# Left end simply supported right end fixed for concentrated intermediate load

## Values for calculation

$l$ $\mathrm{mm}$
$a$ $\mathrm{mm}$
$W$ $\mathrm{N}$
$E$ $\mathrm{MPa}$
$I$ $\mathrm{mm^4}$
$x$ $\mathrm{mm}$

## Calculation

### Vertical end reactions $R_A$

$$R_A=\cfrac{W}{2\cdot l^3}\cdot\left(l-a\right)^2\cdot\left(2\cdot l+a\right)$$

### Reaction end moment $M_A$

$$M_A=0$$

### Angular displacement $θ_A$

$$θ_A=\cfrac{-W\cdot a}{4\cdot E\cdot I\cdot l}\cdot\left(l-a\right)^2$$

### Deflection $y_A$

$$y_A=0$$

### Vertical end reactions $R_B$

$$R_B=\cfrac{W\cdot a}{2\cdot l^3}\cdot\left(3\cdot l^2-a^2\right)$$

### Reaction end moment $M_B$

$$M_B=\cfrac{-W\cdot a}{2\cdot l^2}\cdot\left(l^2-a^2\right)$$

### Angular displacement $θ_B$

$$θ_B=0$$

### Deflection $y_B$

$$y_B=0$$

### Max. moment +

$$M_{max+}=\cfrac{W\cdot a}{2\cdot l^3}\cdot\left(l-a\right)^2\cdot\left(2\cdot l+a\right)$$

### Max. moment -

$$M_{max-}=M_B$$

### Max. deflection

$\text{if }\ a \geq 0.414\cdot l$
$$y_{max}=\cfrac{-W\cdot a}{6\cdot E\cdot I}\cdot\left(l-a\right)^2\cdot\left(\cfrac{a}{2\cdot l+a}\right)^{1/2}$$
$\text{else}$
$$y_{max}=\cfrac{-W\cdot a\cdot\left(l^2-a^2\right)^3}{3\cdot E\cdot I\cdot\left(3\cdot l^2-a^2\right)^2}$$

### Transverse shear

$\text{if }\ x\le a$
$$V=R_A$$
$\text{else}$
$$V=R_A-W$$

### Bending moment

$\text{if }\ x\le a$
$$M=M_A+R_A\cdot x$$
$\text{else}$
$$M=M_A+R_A\cdot x-W\cdot\left(x-a\right)$$

### Slope

$\text{if }\ x\le a$
$$θ=θ_A+\cfrac{M_A\cdot x}{E\cdot I}+\cfrac{R_A\cdot x^2}{2\cdot E\cdot I}$$
$\text{else}$
$$θ=θ_A+\cfrac{M_A\cdot x}{E\cdot I}+\cfrac{R_A\cdot x^2}{2\cdot E\cdot I}-\cfrac{W}{2\cdot E\cdot I}\cdot\left(x-a\right)^2$$

### Deflection

$\text{if }\ x\le a$
$$y=y_A+θ_A\cdot x+\cfrac{M_A\cdot x^2}{2\cdot E\cdot I}+\cfrac{R_A\cdot x^3}{6\cdot E\cdot I}$$
$\text{else}$
$$y=y_A+θ_A\cdot x+\cfrac{M_A\cdot x^2}{2\cdot E\cdot I}+\cfrac{R_A\cdot x^3}{6\cdot E\cdot I}-\cfrac{W}{6\cdot E\cdot I}\cdot\left(x-a\right)^3$$