15. The elastic deflection of a weightless point-loaded cantilever is given by the equation: δ = Fx! 6EI (3L − x) Here, F is the point-loading force, L is the point on the beam where the force is applied, x is the position along the beam ranging from 0 to L, and E and I are constants. Consider a tree branch, which is a natural cantilever. A monkey initially sits on the mid-point of the branch, causing the mid-point to deflect down by some amount 𝛿! from horizontal. The monkey then shifts to the end of the branch, causing the mid-point to deflect to a new amount 𝛿! from horizontal. What is the ratio 𝛿!/𝛿!? a. 1.6 b. 2.5 c. 3.2 d. 5.0

15. The elastic deflection of a weightless point-loaded cantilever is given by the
equation:

δ = Fx!
6EI (3L − x)
Here, F is the point-loading force, L is the point on the beam where the force is
applied, x is the position along the beam ranging from 0 to L, and E and I are
constants. Consider a tree branch, which is a natural cantilever. A monkey initially
sits on the mid-point of the branch, causing the mid-point to deflect down by some
amount 𝛿! from horizontal. The monkey then shifts to the end of the branch, causing
the mid-point to deflect to a new amount 𝛿! from horizontal. What is the ratio 𝛿!/𝛿!?
a. 1.6
b. 2.5
c. 3.2
d. 5.0
The correct ratio of the mid-point deflections for the monkey-on-a-tree-branch cantilever problem is
δ₂/δ₁ = 2.5, which matches option B.

Deflection formula for a point-loaded cantilever

For a weightless cantilever beam of length L with a point load F applied at position x from the fixed end,
the vertical deflection at that same position x is given by:

δ(x) = (F x² / 6 E I) (3L − x)

Here, F is the point-loading force, L is the beam length, E is Young’s modulus, and I is the second moment
of area. For comparing deflections under the same load, the common factors F, E and I cancel, so only the
term x²(3L − x) affects the ratio.

Case 1: Monkey at the mid-point

The tree branch behaves like a horizontal cantilever fixed at the trunk and free at the outer end, and the
mid-point is at a distance L/2 from the fixed end. When the monkey sits at this mid-point, it acts as a
point load F at x = L/2, causing the mid-point to deflect by δ₁.

Substituting x = L/2 into the deflection formula:

δ₁ = F (L/2)² / (6 E I) × (3L − L/2)
= F L² / (24 E I) × (5L / 2)
= 5 F L³ / (48 E I)

Thus, δ₁ is proportional to 5/48 for a given beam and load.

Case 2: Monkey at the free end

When the monkey moves to the free end, the same load F is now applied at a distance a = L from the fixed end,
but the deflection is still observed at the mid-point x = L/2. For a point load at distance a affecting a
section at x ≤ a, the deflection is:

δ(x) = F x² / (6 E I) (3a − x)

Taking a = L and x = L/2:

δ₂ = F (L/2)² / (6 E I) × (3L − L/2)
= F L² / (24 E I) × (5L / 2)
= 25 F L³ / (96 E I)

Therefore, δ₂ is proportional to 25/96, which is larger than δ₁ because moving the load to the free end
increases the bending moment at the mid-point.

Ratio δ₂/δ₁ and correct option

Taking the ratio of the two mid-point deflections eliminates the common factors F, E, I and L³:

δ₂/δ₁ = (25/96) / (5/48)
= (25/96) × (48/5)
= 25/10
= 2.5

Hence the ratio of the mid-point deflections is δ₂/δ₁ = 2.5, so the correct answer is Option B.

Analysis of all options

  • Option A (1.6): Too small; it underestimates the increase in mid-point deflection when the load moves from the mid-point to the free end.
  • Option B (2.5): Matches the exact ratio from the cantilever deflection equation and correctly reflects the stronger bending at the mid-point.
  • Option C (3.2): Too large; it implies a steeper dependence on load position than the x²(3L − x) relationship.
  • Option D (5.0): Unrealistically large and not compatible with the linear-elastic cantilever deflection formulas.

Engineering insight

This monkey-on-a-tree-branch example shows that the elastic deflection of a weightless point-loaded cantilever
is highly sensitive to the location of the applied load. Designers must consider how moving loads toward the
free end rapidly increases deflection and bending moments, which is critical for safe and efficient structural
design.

 

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