Biomechanics: Joint Statics and Gait, Stress and Strain in Bone, and Soft-Tissue Mechanics
2. Forces and stresses in joints; biomechanical analysis of joints
The hip in single-leg stance. With one foot off the ground, the body weight above the stance hip (the whole body minus the stance leg, about five-sixths of body weight) acts some distance medial to the hip joint, and the hip abductors (gluteus medius and minimus), attached to the greater trochanter much closer to the joint, must balance its moment. With a body-weight moment arm about twice the abductor moment arm, the abductor force is about twice that body weight, and the joint reaction — abductor force plus body weight, roughly — comes to about 2.5 to 3 times body weight in standing and more in walking. A cane in the opposite hand adds a long moment arm on the other side and relieves the hip considerably.
- Contact stress in a joint is the joint force divided by the contact area, so the same force is far more damaging on a small area — the argument for congruent joint surfaces and against point-contact implant designs.
- The knee is loaded by the quadriceps through the patella, whose presence lengthens the quadriceps moment arm; patellofemoral forces rise steeply with knee flexion, which is why stair climbing and squatting load it heavily.
- The lumbar spine in lifting: the back extensors act a few centimetres behind the disc while the load acts far in front, so lifting with a bent back and the load held away from the body produces compressive disc forces of several times body weight.
- Synovial joints are lubricated by synovial fluid and by the fluid squeezed from articular cartilage under load, giving friction coefficients of the order of 0.01 or lower — far below most engineering bearings.
3. Gait analysis
The gait cycle runs from one heel strike to the next heel strike of the same foot. At normal walking speed the stance phase (foot on the ground) takes about 60% of the cycle and the swing phase about 40%, with two periods of double support of roughly 10% each. Stance is subdivided into heel strike (initial contact), foot flat (loading response), mid-stance, heel off and toe off; swing into initial, mid- and terminal swing. Running removes double support and introduces a flight phase with neither foot on the ground.
| Parameter | Definition |
|---|---|
| Step length | distance from heel strike of one foot to heel strike of the other |
| Stride length | distance between successive heel strikes of the same foot — two steps |
| Cadence | steps per minute |
| Walking speed | step length × cadence (divide by 60 for per second) = stride length × strides per unit time |
A gait laboratory combines optical motion capture (cameras tracking reflective markers on anatomical landmarks, giving joint angles), force plates (the ground reaction force and its centre of pressure; the vertical component in walking has a double hump peaking a little above body weight, with a trough in mid-stance), and EMG of the main muscle groups. Joint moments and powers are then computed by inverse dynamics — working up from the measured ground reaction force through the foot, shank and thigh using each segment’s mass, inertia and acceleration. Clinical uses include planning surgery in cerebral palsy, fitting and aligning prostheses, and tracking rehabilitation.
4. Stress, strain and the mechanics of bone
Normal stress σ = F/A is force per unit area perpendicular to a surface, and shear stress τ = V/A force per unit area along it; units N/m² = Pa, usually MPa. Normal strain ε = ΔL/L₀ is dimensionless (often quoted in microstrain, 10⁻⁶), and shear strain γ is the change in a right angle, in radians. In the linear elastic range σ = Eε (Hooke’s law) and τ = Gγ; Poisson’s ratio ν = −ε_lateral/ε_axial, and for an isotropic material G = E/[2(1 + ν)]. Engineering stress uses the original area; true stress, the current one.
Bone is a composite: a mineral phase of hydroxyapatite, Ca₁₀(PO₄)₆(OH)₂, which gives stiffness and compressive strength, in a matrix of type I collagen, which gives toughness and tensile strength, plus water. Cortical (compact) bone forms the shafts of long bones and the outer shell of all bones; it is organised into osteons (Haversian systems) — concentric lamellae round a central canal — and has low porosity, roughly 5–10%. Cancellous (trabecular, spongy) bone fills the ends of long bones and the vertebral bodies as a lattice of trabeculae with porosity of roughly 50–90%, its stiffness and strength rising steeply with apparent density.
| Property | Cortical bone (longitudinal) | Cancellous bone |
|---|---|---|
| Elastic modulus | about 17–20 GPa | roughly 0.1–2 GPa, strongly density-dependent |
| Ultimate strength | compression about 190 MPa > tension about 130 MPa > shear about 70 MPa | a few MPa to tens of MPa in compression |
| Porosity | about 5–10% | about 50–90% |
| Behaviour | anisotropic (stiffest along the shaft), viscoelastic, stiffer at higher strain rate | absorbs energy; fails by trabecular buckling and crushing |
Bending of a long bone. Treating the shaft as a beam, the bending stress is σ = My/I, largest at the outer surface; for a hollow circular section of outer diameter D and inner diameter d, I = π(D⁴ − d⁴)/64. Because I depends on the fourth power of distance from the axis, a hollow tube carries nearly the bending stiffness of a solid rod of the same outer diameter at far less weight — the design of every long bone. In torsion, τ = Tr/J with J = π(D⁴ − d⁴)/32, which is why skiing injuries twist the tibia into spiral fractures. Wolff’s law says bone remodels to the loads it carries: it thickens where loaded and is resorbed where not, which is the mechanism of stress shielding when a stiff implant carries the load in bone’s place.
