Mechanical Properties of Solids · Lesson 6 of 7
Elasticity in Engineering and Nature
“Apply elastic stiffness, failure stress and geometry to crane ropes, beams, columns and a mountain-height estimate.”
• Size an idealized crane rope from an allowable stress and interpret a safety margin. • Use beam-sag scaling to compare length, breadth, depth and material stiffness. • Explain I-shaped sections and the role of column end support and instability. • Derive the chapter’s order-of-magnitude mountain-height estimate without confusing it with hydraulic pressure.
Engineering needs stiffness and strength
A structure must carry its loads without excessive deformation or failure. A large Young’s modulus reduces small elastic deflection, but it does not by itself specify the maximum safe load. Yield strength, fracture behaviour, geometry, connections and support conditions also matter. A bridge must carry its own weight and moving traffic while tolerating other forces, such as wind. The simple chapter models explain important design choices; a complete design needs more information than one modulus.
Crane ropes and a margin below yield
Suppose a crane is intended to lift 10 metric tonnes. One tonne is 1000 kg, so the load mass is 10⁴ kg and its static weight is 98,000 N for g = 9.8 m/s². In the source’s illustration, mild steel has a chosen yield strength of 300 MPa. That is a stated design input for this example, distinct from the representative 250 MPa steel entry in the earlier material table.
Problem
Estimate the circular steel area and equivalent radius at 300 MPa, then apply a factor of ten to the static design load.
- 1.At the yield threshold A = 98,000/(300 × 10⁶) = 3.27 × 10⁻⁴ m².
- 2.For a solid equivalent circle r = √(A/π) = 0.0102 m, about 1.0 cm.
- 3.A tenfold load factor requires ten times the area, not ten times the radius. The equivalent radius becomes √10 × 1.02 cm = 3.22 cm, approximately 3 cm.
- 4.The chapter’s illustration shows how a safety allowance changes dimensions. Real ropes must also be checked for acceleration, wear, connections, fatigue and construction details.
A solid wire several centimetres in radius would be difficult to bend and handle. A practical lifting rope uses many thinner wires braided or laid together. This provides a useful combination of flexibility, manufacturability and strength. An equivalent solid area is a first calculation; the actual load-carrying metal area and rope construction must be used for a real rope. The example does not certify a particular commercial rope.
How a beam resists sagging
A beam supported near its ends and carrying a central downward load bends. In ordinary sagging bending, material near the upper surface shortens while material near the lower surface stretches. Material farther from the central layer contributes strongly to resistance because its longitudinal deformation under a given curvature is greater. The resulting deflection depends both on Y and on how the cross-section distributes material.
The source states this result without working through its beam-calculus derivation, so the goal here is to understand its conditions and consequences. The cubic span dependence means doubling an otherwise unchanged span increases sag by eight. Doubling breadth halves sag. Doubling vertical depth reduces sag to one eighth. Doubling Y halves sag. The depth effect explains why orienting a rectangular ruler on edge makes it harder to bend vertically than laying it flat.
Problem
A rectangular beam’s span and material are fixed. Compare doubling its breadth with doubling its vertical depth.
- 1.From δ ∝ 1/b, doubling b gives δnew/δold = 1/2.
- 2.From δ ∝ 1/d³, doubling d gives δnew/δold = 1/8.
- 3.Each change doubles cross-sectional area, but placing the extra dimension in the vertical depth is much more effective for this bending direction.
- 4.If the span also doubled, its factor of 8 would cancel the benefit of doubling depth. Keep all changed variables in the comparison.
Why I sections and supported ends matter
Simply making a bar deeper and thinner can introduce sideways instability. A member can bend or buckle laterally under an unfavourable load arrangement even when its direct axial stress is below material strength. The source illustrates how a thin deep shape can move sideways and motivates the common I-shaped cross-section. Its broad flanges place material where it helps resist bending and provide useful load-bearing surfaces; the web joins them. For a given bending requirement, this arrangement can reduce material and weight compared with a poorly distributed section.
The rectangular-beam formula should not be applied to an I section by pretending all its empty space is solid. The section has a different geometrical bending resistance. The chapter’s qualitative argument is about distribution of material, efficient depth and stability, not an exact design formula for every I beam.
