A cantilever beam AB is 5 m long and fixed at B. A downward point load of 1800 kN acts at the free end A, and a uniformly distributed load of 500 kN/m acts over the 4 m nearest the fixed support (from 1 m to 5 m measured from A). Find the maximum shear force and the maximum bending moment, and describe the shear and bending-moment diagrams.

Step-by-step solution

1. 1

   Recognise where the extremes must occur. On a cantilever carrying only downward loads, both the shear and the bending moment grow monotonically from zero at the free end to their largest magnitudes at the fixed end. So you can compute the two maxima directly at $B$ without hunting for interior turning points.

2. 2

   Replace the distributed load by its resultant. The UDL acts over $4$ m:

   $W=wL_u=500\times 4=2000\ \text{kN}$

   acting at the centroid of its span, which is $2$ m from $B$ (equivalently $3$ m from $A$). This resultant is legitimate for finding reactions and end moments, but the diagrams themselves must still be built from the distributed load.

3. 3

   Find the reactions at the fixed support. Vertical equilibrium gives the reaction force

   $R_B=P+W=1800+2000=3800\ \text{kN}\ (\uparrow)$

   Moment equilibrium about $B$ gives the fixing moment

   $M_B=P\times 5+W\times 2=1800(5)+2000(2)=9000+4000=13000\ \text{kN}\!\cdot\!\text{m}$

4. 4

   Build the shear force diagram from the free end. Measuring $x$ from $A$:

   $0\le x\le 1:\quad V(x)=1800\ \text{kN (constant)}$

   $1\le x\le 5:\quad V(x)=1800+500(x-1)\ \text{(straight line)}$

   So the SFD is a horizontal line at $1800$ kN over the first metre, then a straight ramp rising to $1800+500(4)=3800$ kN at $B$, where the reaction closes the diagram back to zero.

5. 5

   Build the bending moment diagram. Since $M(x)=\int_0^x V\,dx$ in magnitude (hogging on a cantilever):

   $0\le x\le 1:\quad M(x)=1800x\ \text{(linear, reaching }1800\ \text{kN}\!\cdot\!\text{m at }x=1)$

   $1\le x\le 5:\quad M(x)=1800x+250(x-1)^2\ \text{(parabolic)}$

   At $x=5$: $M=9000+250(16)=9000+4000=13000\ \text{kN}\!\cdot\!\text{m}$, matching the fixing moment found from equilibrium.

6. 6

   State the maxima.

   $V_{\max}=3800\ \text{kN},\qquad M_{\max}=13000\ \text{kN}\!\cdot\!\text{m}$

   both at the fixed end $B$; both are zero at the free end $A$.

   $\boxed{V_{\max}=3800\ \text{kN},\quad M_{\max}=13000\ \text{kN}\!\cdot\!\text{m}}$

7. 7

   Cross-check the two routes. The moment at $B$ was obtained twice — once by statics ($P\cdot 5+W\cdot 2$) and once by integrating the shear diagram ($1800x+250(x-1)^2$ at $x=5$) — and both give $13000\ \text{kN}\!\cdot\!\text{m}$. Agreement between the equilibrium value and the integrated diagram is the standard check that the diagrams were drawn consistently.