5. Soft tissues and viscoelastic models
| Tissue | Structure | Function | Mechanical signature |
|---|---|---|---|
| Tendon and ligament | parallel, crimped bundles of type I collagen with few cells | tendon joins muscle to bone; ligament joins bone to bone and guides joints | J-shaped stress–strain curve: a toe region as the crimp straightens, then a stiff linear region (modulus of the order of 1 GPa for tendon), then failure |
| Articular cartilage | type II collagen network holding negatively charged proteoglycans and water; avascular | low-friction, load-spreading bearing surface | biphasic: its stiffness under sudden load comes from trapped fluid, which then flows out slowly (creep) |
| Skeletal muscle | fibres of sarcomeres wrapped in connective tissue | active force generation | Hill’s model: a contractile element with series and parallel elastic elements; force–length and force–velocity relations |
| Skin and blood vessels | elastin and collagen in a hydrated matrix | protection; containing and damping the pressure pulse | elastin carries low strains, collagen is recruited at high strain — non-linear and anisotropic |
Viscoelasticity is response that depends on time and rate: under constant stress a tissue creeps, under constant strain its stress relaxes, and a loading–unloading cycle traces a hysteresis loop whose area is energy dissipated. Three lumped models describe it with springs (stress = Eε) and dashpots (stress = η dε/dt), and their characteristic time is τ = η/E.
| Model | Arrangement | Creep under σ₀ | Relaxation under ε₀ |
|---|---|---|---|
| Maxwell | spring and dashpot in series | instant σ₀/E, then unbounded linear growth: ε = σ₀/E + σ₀t/η | σ = σ₀e−t/τ, decaying to zero |
| Kelvin–Voigt | spring and dashpot in parallel | no instant strain; ε = (σ₀/E)(1 − e−t/τ), bounded | no relaxation — stress stays at Eε₀ once the step is complete |
| Standard linear solid (Zener) | a spring in parallel with a Maxwell element | instant strain, then bounded exponential creep | relaxes exponentially to a non-zero equilibrium stress |
Key takeaways
- Take moments about the joint centre to get the muscle force, then force balance for the joint reaction; third-class levers make muscle forces many times the load.
- Hip reaction in single-leg stance is about 2.5–3 × body weight; contact stress = force/area.
- Gait: stance about 60%, swing 40%, double support about 10% twice; speed = step length × cadence.
- Cortical bone: E about 17–20 GPa, strongest in compression, weakest in shear, anisotropic; σ = My/I with I = π(D⁴ − d⁴)/64 for a hollow shaft.
- Maxwell relaxes σ₀e−t/τ but creeps without bound; Kelvin–Voigt creeps (σ₀/E)(1 − e−t/τ) but does not relax; τ = η/E.
Practice questions (13)
Attempt each one before opening the answer. Every explanation names the tempting wrong option as well as the right one, because that is where marks are lost.
Flexing the elbow with the biceps while holding a weight in the hand is an example of a lever of the
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Answer: A — third class
The elbow is the fulcrum, the biceps inserts a few centimetres away, and the load is at the hand further out, so the effort lies between fulcrum and load: third class, mechanical advantage below 1. Rising on the toes is second class; nodding the head on the atlas is first class.A horizontal forearm weighs 15 N acting 15 cm from the elbow, and the hand holds 50 N at 35 cm. The biceps acts vertically at 4 cm from the elbow. The biceps force (in N), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 493.75
Moments about the elbow: 4F = 15 × 15 + 50 × 35 = 225 + 1750 = 1975, so F = 493.75 N, and the joint reaction is 493.75 − 65 = 428.75 N. Leaving out the forearm’s own weight gives 437.5 N; adding the loads without their moment arms gives 65 N.A compressive load of 3000 N acts on a bone cross-section of 3 cm². The compressive stress (in MPa) is ____.
Numerical answer — type the value.
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Answer: 10
σ = F/A = 3000/(3 × 10⁻⁴ m²) = 10⁷ Pa = 10 MPa, well below the roughly 190 MPa compressive strength of cortical bone. Forgetting that 1 cm² = 10⁻⁴ m² gives 1000 Pa, off by a factor of 10⁴.A femoral shaft is modelled as a hollow circular tube of outer diameter 30 mm and inner diameter 15 mm under a bending moment of 100 N·m. The maximum bending stress (in MPa), to one decimal place, is ____.