Columns are often compressed rather than stretched. Their ends and restraints affect whether they remain straight under load. The source contrasts a pillar with rounded ends with one having broader distributed ends. Broadening and appropriate support can improve load transfer and restraint, but a drawing of end shape alone cannot determine a column’s exact critical load. Length, section shape, straightness, material and how rotation/sideways motion are constrained also matter. Do not confuse a low FL/(AY) shortening with proof that a slender column cannot buckle.
An approximate upper scale for mountain height
A tall mountain loads the rock beneath it. Its base is not subjected to an equal fluid pressure on every side: the overburden acts mainly vertically while lateral constraints differ. The stress state therefore includes shear that can permit rock to deform and flow. The chapter estimates the shear scale by the weight per unit base area, of order ρgh. It is an order-of-magnitude model, not a precise solution for the shape and geology of every mountain.
Problem
Use a representative critical rock stress of 30 × 10⁷ Pa, density 3 × 10³ kg/m³ and g = 10 m/s².
- 1.Set ρgh ≈ 30 × 10⁷ Pa = 3.0 × 10⁸ Pa.
- 2.h ≈ (3.0 × 10⁸)/(3.0 × 10³ × 10) = 1.0 × 10⁴ m = 10 km.
- 3.The result is a plausible terrestrial mountain-height scale, a little above Everest’s height. It is not a sharp universal ceiling; rock type, geometry, time-dependent deformation and geological support are not fully represented.
- 4.The free sides and nonuniform support make this a shear/failure argument, not an application of water’s bulk modulus.
A hill or heap tends to spread over a broad base rather than stand as an arbitrarily tall narrow stack. For the rock-mountain model, growing height raises the basal stress until irreversible deformation can occur. A loose sand heap is governed by grain friction and rearrangement as well, so the mountain formula should not be presented as a full theory of granular slopes. The chapter’s shared lesson is that shape and attainable height are constrained by how material carries loads.
Quick Quiz
Quiz
If a rope’s required circular metal area increases by a factor of 10, its equivalent radius increases by
Area is πr², so radius scales with the square root of area.
Under the rectangular-beam model, what happens to sag if both span l and depth d double while W, b and Y stay fixed?
The factor 2³ from l³ cancels the factor 2³ from d³.
Why does a small computed axial compression fail to prove a slender column is safe?
Material strain and stability are separate checks; support conditions and geometry affect buckling.
In the mountain estimate, what is ρgh?
Density × gravity × height has units Pa. The model compares this stress scale with a representative rock failure/flow stress.
Which quantity directly controls the initial elastic sag of the specified beam, while yield strength controls a different failure limit?
The small-deflection formula contains Y. High Y does not automatically imply a high yield or fracture stress.
Practice Problems
- In an ideal static stress-limited model, a rope supports a 3000 kg load with allowable working stress 120 MPa. Find the required load-carrying metal area and equivalent solid circular radius using g = 9.8 m/s². State the difference between an allowable working stress and a material’s yield stress.
- A design change requires four times the load-carrying area for an otherwise equivalent circular rope model. Find the factor by which its equivalent radius changes. Explain why a bundle of fine wires is useful even when a large total metal area is required.
- For the rectangular-beam model in the lesson, compare the sag reduction obtained by (a) doubling breadth and (b) increasing vertical depth by 50%, with load, span and material unchanged. Then determine how sag changes if both span and depth increase by 50% while breadth stays fixed.
- Two hypothetical materials have (Y, yield stress) values (200 GPa, 150 MPa) and (100 GPa, 300 MPa). At an applied axial stress of 100 MPa, find each linear-model strain and check the stated yield thresholds. Explain which material is stiffer and which has the higher yield strength, without treating these as the same property.
- Estimate a mountain-height scale using critical rock stress 1.8 × 10⁸ Pa, density 3.0 × 10³ kg/m³ and g = 10 m/s². How would this estimate change if g were 5 m/s² with the same assumed density and critical stress? Explain why the estimate is not a uniform bulk-compression calculation or an exact universal height limit.
Key Takeaways
• Use an allowable stress and explicit load assumptions when estimating a rope area; safety factors act on area through the chosen model. • For the specified rectangular beam, sag scales as l³/(bd³Y). • I sections distribute material efficiently, while beams and columns still require stability and support checks. • The mountain argument compares ρgh with a representative rock stress and gives a height scale, not an exact universal maximum.