$ .

$$Q_{K;1}= Q_{K;2} rightarrow -F+R_A = -R_B $$

• The bending moments .

$$M_1(K;2) = M_{K;2} rightarrow -M_A +R_A a= -M_B+ R_Bcdot(L-a)$$

• The angle at $x=a^+$ .

$$dot{y}_1(a^+) = dot{y}_2(a^+)$$

$$frac{1}{EI}left(-M_A a +frac{R_A}{2} a^2right) = -frac{1}{EI}left(left( -M_B L+ R_Bcdot(L^2-frac{L^2}{2}) right) - left( -M_B a+ R_Bcdot(L a-frac{a^2}{2}) right)right)$$

• The displacement at $x=a^+$ for problem 1 (i.e. $y_1(a^+)$) and problem 2 (i.e. $y_2(a^+)$) should be equal.

$$y_1(a^+) = y_2(a^+)rightarrow$$

$$ frac{1}{EI} left( -frac{M_A}{2} a^2 +frac{R_A}{6} a^3 right) = text{what ever you derive from the integration for }y_2(a^+) $$

So finally you have 5 equations with 5 unknowns ($Q_{K;1}$ is not used) and you solve the system.

final solution

What you should get (if all the above is done correctly):