Strength of Materials
Strength of Materials
Simple stress is the internal resisting force per unit area acting on a body.
Simple Stress
Where: \(P\) = applied load, \(A\) = cross-sectional area
1. Simple Stress
Simple stress expresses the intensity of internal force per unit resisting area.
1.1 Normal Stress
Normal stress is caused by forces acting perpendicular to the resisting area.
- Tensile stress: pulls or elongates member
- Compressive stress: pushes or crushes member
Normal Stress
ME Board April 1998
Example. A force of \(10\ \text{N}\) is applied to one end of a \(10\ \text{inch}\) diameter circular rod. Calculate the stress.
Solution. Convert diameter to meters and solve for stress:
Answer: \(0.20\ \text{kPa}\)
1.2 Shearing Stress
Shearing stress is caused by forces acting parallel to the resisting area.
- Single shear: force resisted by one section
- Double shear: force resisted by two sections
- Punching shear: force cutting through plate perimeter
Shearing Stress
Where: \(V\) = shearing force, \(A\) = shear area
Punching Shear Force
Where: \(d\) = hole diameter, \(t\) = plate thickness
ME Board April 1996
Example. If the ultimate shear strength of a steel plate is \(42{,}000\ \text{psi}\), what force is necessary to punch a \(0.75\ \text{inch}\) diameter hole in a \(0.625\ \text{inch}\) thick plate?
Solution. Calculate punching force using cylindrical shear area:
Answer: \(61{,}850\ \text{lbs}\)
1.3 Bearing Stress
Bearing stress is the contact pressure developed between separate bodies in contact.
- Contact pressure between distinct surfaces
- Differs from internal compressive stress
- Projected area equals diameter times thickness
Bearing Stress
Where: \(P_b\) = contact load, \(A_b\) = projected contact area
2. Thin-Walled Pressure Vessels
Thin-walled pressure vessel is a tank or pipe with wall thickness small relative to diameter carrying internal fluid pressure.
2.1 Cylindrical Vessel
Cylindrical vessel under internal pressure develops tangential and longitudinal tensile stresses.
- Tangential stress: circumferential, girth, or hoop stress
- Longitudinal stress: acts along transverse sections
- Ratio: longitudinal stress is half tangential stress
Tangential Stress
Where: \(p\) = internal pressure, \(D\) = inside diameter, \(t\) = wall thickness
Longitudinal Stress
EE Board April 1996
Example. A cylindrical water tank is \(8\ \text{m}\) in diameter and \(12\ \text{m}\) high. If the tank is to be completely filled, determine the minimum thickness of the tank plating if the stress is limited to \(40\ \text{MPa}\).
Solution. Find maximum hydrostatic pressure at bottom and solve for thickness:
Answer: \(11.77\ \text{mm}\)
2.2 Spherical Vessel
Spherical vessel under internal pressure experiences uniform stress in all tangential directions.
Spherical Wall Stress
Where: \(p\) = internal pressure, \(D\) = diameter, \(t\) = thickness
EE Board October 1996
Example. What is the stress in a thin-walled spherical shell of diameter D and a wall thickness t when subjected to internal pressure p?
Solution. Balance internal pressure force against wall area:
Answer: \(\sigma = \dfrac{p D}{4t}\)
3. Simple Strain and the Stress-Strain Diagram
Simple strain is the ratio of total elongation or deformation to original length.
Simple Strain
Where: \(\delta\) = elongation, \(L\) = original length
Stress-strain diagram is a plot of normal stress vs unit strain obtained from tension testing.
- Proportional limit: linear relation limit
- Elastic limit: max stress without permanent set
- Yield point: yielding without added load
- Ultimate strength: maximum ordinate stress
- Rupture strength: stress at failure
- Allowable stress: yield or ultimate divided by factor of safety
- Modulus of resilience: elastic energy absorption capacity
- Modulus of toughness: total fracture energy capacity
4. Hooke’s Law
Hooke’s Law states that within the elastic limit, stress is directly proportional to strain.
