Mechanics of Materials · Chapter 8

Combined Loadings

Stresses in pressure vessels, and superposition of axial, shear, bending, and torsional stresses at a point.

Learning objectives

Derive the hoop and longitudinal stresses in thin-walled cylindrical and spherical pressure vessels.
Apply the method of superposition to find the complete state of stress at a point in a member carrying several internal loadings at once.

8.1Thin-Walled Pressure Vessels

The Thin-Wall Assumption

Boilers, gas tanks, and pipes are treated as thin-walled when the inner-radius-to-thickness ratio satisfies \(r/t \ge 10\); at exactly r/t = 10 the simple analysis underestimates the true maximum stress by only about 4 %, and the error shrinks for thinner walls. The analysis uses gage pressure (pressure above atmospheric), since atmosphere acts equally inside and out before pressurization, and assumes the stress is uniform through the wall.

Cylindrical Vessel — Hoop and Longitudinal Stresses

\[ \sigma_1 = \frac{p\,r}{t} \qquad\qquad \sigma_2 = \frac{p\,r}{2t} \]

\(\sigma_1\) = circumferential (hoop) stress; \(\sigma_2\) = longitudinal (axial) stress — both tensile and uniform through the wall
\(p\) = internal gage pressure; \(r\) = inner radius; \(t\) = wall thickness
The hoop stress is twice the longitudinal stress — so longitudinal welded joints must be designed for twice the load of circumferential joints, and over-pressurized pipes rupture along their length.

Spherical Vessel

\[ \sigma_2 = \frac{p\,r}{2t} \quad\text{(in every direction)} \]

Sectioning a sphere in any direction gives the same result — the wall carries equal biaxial tension, making the sphere the most efficient (though hardest to fabricate) pressure vessel.

Biaxial stress state on an element of a thin-walled cylindrical pressure vessel: hoop stress σ1 = pr/t and longitudinal stress σ2 = pr/2t.

FIG. 8-1 (original diagram)

Limitations

The pressure also produces a radial stress \(\sigma_3\), maximum (= p, compressive) at the inner wall and zero at the outer surface. For \(r/t \ge 10\), \(\sigma_1\) and \(\sigma_2\) are at least 10 and 5 times larger than \((\sigma_3)_{max}\), so it is neglected — the wall element is treated as biaxial. Note also that external pressure produces compressive wall stresses that can cause sudden inward buckling rather than material failure.

8.2State of Stress Caused by Combined Loadings

Real members rarely see a single internal loading. When a cross section carries some combination of normal force, shear force, bending moment, and torque, the resultant stress at a point is found by the method of superposition — valid for homogeneous, linear elastic material at points away from discontinuities (Saint-Venant).

Procedure for Analysis — Combined Loadings

1. Internal loadings. Section the member through the point; use equilibrium to find the resultant N, V, M, and T, with forces acting through the centroid and moments about centroidal principal axes.

2. Stress from each loading. Normal force: \(\sigma = N/A\) (uniform). Shear force: \(\tau = VQ/It\). Bending: \(\sigma = -My/I\) (linear; for curved members \(\sigma = My/[Ae(R-y)]\)). Torque: \(\tau = T\rho/J\) (circular sections). Pressure vessel: \(\sigma_1 = pr/t\), \(\sigma_2 = pr/2t\).

3. Superpose. Add the normal-stress components algebraically; add the shear-stress components vectorially in the section plane. Show the result on a small volume element at the point.

Physical meaning

Internal resultants create local stress through member geometry; use a cut free-body and stated sign convention before superposition.

Problem-solving strategy

Obtain signed internal resultants from equilibrium.
Choose axes and identify the precise material point.
Apply only relations whose assumptions fit the geometry.
Audit units, direction, and the governing failure mode.

EXAMPLE 1Receiver membrane stresses

A closed cylindrical air receiver has \(r=250\ \text{mm}\), \(t=8\ \text{mm}\), and internal gauge pressure \(p=1.20\ \text{MPa}\). Find hoop and longitudinal stresses and assess the thin-wall assumption.

Recognition strategy

Cylinder pressure and wall stresses select membrane equilibrium, producing biaxial normal stress rather than beam stress.

Given information

Use gauge pressure, closed ends, uniform thickness, and no external pressure.

Model and interpretation

Tension is positive. Diametral and end-cap free bodies respectively establish hoop and longitudinal equilibrium.

Step-by-step solution

1. Plan. The ratio \(r/t=31.25\) exceeds 10, so the thin-wall idealization is appropriate for this educational calculation.

2. Formulate. Use \(\sigma_h=pr/t\) and \(\sigma_l=pr/(2t)\), from the two independent equilibrium cuts.

3. Calculate. Substitution gives \(\sigma_h=1.20(250)/8=37.5\ \text{MPa}\) and \(\sigma_l=1.20(250)/16=18.8\ \text{MPa}\).

4. Resolve. Both components are tensile normal stresses in perpendicular wall directions, with hoop stress twice longitudinal stress.

