Table 1: Cantilever Beams

Shear, bending moments, slopes, and deflections of cantilever beams under various loadings.

Diagrams	Equations
1. Moment load at the free-end	\( M_A = -M \) \( \theta_\text{max} = -\dfrac{ML}{EI} \) \( \theta_B = \theta_\text{max} \) \( \delta_\text{max} = -\dfrac{ML^2}{2EI} \) \( \delta_B = \delta_\text{max} \) At any point: \[ \theta = -\dfrac{Mx}{EI} \] \[ y = -\dfrac{Mx^2}{2EI} \]
2. Moment load on the span	\( M_A = -M \) \( \theta_\text{max} = -\dfrac{Ma}{EI} \) \( \theta_B = \theta_C = \theta_\text{max} \) \( \delta_\text{max} = -\dfrac{Ma}{2EI} (2L - a) \) \( \delta_B = \delta_\text{max} \) \( 0 \le x \le a \): \[ \theta = -\dfrac{Mx}{EI} \] \[ y = -\dfrac{Mx^2}{2EI} \] \( a \le x \le L \): \[ \theta = -\dfrac{Ma}{EI} \] \[ y = -\dfrac{Ma}{2EI} (2x - a) \]
3. Concentrated load at the free-end	\( R_A = P \) \( M_A = -PL \) \( \theta_\text{max} = -\dfrac{PL^2}{2EI} \) \( \theta_B = \theta_\text{max} \) \( \delta_\text{max} = -\dfrac{PL^3}{3EI} \) \( \delta_B = \delta_\text{max} \) At any point: \[ \theta = -\dfrac{P}{2EI} \left( 2Lx - x^2 \right) \] \[ y = -\dfrac{P}{6EI} \left( 3Lx^2 - x^3 \right) \]
4. Concentrated load on the span	\( R_A = P \) \( M_A = -Pa \) \( \theta_\text{max} = -\dfrac{Pa^2}{2EI} \) \( \theta_B = \theta_C = \theta_\text{max} \) \( \delta_\text{max} = -\dfrac{Pa^2}{6EI} (3L - a) \) \( \delta_B = \delta_\text{max} \) \( 0 \le x \le a \): \[ \theta = -\dfrac{P}{2EI} \left( 2ax - x^2 \right) \] \[ y = -\dfrac{P}{6EI} \left( 3ax^2 - x^3 \right) \] \( a \le x \le L \): \[ \theta = -\dfrac{Pa^2}{2EI} \] \[ y = -\dfrac{Pa^2}{6EI} \left( 3x - a \right) \]
5. Uniform load over the entire span	\( R_A = wL \) \( M_A = -\dfrac{wL^2}{2} \) \( \theta_\text{max} = -\dfrac{wL^3}{6EI} \) \( \theta_B = \theta_\text{max} \) \( \delta_\text{max} = -\dfrac{wL^4}{8EI} \) \( \delta_B = \delta_\text{max} \) At any point: \[ \theta = -\dfrac{w}{6EI} \left( x^3 - 3Lx^2 + 3L^2x \right) \] \[ y = -\dfrac{w}{24EI} \left( x^4 -4Lx^3 + 6L^2 x^2 \right) \]
6. Uniform load applied partially along the span	\( R_A = wa \) \( M_A = -\dfrac{wa^2}{2} \) \( \theta_\text{max} = -\dfrac{wa^3}{6EI} \) \( \theta_B = \theta_C = \theta_\text{max} \) \( \delta_\text{max} = -\dfrac{wa^3}{24EI} (4L - a) \) \( \delta_B = \delta_\text{max} \) \( 0 \le x \le a \): \[ \theta = -\dfrac{w}{6EI} \left( x^3 - 3ax^2 + 3a^2x \right) \] \[ y = -\dfrac{w}{24EI} \left( x^4 - 4ax^3 + 6a^2x^2 \right) \] \( a \le x \le L \): \[ \theta = -\dfrac{wa^3}{6EI} \] \[ y = -\dfrac{wa^3}{24EI} (a - 4x) \]
7. Triangular load over the entire span	\( R_A = \dfrac{wL}{2} \) \( M_A = -\dfrac{wL^2}{6} \) \( \theta_\text{max} = -\dfrac{wL^3}{24EI} \) \( \theta_B = \theta_\text{max} \) \( \delta_\text{max} = -\dfrac{wL^4}{30EI} \) \( \delta_B = \delta_\text{max} \) At any point: \[ \theta = -\dfrac{w}{24EIL} \left( 4L^3x - 6L^2x^2 + 4Lx^3 - x^4 \right) \] \[ y = -\dfrac{w}{120EIL} \left( 10L^3x^2 - 10L^2x^3 + 5Lx^4 - x^5 \right) \]
8. Triangular load applied partially along the span	\( R_A = \dfrac{wa}{2} \) \( M_A = -\dfrac{wa^2}{6} \) \( \theta_\text{max} = -\dfrac{wa^3}{24EI} \) \( \theta_B = \theta_C = \theta_\text{max} \) \( \delta_\text{max} = -\dfrac{wa^3}{120EI} (5L - a) \) \( \delta_B = \delta_\text{max} \) \( 0 \le x \le a \): \[ \theta = -\dfrac{w}{24EIa} \left( 4a^3x - 6a^2x^2 + 4ax^3 - x^4 \right) \] \[ y = -\dfrac{w}{120EIa} \left( 10a^3x^2 - 10L^2x^3 + 5ax^4 - x^5 \right) \] \( a \le x \le L \): \[ \theta = -\dfrac{wa^3}{24EI} \] \[ y = -\dfrac{wa^3}{120EI} (a - 5x) \]

Table 1.1: Equations according to degree \( n \) of \( M/EI \) Diagram

Degree of \( M / EI \), \( n \)
Maximum rotation \( EI \ \theta = \dfrac{M_R \ L}{n + 1} \)
Moment reaction, \( M_R \)
Maximum deflection, \( EI \ \delta = \dfrac{M_R \ L^2}{n + 2} \)

Case	\( n \)	\( M_R \)	\( EI \ \theta \)	\( EI ~ \delta \)
1. Moment Load	\( 0 \)	\( M \)	\( \dfrac{M_R \ L}{1} \)	\( \dfrac{M_R \ L^2}{2} \)
2. Concentrated Load	\( 1 \)	\( PL \)	\( \dfrac{M_R \ L}{2} \)	\( \dfrac{M_R \ L^2}{3} \)
3. Uniform Load	\( 2 \)	\( \dfrac{w \ L^2}{2} \)	\( \dfrac{M_R \ L}{3} \)	\( \dfrac{M_R \ L^2}{4} \)
4. Triangular Load	\( 3 \)	\( \dfrac{w \ L^2}{6} \)	\( \dfrac{M_R \ L}{4} \)	\( \dfrac{M_R \ L^2}{5} \)