Numerical answer — type the value.
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Answer: 40.2
I = π(30⁴ − 15⁴)/64 = π(810 000 − 50 625)/64 = 37 276 mm⁴ and y = 15 mm, so σ = My/I = 100 000 × 15/37 276 = 40.2 MPa. Treating the bone as a solid rod of the same outer diameter gives I = 39 761 mm⁴ and 37.7 MPa — only slightly less stress for twice the material, which is the point of the hollow design.Cortical bone with E = 17 GPa carries a uniaxial stress of 34 MPa in its linear range. The strain (in microstrain) is ____.
Numerical answer — type the value.
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Answer: 2000
ε = σ/E = 34 × 10⁶/17 × 10⁹ = 0.002 = 2000 microstrain, a typical peak strain in vigorous activity. Quoting 0.002 answers in plain strain rather than the microstrain asked for; mixing MPa with GPa unconverted gives 2.A ligament modelled as a Maxwell element (τ = η/E = 50 s) is suddenly stretched and held, the initial stress being 2 MPa. The stress (in MPa) after 50 s, to three decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.736
Stress relaxation in a Maxwell element is σ = σ₀e−t/τ = 2e−1 = 2 × 0.3679 = 0.736 MPa. Writing 2(1 − e−1) = 1.264 MPa uses the creep form of the Kelvin–Voigt model, which describes a different test.A tissue modelled as a Kelvin–Voigt element has E = 10 MPa and η = 100 MPa·s. A constant stress of 1 MPa is applied at t = 0. The strain at t = 10 s, to four decimal places, is ____.
Numerical answer — type the value.
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Answer: 0.0632
τ = η/E = 10 s, and ε = (σ₀/E)(1 − e−t/τ) = 0.1 × (1 − e−1) = 0.1 × 0.6321 = 0.0632, rising towards 0.1. The Maxwell creep formula σ₀/E + σ₀t/η = 0.1 + 0.1 = 0.2 applies to a spring and dashpot in series, not in parallel.Which viscoelastic model shows no stress relaxation when held at a constant strain?
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Answer: A — Kelvin–Voigt
In the parallel Kelvin–Voigt model, once the strain is constant the dashpot carries no stress and the spring carries Eε₀ indefinitely. The Maxwell model relaxes to zero and the standard linear solid relaxes to a non-zero plateau, so only one of the three shows no relaxation.Which statements about bone are correct?
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Answer: A — cortical bone is stronger in compression than in tension; B — bone is mechanically anisotropic; C — cancellous bone has a much higher porosity than cortical bone
Cortical bone’s strength ranks compression (about 190 MPa) above tension (about 130 MPa) above shear (about 70 MPa), so it is weakest in shear, not compression; its osteons make it stiffest along the shaft. Cancellous porosity of 50–90% dwarfs cortical bone’s 5–10%.At normal walking speed, the stance phase occupies approximately what fraction of the gait cycle?
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Answer: A — 60%
Stance is about 60% and swing about 40%, the overlap of the two feet’s stance phases giving two double-support periods of about 10% each. 40% is the swing phase; 50% would leave no double support, which is the boundary between walking and running.A subject walks with a step length of 0.7 m at a cadence of 110 steps per minute. The walking speed (in m/s), to two decimal places, is ____.
Numerical answer — type the value.
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Answer: 1.28
Speed = step length × cadence = 0.7 × 110/60 = 1.283 m/s. Treating 0.7 m as the stride length and halving gives 0.64 m/s; forgetting the conversion from per minute gives 77 m/s.Which statements about the tensile behaviour of tendon are correct?
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Answer: A — the initial toe region reflects straightening of crimped collagen fibres; B — loading and unloading curves form a hysteresis loop; C — its apparent stiffness increases with strain rate
The J-shaped curve starts with a compliant toe region as the crimp is pulled out, then stiffens as fibres carry load directly; as a viscoelastic tissue it dissipates energy in each cycle and resists fast loading more than slow. The toe region is exactly why it is not linear from zero.In single-leg stance, the body weight above the stance hip is 600 N acting 10 cm medial to the hip joint centre, and the abductor muscles act vertically at 5 cm lateral to it. Treating all forces as vertical, the hip joint reaction force (in N) is ____.
Numerical answer — type the value.
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Answer: 1800
Moments about the hip: F_ab × 5 = 600 × 10, so F_ab = 1200 N; both the abductor pull and the body weight act downward on the pelvis relative to the femoral head, so the joint reaction is 1200 + 600 = 1800 N, three times the supported weight. Taking the difference, 600 N, treats the abductors as if they lifted the pelvis off the joint.