Hooke’s Law
Where: \(E\) = Young’s modulus of elasticity
ME Board October 1995
Example. What is the modulus of elasticity if the stress is \(44{,}000\ \text{psi}\) and unit strain of \(0.00105\)?
Solution. Apply Hooke’s law:
Answer: \(41.905 \times 10^6\ \text{psi}\)
5. Axial Deformation
Axial deformation is the total elongation or compression of a member under axial loading.
Axial Deformation
Where: \(P\) = axial force, \(L\) = length, \(A\) = area, \(E\) = Young’s modulus
Self-Weight Deformation
Where: \(\rho\) = mass density, \(M\) = total mass
Example. A \(30\ \text{m}\) long aluminum bar is subjected to a tensile stress of \(172\ \text{MPa}\). Find the elongation if \(E = 69{,}116\ \text{MPa}\).
Solution. Calculate deformation from stress and length:
Answer: \(7.46\ \text{cm}\)
6. Stiffness
Stiffness is the ratio of applied steady force to resulting elastic displacement.
Axial Stiffness
Where: \(P\) = force, \(\delta\) = displacement, \(A\) = area, \(E\) = modulus, \(L\) = length
7. Shearing Deformation
Shearing deformation is the angular change in shape caused by shear forces without length change.
- Shear strain: angular distortion in radians
- Modulus of rigidity: ratio of shear stress to shear strain
Modulus of Rigidity
Where: \(\tau\) = shear stress, \(\gamma\) = shear strain
Shearing Deformation
Where: \(V\) = shearing force, \(A_s\) = shear area, \(G\) = shear modulus
8. Poisson’s Ratio
Poisson’s ratio is the ratio of lateral strain to longitudinal strain under axial loading.
- Negative ratio indicates lateral contraction under tension
- Typical value for steel: \(0.25\) to \(0.30\)
- Typical value for concrete: \(0.20\)
Poisson’s Ratio
Where: \(\varepsilon_y\) = lateral strain, \(\varepsilon_x\) = longitudinal strain
Relation Between Elastic Constants
Where: \(E\) = Young’s modulus, \(G\) = shear modulus, \(\nu\) = Poisson’s ratio
Example. Compute the value of the shear modulus G of steel whose modulus of elasticity E is \(200\ \text{GPa}\) and Poisson’s ratio is \(0.30\).
Solution. Substitute E and Poisson’s ratio into relation formula:
Answer: \(76{,}923\ \text{MPa}\)
9. Thermal Stress
Thermal stress is the stress developed in a constrained body caused by internal forces due to a temperature change.
- Restrained deformation: internal thermal stress is induced
- Unrestrained deformation: body expands freely without stress
- Temperature rise: causes compressive stress in fixed members
- Temperature drop: causes tensile stress in fixed members
Thermal Deformation
Where: \(\alpha\) = coefficient of thermal expansion, \(L\) = length, \(\Delta T\) = temperature change
Thermal Stress
Where: \(\alpha\) = coefficient of thermal expansion, \(E\) = modulus of elasticity, \(\Delta T\) = temperature change
Example. Calculate the thermal deformation of a \(200\ \text{ft}\) steel rail subjected to a temperature change from \(0^\circ\text{F}\) to \(100^\circ\text{F}\) if \(\alpha = 10 \times 10^{-6}\ /\text{}^\circ\text{F}\).
Solution. Substitute values into the thermal deformation equation:
Answer: \(0.20\ \text{ft}\)
10. Torsion
Torsion refers to the twisting of circular shafts caused by applied torque.
10.1 Shearing Stress, Maximum Shearing Stress, Angular Deformation, Power Transmitted
Torsional shearing stress is the internal shear stress developed at any radial distance from the longitudinal axis.
