Final mechanics synthesis workshop

Six steps, always in this order. (1) Draw the free-body diagram and find internal resultants $P$, $T$, $V$, $M$ at the section of interest using equilibrium. (2) Identify candidate critical points on the cross-section — extreme fibres for bending, outer surface for torsion, and any point where effects could combine unfavourably; check more than one point if signs might combine differently. (3) Compute each stress component separately with the matching formula. (4) Superpose stresses acting on the same plane algebraically (normal with normal, shear with shear) to assemble the stress element $(\sigma_x,\sigma_y,\tau_{xy})$. (5) Apply transformation or Mohr's-circle formulas to find principal stresses if asked. (6) For a slender member in net compression, separately check Euler buckling and verify the critical stress stays below yield.

The component formulas you are choosing between.

where: $P$ is axial force, $A$ cross-sectional area, $M$ bending moment, $c$ distance from the neutral axis, $I$ second moment of area, $T$ applied torque, $J$ polar second moment of area, $V$ shear force, $Q$ first moment of the area beyond the point of interest about the neutral axis, and $b$ the section width there. For a thin-walled pressure vessel (valid once $r/t\gtrsim10$) the biaxial pair is $\sigma_h=pr/t$ (hoop) and $\sigma_l=pr/2t$ (longitudinal, cylinders) or $\sigma=pr/2t$ (spheres), with $p$ the internal gauge pressure and $r$ the mean radius. A pressurized shaft or a pipe under pressure plus torque is the natural capstone case: the pressure alone already gives a genuine biaxial $(\sigma_x,\sigma_y)$ pair — unlike a plain shaft, where $\sigma_y=0$ — so any added torque or axial load superposes onto it before transformation.

Once the element is assembled, principal stresses and the maximum in-plane shear follow from one pair of formulas built around the average and radius of Mohr's circle:

where: $\sigma_1\ge\sigma_2$ are the principal (extreme) normal stresses, acting on planes where shear vanishes, and $\tau_{max}$ is the maximum in-plane shear stress, which acts on planes rotated $45°$ from the principal planes. A useful invariant for checking arithmetic: $\sigma_1+\sigma_2=\sigma_x+\sigma_y$, since rotating the element cannot change the sum of the normal stresses on perpendicular planes.

Buckling is a separate failure mode, not a stress limit:

where: $E$ is Young's modulus, $I$ the second moment of area about the weaker axis, and $L_e$ the effective length, which depends on end conditions ($L_e=L$ for pinned–pinned; shorter for fixed ends, longer for a fixed-free cantilever column). A member can pass a stress check comfortably and still buckle, so slender compression members always need both checks run independently.

A closed-ended thin-walled cylindrical pressure vessel, mean radius $r=300\text{ mm}$, wall thickness $t=6\text{ mm}$, carries internal gauge pressure $p=2.5\text{ MPa}$ and an external torque $T=15\text{kN}\,\cdot\,\text{m}$ about its axis. Find the principal stresses at a point on the outer surface.

Step 1. Check thin-wall validity: $r/t=300/6=50\gg10$, so $\sigma_h=pr/t$ and $\sigma_l=pr/2t$ apply.

Step 2. Pressure stresses:

Step 3. Torsional shear stress, treating the thin wall with $J\approx2\pi r^3t$:

Step 4. Assemble the stress element with $x$ = longitudinal, $y$ = hoop: $\sigma_x=62.5\text{ MPa}$, $\sigma_y=125\text{ MPa}$, $\tau_{xy}=4.42\text{ MPa}$.

Step 5. Apply the derivation's formula. Average $=93.75\text{ MPa}$, and $R=\sqrt{(-31.25)^2+4.42^2}=\sqrt{976.6+19.5}=31.56\text{ MPa}$, so

Check: $\sigma_1+\sigma_2=187.5\text{ MPa}=\sigma_x+\sigma_y$ ✓. The principal stresses are 125.3 MPa and 62.2 MPa — the torque barely perturbs the pressure-only values (125 and 62.5 MPa) because $\tau_{xy}=4.42\text{ MPa}$ is small next to $(\sigma_x-\sigma_y)/2=31.25\text{ MPa}$, but the full check is exactly what a design engineer must perform rather than assume.

Competitive stretch

Now add the independent buckling check style of problem. A vertical pin-ended steel strut, $L=2.5\text{ m}$, solid circular section $d=40\text{ mm}$, carries $P=20\text{ kN}$ with eccentricity $e=5\text{ mm}$; $E=200\text{ GPa}$, $\sigma_{yield}=250\text{ MPa}$. Section properties: $A=1.257\times10^{-3}\text{ m}^2$, $I=1.257\times10^{-7}\text{ m}^4$, $c=0.02\text{ m}$. Combined stress: $\sigma_{axial}=20\times10^3/A=15.91\text{ MPa}$, $M=Pe=100\text{N}\,\cdot\,\text{m}$, $\sigma_{bending}=Mc/I=15.91\text{ MPa}$, so $\sigma_{max(comp)}=-(15.91+15.91)=-31.8\text{ MPa}$. Separately, with $L_e=L$ (pinned–pinned): $P_{cr}=\pi^2EI/L_e^2\approx39.7\text{ kN}$, giving $\sigma_{cr}=P_{cr}/A=31.6\text{ MPa}$. Both checks pass ($P=20$ kN well below $P_{cr}=39.7$ kN; 31.8 MPa well below yield of 250 MPa), but note how close 31.6 MPa (buckling) sits to 31.8 MPa (working stress) as raw numbers — a reminder that the two checks are numerically unrelated even when they land near each other.

• Treating buckling and stress checks as one calculation. They are independent failure modes; a member can pass a stress check and still buckle, so both must be checked explicitly.
• Forgetting the thin-wall validity ratio $r/t\gtrsim10$ before using $\sigma_h=pr/t$; thicker vessels need Lamé thick-cylinder theory, which is out of scope here.
• Losing track of which stresses act on which plane. Build a table of $\sigma_x$, $\sigma_y$, $\tau_{xy}$ contributions, filling it in one loading type at a time, to avoid sign and plane-mismatch errors.
• Skipping the "identify the critical point" step under time pressure. Jumping straight to formulas without sketching the section first is the most common source of using the wrong numbers.

Ask the tutor: ask for a mixed practice set spanning several loading types in one problem, or to walk through where you got stuck applying this six-step workflow to a specific question.