Stresses in a curved beam under combined thrust and moment: 22-Mec-B9 Advanced Engineering Structures, December 2019, Q7

Question 7: Stresses in a curved beam under combined thrust and moment (20 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. A sharply curved bar of rectangular section, built in at one end and loaded at the free end by a tangential force and a couple.

Curved-beam data
Quantity	Value
Inner radius r_i / outer radius r_o	150 mm / 200 mm
Radial depth h × width b	50 × 30 mm
Axial (tangential) force N	200 kN, tensile
Applied moment M	150 N·m (see the note below on the printed magnitude)
Yield strength / modulus	250 MPa / 200 GPa

Find. The circumferential stress distribution σ_θθ and the radial stress σ_rr, then whether the section yields and, if not, the factor of safety.

Figure 7.1 — The curved bar with its centroidal radius and the neutral axis, which lies 1.20 mm inside the centroid, together with the hyperbolic hoop-stress distribution across the depth. The stress is highest at the bore.

Check: the printed moment is three orders of magnitude too large for this section. A moment of 150 kN·m on a 50 × 30 mm bar produces a bore stress of about 13 400 MPa — fifty times the quoted yield strength and well beyond any engineering material — which would make the closing instruction "if not, determine the factor of safety" unreachable. The reading adopted here is M = 150 N·m (equivalently 150 kN·mm), which is the value that makes the section elastic and the question answerable as posed. The literal reading is carried in the results table for completeness. Note also that the question labels the modulus G; at 200 GPa it is plainly Young's modulus E, and in any case no elastic constant enters the stress answer, which depends on geometry alone.

Approach. Compute the section constants of Winkler curved-beam theory, superpose the uniform stress from the thrust on the hyperbolic stress from the moment, obtain the radial stress from equilibrium of the inner portion of the section, and compare the peak with the yield strength.

Compute the section constants. With $r_i=150$ mm and $r_o=200$ mm, the area and centroidal radius are $A=30(50)=1500\ \text{mm}^2$ and $\bar R=175$ mm. For a rectangle the neutral-axis radius follows from $$r_n=\frac{h}{\ln\left(r_o/r_i\right)}=\frac{50}{\ln(200/150)}=173.803\ \text{mm}$$ so the eccentricity between centroid and neutral axis is $$e=\bar R-r_n=175-173.803=1.1970\ \text{mm}$$ The small size of $e$ relative to the depth is the whole reason curved-beam theory needs care: it appears in the denominator of the stress formula, so a three-figure value of $r_n$ is not accurate enough.
Write the circumferential stress. Superposing the uniform stress from the thrust on the Winkler bending distribution, $$\sigma_{\theta\theta}(r)=\frac{N}{A}+\frac{M\left(r_n-r\right)}{A\,e\,r}$$ The first term is $200\,000/1500=133.33$ MPa everywhere. The second is hyperbolic in $r$, vanishing at $r=r_n$ and reaching its extremes at the two surfaces.
Evaluate at the two surfaces. At the bore, $r=150$ mm and $r_n-r=+23.80$ mm, giving a bending contribution of $+13.26$ MPa; at the outside, $r=200$ mm and $r_n-r=-26.20$ mm, giving $-10.94$ MPa. Adding the thrust, $$\boxed{\sigma_{\theta\theta}(r_i) = +146.6\ \text{MPa},\qquad \sigma_{\theta\theta}(r_o) = +122.4\ \text{MPa}}$$ Both fibres are in tension because the thrust dominates. The bending part is asymmetric — 13.26 against 10.94 MPa — which is the signature of curvature; a straight beam of the same section would give $\pm12.0$ MPa, so the bore stress is magnified by 10.5 per cent and the outer fibre relieved by a similar amount.
Derive the radial stress. Radial equilibrium of the material between the bore and any radius $r$, over an element subtending $d\phi$, requires the radial stress on the cut at $r$ to balance the inward resultant of the hoop stresses acting on that strip: $$\sigma_{rr}(r)=\frac{1}{r}\int_{r_i}^{r}\sigma_{\theta\theta}^{(M)}\,d\rho=\frac{M}{A\,e\,r}\left[r_n\ln\!\left(\frac{r}{r_i}\right)-\left(r-r_i\right)\right]$$ The bracket vanishes at $r=r_i$ by inspection and at $r=r_o$ because $r_n\ln(r_o/r_i)=h$ exactly, so the expression satisfies the traction-free condition on both surfaces without any adjustment.
Locate and evaluate the peak radial stress. Differentiating and setting the result to zero gives $\ln(r^{*}/r_i)=1-r_i/r_n$, hence $$r^{*}=r_i\exp\!\left(1-\frac{r_i}{r_n}\right)=172.0\ \text{mm}$$ $$\boxed{\sigma_{rr,\max} = 0.87\ \text{MPa at } r = 172.0\ \text{mm}}$$ That is six-tenths of one per cent of the hoop stress. For a solid rectangular section the radial stress is always negligible; it becomes a design driver only in I- and T-sections, where a thin web must carry the same radial resultant that a wide flange generates.
Check against yield. The governing stress is the peak hoop stress at the bore, and the radial stress neither coincides with it in location nor is large enough to alter the principal values appreciably, so $$\text{FoS}=\frac{\sigma_Y}{\sigma_{\theta\theta,\max}}=\frac{250}{146.6}$$ $$\boxed{\text{The beam does not yield; FoS} = 1.71}$$
Test the sensitivity to the sense of the moment. The figure does not settle unambiguously whether the couple opens or closes the curvature. Reversing it moves the peak to the outer fibre, at 144.3 MPa, and the factor of safety to 1.73 — a change of one per cent. The conclusion is therefore robust: whichever way the moment acts, the thrust of 133.3 MPa dominates and the section is elastic with a margin of roughly seventy per cent.

