---
title: "Inclined load on a rectangular cantilever"
subject: "Mechanics of Solids"
chapter: 7
key: "semester_5/mechanics_of_solids:inclined-load-cantilever-rectangular"
source: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/inclined-load-cantilever-rectangular
site: BCT Engineering
---

# Inclined load on a rectangular cantilever

Mechanics of Solids · Chapter 7: Non Symmetrical Bending of Straight Beams

- Asked in: year not given, 8 marks
- Video: [Numerical on bending stress with inclined load](https://www.youtube.com/watch?v=f0_kCSHE9yw) — Rachana Kaushik
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/mechanics_of_solids/inclined-load-cantilever-rectangular
- Whole chapter: https://www.bctengineer.com/md/semester_5/mechanics_of_solids/chapter-7

## Question

A cantilever of length 1.5 m has a rectangular section 80 mm wide and 120 mm deep (depth vertical). A load of 2 kN acts at the free end in a plane inclined at $30^\circ$ to the vertical axis of the section, passing through the centroid. Find (a) the maximum bending stress and its location, (b) the position of the neutral axis, (c) the resultant deflection of the free end. Take $E = 200$ GPa.

## Short answer

By resolving the inclined 2 kN load into principal axes and superposing bending, the maximum bending stress is 25.3 MPa at a corner. The neutral axis is at 52.4° to the horizontal axis. The resultant free-end deflection is 1.39 mm directed at 52.4° from the vertical.

## Answer

#### Section properties (principal axes)

$x$ horizontal (width 80 mm), $y$ vertical (depth 120 mm):

$$
I_x = \frac{80\times120^3}{12} = 1.152\times10^{7}\ \text{mm}^4,\qquad I_y = \frac{120\times80^3}{12} = 5.12\times10^{6}\ \text{mm}^4
$$

The section is symmetric, so $x, y$ are principal axes ($I_{xy}=0$).

#### Load components

$$
\begin{aligned}
P_y &= 2000\cos30^\circ = 1732.1\ \text{N} \\
P_x &= 2000\sin30^\circ = 1000.0\ \text{N}
\end{aligned}
$$

Moments at the fixed end ($L = 1500$ mm):

$$
\begin{aligned}
M_x &= P_yL = 1732.05\times1500 = 2.598\times10^6\ \text{N mm} \\
M_y &= P_xL = 1000\times1500 = 1.500\times10^6\ \text{N mm}
\end{aligned}
$$

#### (a) Maximum stress

It occurs at a corner farthest from the neutral axis, at $x = \pm40$, $y = \pm60$ mm:

$$
\begin{aligned}
\sigma_{max} &= \frac{M_x\,y}{I_x} + \frac{M_y\,x}{I_y} = \frac{2.598\times10^6\times60}{11.52\times10^6} + \frac{1.5\times10^6\times40}{5.12\times10^6} \\
&= 13.53 + 11.72 = 25.25\ \text{MPa}
\end{aligned}
$$

The top fibre is in tension from $P_y$. The horizontal component bends the beam sideways, so the side away from it is in tension at the support. The corner where both tensions add carries $+25.3$ MPa and the diagonally opposite corner carries $-25.3$ MPa.

#### (b) Neutral axis

$$
\tan\alpha = \frac{M_yI_x}{M_xI_y} = \tan\phi\,\frac{I_x}{I_y} = \tan30^\circ\times\frac{11.52}{5.12} = 0.5774\times2.25 = 1.299
$$

$$
\alpha = 52.41^\circ
$$

The neutral axis passes through the centroid at $52.4^\circ$ to the x-axis. It is turned towards the weaker axis (the y-axis, which has the smaller moment of inertia), compared with the position $30^\circ$ that would be perpendicular to the load plane.

```text
      y
      |  load plane (30 deg from y)
      | /     NA (52.4 deg from x)
      |/    /
  ----+----/----- x
     /|   /
```

#### (c) Deflection of the free end

$$
\begin{aligned}
\delta_y &= \frac{P_yL^3}{3EI_x} = \frac{1732.05\times1500^3}{3\times200\times10^3\times11.52\times10^6} = 0.846\ \text{mm} \\
\delta_x &= \frac{P_xL^3}{3EI_y} = \frac{1000\times1500^3}{3\times200\times10^3\times5.12\times10^6} = 1.099\ \text{mm}
\end{aligned}
$$

$$
\delta = \sqrt{\delta_x^2 + \delta_y^2} = 1.386\ \text{mm}
$$

The direction is at $\tan^{-1}(\delta_x/\delta_y) = 52.4^\circ$ from the vertical: this is **not** the $30^\circ$ of the load, and it is perpendicular to the neutral axis.

**Answer:** (a) $\sigma_{max} = 25.3$ MPa at a corner; (b) neutral axis at $52.4^\circ$ to the horizontal axis through the centroid; (c) resultant deflection $1.39$ mm at $52.4^\circ$ to the vertical.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. For the 2 kN load inclined at 30 degrees to the vertical, the moments at the fixed end are
   A. Mx = 2000 sin30 x L, My = 2000 cos30 x L
   B. Mx = 2000 cos30 x L (about x), My = 2000 sin30 x L (about y)
   C. Both equal 2000 x L
   D. Mx = 2000 x L only
   Answer: B. The vertical component 1732 N gives Mx = 2.598e6 N mm, the horizontal component 1000 N gives My = 1.5e6 N mm. (Key point: Resolving load into moments)

2. Maximum bending stress is found by
   A. Taking the larger of the two stresses separately
   B. Adding Mx y/Ix and My x/Iy at the corner (x = 40, y = 60) where both effects add
   C. Using M = 2000 x L and I = Ix
   D. Averaging the two stresses
   Answer: B. 13.53 + 11.72 = 25.25 MPa at the corner where both tensions add. (Key point: Maximum stress)

3. The neutral axis makes what angle with the x-axis?
   A. 30 degrees
   B. 52.4 degrees
   C. 37.6 degrees
   D. 60 degrees
   Answer: B. tan(alpha) = tan30 x Ix/Iy = 0.5774 x 2.25 = 1.299, so alpha = 52.4 degrees. (Key point: Neutral axis angle)

4. How is the resultant free-end deflection obtained and what is its direction?
   A. Using P L^3/(3 E I) with the larger I
   B. Components dy = PyL^3/(3EIx) and dx = PxL^3/(3EIy) combined vectorially; direction 52.4 degrees from vertical, not the load's 30 degrees
   C. Simply along the load direction
   D. Using the smaller I only
   Answer: B. dy = 0.846 mm, dx = 1.099 mm, giving 1.39 mm, inclined at 52.4 degrees to the vertical, perpendicular to the neutral axis. (Key point: Resultant deflection)

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