Verification

Units: N and mm give \(\text{N}/\text{mm}^2\), or MPa, after the stated section-property substitution; the calculation is dimensionally consistent.

Signs and direction: The stated direction follows the cut-face convention. Reversing the internal resultant reverses the traction direction, but not the reported magnitude.

Plausibility: The computed magnitude has the expected limiting behavior and is independently consistent with the physical location of the maximum response.

Final result: \(\boxed{\sigma_h=37.5\ \text{MPa},\quad\sigma_l=18.8\ \text{MPa}}\)

Engineering interpretation

Membrane stress does not certify a pressure vessel. Code allowables, weld efficiency, corrosion allowance, openings, cycles, temperature, inspection, and relief protection require qualified design review.

EXAMPLE 2Eccentric compression

A 120 mm by 200 mm short post carries \(90\ \text{kN}\) compression 30 mm from its centroid along the depth. Find normal stresses at both depth edges.

Recognition strategy

An eccentric force equals centroidal axial force plus bending moment, so uniform and linear normal stresses are superposed.

Given information

Use \(P=90\ \text{kN}\), \(e=30\ \text{mm}\), \(b=120\ \text{mm}\), and \(h=200\ \text{mm}\).

Model and interpretation

Take tension positive and neglect buckling only for this local linear-elastic stress calculation.

Step-by-step solution

1. Plan. The section free-body gives \(N=-90\,000\ \text{N}\) and \(M=Pe=2.70\times10^6\ \text{N}\cdot\text{mm}\).

2. Formulate. Section properties are \(A=24\,000\ \text{mm}^2\), \(I=80.0\times10^6\ \text{mm}^4\), and \(c=100\ \text{mm}\).

3. Calculate. Direct stress is \(-3.75\ \text{MPa}\), while the bending edge magnitude is \(Mc/I=3.375\ \text{MPa}\).

4. Resolve. Superposition gives \(-7.13\ \text{MPa}\) at the loaded edge and \(-0.375\ \text{MPa}\) at the far edge.

Verification

Units: N and mm give \(\text{N}/\text{mm}^2\), or MPa, after the stated section-property substitution; the calculation is dimensionally consistent.

Signs and direction: The stated direction follows the cut-face convention. Reversing the internal resultant reverses the traction direction, but not the reported magnitude.

Plausibility: The computed magnitude has the expected limiting behavior and is independently consistent with the physical location of the maximum response.

Final result: \(\boxed{\sigma_{\rm loaded}=-7.13\ \text{MPa},\quad\sigma_{\rm far}=-0.375\ \text{MPa}}\)

Engineering interpretation

The far edge is nearly unloaded, so small placement errors can create tension. Real compression members also require buckling, bearing, cracking, and load-combination checks.

EXAMPLE 3Combined bending and torsion

A solid 60 mm diameter drive shaft has \(M=1.20\ \text{kN}\cdot\text{m}\) and \(T=1.80\ \text{kN}\cdot\text{m}\) at a critical cut. Find stress components at the outer point in bending tension.

Recognition strategy

Moment and torque at one section require component-by-component superposition at the same material point.

Given information

Use \(c=30\ \text{mm}\), \(M=1.20\times10^6\ \text{N}\cdot\text{mm}\), and \(T=1.80\times10^6\ \text{N}\cdot\text{mm}\).

Model and interpretation

Use centroidal shaft axes, positive normal tension, and the right-hand rule for torsional shear direction.

Step-by-step solution

1. Plan. The cut free-body establishes \(M\) and \(T\); no axial or transverse-shear component is specified at this surface point.

2. Formulate. For a solid circle, \(I=\pi d^4/64=636\,173\ \text{mm}^4\) and \(J=\pi d^4/32=1.272\times10^6\ \text{mm}^4\).

3. Calculate. Bending gives \(\sigma_x=Mc/I=56.6\ \text{MPa}\), while torsion gives \(\tau_{x\theta}=Tc/J=42.4\ \text{MPa}\).

4. Resolve. Place normal and complementary shear tractions on one stress element before transforming components or assessing failure.

Verification

Units: N and mm give \(\text{N}/\text{mm}^2\), or MPa, after the stated section-property substitution; the calculation is dimensionally consistent.

Signs and direction: The stated direction follows the cut-face convention. Reversing the internal resultant reverses the traction direction, but not the reported magnitude.

Plausibility: The computed magnitude has the expected limiting behavior and is independently consistent with the physical location of the maximum response.

Final result: \(\boxed{\sigma_x=+56.6\ \text{MPa},\quad\tau_{x\theta}=42.4\ \text{MPa}}\)

Engineering interpretation

A shaft decision needs an appropriate failure theory plus fatigue, keyway concentration, deflection, bearings, manufacturing, and service-load checks.

Chapter recap

Use valid pressure-vessel relations and superpose stress components, never force resultants, at the same point.
Before moving on, draw the cut free-body, declare signs and units, and state the applicable failure or serviceability check.