22-Mec-A7 Advanced Strength of Materials · December 2014
Question 6 of 8: Circular ring with a rigid diametral bar under a tensile pull
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — December 2014, 07-Mec-A7 Advanced Strength of Materials. Open-book, 3 hours; any five of the eight problems constitute a complete paper and all problems are of equal value. All eight problems are solved as a study resource.
Given. Mean radius $R=200$ mm; section $b=75$ mm wide by $h=18$ mm (radial) deep; a rigid bar spans the horizontal diameter; vertical pull $P$ up and down; $\sigma_{allow}=250$ MPa.
Given data
Mean radius R
200 mm
Section width b / depth h
75 mm / 18 mm
Section modulus data: I = bh³/12
36,450 mm4
Fibre distance c = h/2
9 mm
Allowable stress
250 MPa
Find. the maximum tensile force P.
Ring pulled apart vertically by P; the rigid horizontal bar holds the horizontal diameter constant, so it feeds a horizontal reaction Q into the ring at the side sections.
Approach. Treat as a thin ring: superpose the vertical pull P and the (unknown) horizontal bar reaction Q, choose Q so the horizontal diameter does not change (rigid bar), then size P from the peak bending stress.
Thin-ring diametral coefficients. For a pair of opposite radial forces, the bending moment is $+0.1817\,PR$ at 90° from the loads and $-0.3183\,PR$ at the load points; the loaded diameter grows by $0.1488\,PR^3/EI$ and the perpendicular diameter shrinks by $0.1366\,PR^3/EI$.
Rigid-bar compatibility. The vertical pull would shrink the horizontal diameter by $0.1366\,PR^3/EI$; the bar force Q (acting outward at the sides) must restore it:
$$0.1488\,Q=0.1366\,P\;\Rightarrow\;\boxed{Q=0.918\,P}$$
Superposed bending moments. Adding the two load cases at the key sections,
$$M_{\text{top}}=(-0.3183+0.1817\cdot0.918)PR=-0.1515\,PR,\quad M_{\text{side}}=(0.1817-0.3183\cdot0.918)PR=-0.1106\,PR.$$
The maximum magnitude, $|M|_{max}=0.1515\,PR$, occurs at the load points (top and bottom), where the axial force is essentially zero.
Size the load. With $I=bh^3/12=36\,450\ \text{mm}^4$ and $c=9$ mm, set the peak bending stress to the allowable:
$$\sigma=\frac{M_{max}c}{I}=\frac{0.1515\,PR\,c}{I}=250\ \text{MPa}\;\Rightarrow\;P=\frac{250\,I}{0.1515\,R\,c}=\boxed{33.4\ \text{kN}}$$
(The side section, carrying $0.1106\,PR$ plus a small $P/2$ axial force, is less stressed, so the load points govern.)
Check: Thin-ring theory is used ($R/h\approx11$); a curved-beam (Winkler) correction would raise the inner-fibre stress a few percent, making 33.4 kN a slightly non-conservative upper bound. The rigid bar is modelled as holding the horizontal diameter exactly constant.
