Shear and Moment Functions for a Cantilever: 04-BS-6, May 2015, Q7

Question 7: Shear and Moment Functions for a Cantilever (20 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Cantilever: 4 kN point load at x=2 m, w=1.5 kN/m over x∈[2,6] m; resulting shear-force and bending-moment diagrams.

Given.

Quantity	Value
Fixed support	at x = 0; free end at x = 6 m
Concentrated load	4 kN, downward, at x = 2 m
UDL w	1.5 kN/m over x ∈ [2, 6] m

Find. V(x) and M(x) as explicit functions over the whole span; the shear and moment diagrams with all critical points labelled.

Approach. For a cantilever it is simplest to cut at a general x and sum forces and moments of everything to the RIGHT of the cut (toward the free end), which never requires the fixed-end reactions explicitly; use $V=dM/dx$ as a running check between the two regions.

Region 1 (0 ≤ x ≤ 2 m) — both the point load and the full UDL are to the right of the cut. UDL resultant $=1.5(4)=6$ kN at its centroid $x=4$ m: $$V(x) = -[4+6] = \boxed{-10\ \text{kN (constant)}}$$ $$M(x) = -[4(2-x)+6(4-x)] = \boxed{10x-32}\ \ [\text{kN}\cdot\text{m; x in m}]$$ Check: $M(0)=-32$ kN·m, matching the fixed-end reaction moment by statics ($4(2)+6(4)=8+24=32$ kN·m, hogging).
Region 2 (2 ≤ x ≤ 6 m) — only the remaining UDL is to the right. UDL resultant over $[x,6]$ is $1.5(6-x)$ at its own centroid, distance $(6-x)/2$ from the cut: $$V(x) = -1.5(6-x) = \boxed{1.5x-9}\ \ [\text{kN}]$$ $$M(x) = -0.75(6-x)^2 = \boxed{-0.75x^2+9x-27}\ \ [\text{kN}\cdot\text{m}]$$ Continuity check at x=2: $V(2^-)=-10$, $V(2^+)=1.5(2)-9=-6$ (a jump of $+4$ kN, exactly the point load, as required); $M(2^-)=-12$, $M(2^+)=-0.75(4)+18-27=-12$ (continuous, as required since a point load creates a shear jump but not a moment jump). At the free end, $V(6)=0$, $M(6)=-27+54-27=0$, both correctly zero.
Critical points. $|V|$ is maximum at the fixed end, $V(0)=-10$ kN; $|M|$ is also maximum at the fixed end, $M(0)=-32$ kN·m (hogging throughout — a cantilever under downward loads only never changes moment sign, so there is no inflection point).

Final Results.

Quantity	Value
V(x), 0≤x&le2	−10 kN (constant)
M(x), 0≤x&le2	10x−32 kN·m
V(x), 2≤x&le6	1.5x−9 kN
M(x), 2≤x&le6	−0.75x²+9x−27 kN·m
Maximum \|V\|, \|M\|	10 kN, 32 kN·m, both at the fixed support (x=0)

Topic: Direct free-body derivation of shear/moment functions for a cantilever with a point load and a partial UDL, cutting from the free end.

Key relations: $V(x)=dM/dx$; summing loads to the right of a cut avoids ever needing the fixed-end reactions; a concentrated load produces a JUMP in V but not in M, while a concentrated couple (as in Question 2) produces the opposite.

Why this works: For a cantilever, the free (unsupported) end is the simpler side to reference because no unknown reactions act there — every quantity needed for the cut is just the sum of the KNOWN applied loads between the cut and that free end.

Common pitfall: Forgetting that the UDL only spans part of the member (2 to 6 m, not the whole 6 m), which changes both which loads are "to the right" of a Region-1 cut and the centroid distance used for the moment.

Source: Hibbeler, Mechanics of Materials, 10th ed., Ch. 6 (shear and moment diagrams).

Question 7 of 8: Shear and Moment Functions for a Cantilever

Question 7: Shear and Moment Functions for a Cantilever (20 marks)