Composite Concrete/Steel Column Under Axial Load: 04-BS-6, December 2019, Q1

Question 1: Composite Concrete/Steel Column Under Axial Load (13+4+3 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.

Given.

Quantity	Value
Concrete core	100 mm × 100 mm ($A_c=10{,}000\text{ mm}^2$)
Steel sleeve wall	2 mm ($104\times104$ outer, $A_s=816\text{ mm}^2$)
Applied load	$P=250\text{ kN}$ (via rigid cap plate)
Length	$L=500\text{ mm}$
Steel	$E_s=200\text{ GPa}$, $\sigma_{allow,s}=175\text{ MPa}$
Concrete	$E_c=16\text{ GPa}$, $\sigma_{allow,c}=20\text{ MPa}$

Find. (1) $F_c,F_s$; (2) column shortening $\delta$; (3) maximum allowable $P$.

Composite square column: 100×100 mm concrete core inside a 2 mm steel sleeve, rigid cap plate, 500 mm length.

Approach. The rigid cap plate forces the concrete and the sleeve to shorten by the same amount, so equal-strain compatibility splits the load between the two materials in proportion to $A\cdot E$; the governing material for part (3) is whichever reaches its own allowable stress first.

Compatibility — equal strain. Both materials shorten together: $\varepsilon_c=\varepsilon_s \Rightarrow \sigma_s=\sigma_c(E_s/E_c)=n\,\sigma_c$, with modular ratio $n=200/16=12.5$.
Equilibrium — split the load. $P=\sigma_c A_c+\sigma_s A_s=\sigma_c(A_c+nA_s)=\sigma_c(10{,}000+12.5\times816)$, giving $\sigma_c=250{,}000/20{,}200=\boxed{12.38\text{ MPa}}$ and $\sigma_s=12.5\times12.38=\boxed{154.7\text{ MPa}}$ (both below their allowables, so the applied 250 kN is safely carried).
Part (1) — forces. $F_c=\sigma_c A_c=12.38\times10{,}000=\boxed{123.8\text{ kN}}$; $F_s=\sigma_s A_s=154.7\times816=\boxed{126.2\text{ kN}}$ (sum $=250.0\text{ kN}$, checks against $P$).
Part (2) — shortening. $\delta=\varepsilon_c L=(\sigma_c/E_c)L=(12.38/16{,}000)\times500=\boxed{0.387\text{ mm}}$ (matches $(\sigma_s/E_s)L$ identically, confirming compatibility).
Part (3) — governing material. If concrete reached $\sigma_{allow,c}=20\text{ MPa}$ first, steel would sit at $n\times20=250\text{ MPa}$ — already past its own $175\text{ MPa}$ allowable, so steel governs. Set $\sigma_s=175\text{ MPa}\Rightarrow\sigma_c=175/12.5=14.0\text{ MPa}$ ($<20$, OK). $P_{max}=\sigma_c A_c+\sigma_s A_s=14.0\times10{,}000+175\times816=140{,}000+142{,}800=\boxed{282.8\text{ kN}}$.

Final results
Quantity	Value
$F_c$	123.8 kN
$F_s$	126.2 kN
$\delta$	0.387 mm
Governing material	steel (reaches 175 MPa first)
$P_{max}$	282.8 kN

Topic: Composite (parallel) axial members under a rigid cap

Key relations: Equal-strain compatibility $\varepsilon_c=\varepsilon_s$; load split $P=\sigma_c(A_c+nA_s)$, $n=E_s/E_c$; shortening $\delta=(\sigma_c/E_c)L$.

Why this works: A rigid, weightless cap plate guarantees both materials shorten identically, so the two-material system reduces to ONE compatibility equation plus equilibrium — no need to solve a coupled system, since the stiffer material (larger $E$) always carries proportionally more stress at the same strain.

Common pitfall: Splitting the load by AREA alone (as if both materials had the same $E$) instead of by $A\cdot E$; and checking only one material's allowable stress in part (3) without checking whether the OTHER material has already been exceeded at that same load.

Source: Hibbeler, Mechanics of Materials, 10th ed., §4.2–4.3 Elastic Deformation / Composite Axially Loaded Members

Question 1 of 8: Composite Concrete/Steel Column Under Axial Load

Question 1: Composite Concrete/Steel Column Under Axial Load (13+4+3 marks)