Question 7 — results
Quantity	Value
Area / centroidal radius	1500 mm² / 175 mm
Neutral-axis radius r_n / eccentricity e	173.803 mm / 1.1970 mm
Stress from the thrust N/A	+133.33 MPa
Bending stress, bore / outside	+13.26 / −10.94 MPa
σ_θθ at the bore / at the outside	+146.6 / +122.4 MPa
σ_rr maximum (at r = 172.0 mm)	0.87 MPa
Yield verdict / factor of safety	Does not yield; FoS = 1.71
With the moment reversed	Peak 144.3 MPa at the outside; FoS = 1.73
At the literal printed moment of 150 kN·m	Bore stress 13 390 MPa; FoS = 0.019 (not physically attainable)

Topic. Winkler curved-beam theory: hyperbolic hoop stress, the shifted neutral axis, and the radial stress that accompanies it.

Key relations. For a rectangle $r_n=h/\ln(r_o/r_i)$ and $e=\bar R-r_n$; hoop stress $\sigma_{\theta\theta}=N/A+M(r_n-r)/(A\,e\,r)$; radial stress $\sigma_{rr}=\dfrac{M}{A\,e\,r}\left[r_n\ln(r/r_i)-(r-r_i)\right]$, peaking at $r^{*}=r_i\exp(1-r_i/r_n)$.

Why this works. In a curved bar the fibres on the inside are shorter than those on the outside, so a rotation of one cross-section relative to the next produces a strain that varies as $1/r$ rather than linearly. The stress distribution is therefore hyperbolic, and the axis of zero stress shifts towards the centre of curvature. Equilibrium of moments then forces the small eccentricity $e$ into the denominator. The radial stress is not an independent assumption: it is whatever is needed to hold the inner part of the section in radial equilibrium against the hoop stresses acting on it, which is why it must vanish on both free surfaces.

Common pitfall. Rounding $r_n$ too early. Here $r_n$ and $\bar R$ differ by only 1.2 mm, so quoting $r_n$ to three significant figures throws the eccentricity out by tens of per cent and the bending stress with it — carry at least six figures through that subtraction. The second pitfall is treating $\bar R$ as the neutral radius, which is the straight-beam habit and removes the entire effect the question is testing.

Source. Boresi and Schmidt, Advanced Mechanics of Materials, 6th ed., ch. 9 (curved beams; radial stresses); Ugural and Fenster, Advanced Mechanics of Materials and Applied Elasticity, 5th ed., §5.9–5.11.

Question 7 of 7: Stresses in a curved beam under combined thrust and moment

Question 7: Stresses in a curved beam under combined thrust and moment (20 marks)