Question 4 of 8: Eccentric column load — combined stress at the base
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — May 2013 — 04-BS-6 Mechanics of Materials. Three-hour, closed-book exam (one double-sided hand-written aid sheet permitted, approved Casio/Sharp calculator only). Format: eight questions offered; any five constitute a complete paper and only the first five appearing in the answer book are marked. All eight are solved below for completeness. No credit is given in the original exam for solutions built by superposing existing standard solutions on Questions 1 and 6 — both are solved here by direct integration/equilibrium as required.
Reference texts: Hibbeler, Mechanics of Materials (10th ed., Pearson) — beam deflection by integration (Ch. 12), stress transformation and Mohr's circle (Ch. 9), columns and buckling (Ch. 13), combined axial & bending loading (Ch. 8), torsion of circular shafts (Ch. 5), transformed-section method for composite beams (Ch. 6), shear and moment diagrams (Ch. 6-7); Beer, Johnston & DeWolf, Mechanics of Materials (7th ed., McGraw-Hill) — cross-check reference for the same topics.
Question 4: Eccentric column load — combined stress at the base (20 marks)
Column elevation: 700 kN at 40° above horizontal, applied 1050 mm from centreline, 2000 mm above the base.
Section A-A: H-shaped column (flanges vertical, in the plane of bending).
Given.
Quantity
Value
Applied load
700 kN at 40° above horizontal (up and outward)
Eccentricity (horizontal)
1050 mm from column centreline
Height (vertical)
2000 mm above the base
Section (H-shape)
flanges 300 mm tall × 10 mm thick (vertical, in bending plane), web 12 mm thick, 500 mm clear span → overall depth 520 mm
$F_y$, $E$
400 MPa, 200 GPa
Find. (a) The normal stress distribution across the base; (b) the maximum shear stress at the base.
Approach. Resolve the 700 kN load into horizontal and vertical components; take the free body above the base to get the axial force $N$, shear $V$, and moment $M$ transmitted at the base; superpose $\sigma=N/A+Mx/I$ for the normal-stress distribution, and use $\tau=VQ/(It)$ at the neutral axis for the shear.
Resolve the load and locate it. $F_x=700\cos40^\circ=536.2$ kN, $F_z=700\sin40^\circ=450.0$ kN, applied at $(x_p,z_p)=(1050,2000)$ mm from the base centre.
Section properties. With flanges spanning the full "depth" ($D=520$ mm, $b_f=300$ mm, $t_f=10$ mm) and web ($t_w=12$ mm, clear span 500 mm) providing the bending resistance in the plane of the load:
$$A=2(300)(10)+12(500)=12\,000\ \text{mm}^2,\qquad I=\frac{300(520)^3}{12}-\frac{(300-12)(500)^3}{12}=5.152\times10^{8}\ \text{mm}^4$$
Internal axial force and moment at the base. Taking the free body above the base cut (only the applied load acts on it):
$$N=F_z=450.0\ \text{kN (tension)},\qquad M=x_pF_z-z_pF_x=(1.05)(450.0)-(2.0)(536.2)=\boxed{-600.0\ \text{kN}\!\cdot\!\text{m}}$$
The horizontal component acting at the 2 m height dominates the eccentric pull, so the net moment puts the far side of the column (away from the load) into tension.
Normal stress distribution (part a). With $x$ measured from the centroid ($\pm260$ mm at the flange outer faces):
$$\sigma(x)=\frac{N}{A}+\frac{Mx}{I}\ \Rightarrow\ \sigma(+260)=\boxed{-265.3\ \text{MPa (compression)}},\quad \sigma(-260)=\boxed{+340.3\ \text{MPa (tension)}}$$
The stress varies linearly between these extremes, crossing zero at $x_0=-NI/(MA)=+32.2$ mm from the centroid (slightly on the load side, because the uniform tensile term shifts the neutral axis away from a pure-bending centerline). Both peaks are below the 400 MPa yield stress.
Maximum shear stress (part b). $V=F_x=536.2$ kN acts through the section; at the neutral axis, $Q=(t_w)(h_w/2)(h_w/4)+(b_ft_f)\!\left(\dfrac{h_w}{2}+\dfrac{t_f}{2}\right)=1.140\times10^{6}\ \text{mm}^3$ (web strip plus one flange), so
$$\tau_{max}=\frac{VQ}{It_w}=\frac{(536\,231)(1.140\times10^6)}{(5.152\times10^8)(12)}=\boxed{98.9\ \text{MPa}}$$
occurring at the web, at the centroidal axis.