- Acts tangentially across circular cross sections
- Varies linearly from zero at axis to maximum at surface
- Polar moment of inertia measures torsional resistance
Torsional Shearing Stress
Where: \(T\) = applied torque, \(\rho\) = radial distance from center, \(J\) = polar moment of inertia
Solid Shaft Maximum Shear Stress
Where: \(T\) = torque, \(d\) = shaft diameter
Hollow Shaft Maximum Shear Stress
Where: \(D\) = outer diameter, \(d\) = inner diameter
Angle of Twist
Where: \(T\) = torque, \(L\) = length, \(J\) = polar moment of inertia, \(G\) = shear modulus
Power Transmitted in Watts
Where: \(f\) = rotational frequency in \(\text{rev/s}\), \(T\) = torque in \(\text{N-m}\)
CE Board May 1996
Example. Find the maximum allowable torque in \(\text{kN-m}\) for a \(50\ \text{mm}\) diameter steel shaft when the allowable shearing stress is \(81.5\ \text{MPa}\).
Solution. Solve for torque using solid shaft shear formula:
Answer: \(2.0\ \text{kN-m}\)
11. Helical Springs
Helical spring is a wire bar wound into a helix to absorb energy or exert forces under axial load.
11.1 Maximum Shearing Stress
Maximum shearing stress in a helical spring includes combined direct shear and torsional shear forces.
Light Helical Spring Shear Stress
Where: \(P\) = axial load, \(R\) = mean coil radius, \(d\) = wire diameter
Heavy Helical Spring Shear Stress (Wahl Formula)
Where: \(m = \dfrac{2R}{d}\) = spring index
ECE Board November 1997
Example. Determine the maximum shearing stress in a helical steel spring composed of \(20\) turns of \(20\ \text{mm}\) diameter wire on a mean radius of \(80\ \text{mm}\) when supporting a load of \(2\ \text{kN}\).
Solution. Compute spring index \(m\), Wahl factor \(K\), and shear stress:
Answer: \(120.6\ \text{MPa}\)
11.2 Deformation
Spring deformation is the axial deflection produced along the axis of a helical spring under load.
Helical Spring Deflection
Where: \(P\) = axial load, \(R\) = mean coil radius, \(n\) = active turns, \(G\) = shear modulus, \(d\) = wire diameter
11.3 Springs in Series
Springs in series are connected end-to-end such that each spring transmits the total axial force.
- All springs carry equal axial load
- Total deflection equals sum of individual deflections
- Equivalent spring stiffness is smaller than individual stiffnesses
Springs in Series Stiffness
Where: \(k_1, k_2, \dots, k_n\) = stiffness of individual springs
11.4 Springs in Parallel
Springs in parallel are connected side-by-side to share the total load with equal deflections.
- All parallel springs undergo identical axial deflection
- Total load equals sum of individual spring forces
- Equivalent stiffness equals sum of individual stiffnesses
Springs in Parallel Stiffness
Where: \(k_1, k_2, \dots, k_n\) = stiffness of individual parallel springs
12. Shear and Moment in Beams
Shear and moment in beams describe internal vertical forces and bending couples developed along a transverse beam element under load.
12.1 Statically Determinate Beams
Statically determinate beams are beams whose support reactions can be calculated completely using static equilibrium equations alone.
- Cantilever beam: fixed at one end, free at other
- Simple beam: supported at ends by roller and pin
- Overhanging beam: extends beyond one or both supports
Static Equilibrium Equations
12.2 Statically Indeterminate Beams
Statically indeterminate beams possess more support reactions than available equations of static equilibrium.
- Propped beam: fixed at one end, supported at other
- Restrained beam: fixed rigidly at both end supports
- Continuous beam: supported across three or more supports
Degree of Indeterminacy
Where: \(R\) = number of unknown reactions, \(N_e\) = static equilibrium equations
12.3 Types of Loading
Types of loading describe external loads and moments applied along a beam span.
- Concentrated load: single force applied at a point
- Uniformly distributed load: constant load per unit length
- Uniformly varying load: linearly changing force intensity
- Applied couple: concentrated moment applied to span
12.4 Shear and Moment Diagrams
Shear and moment diagrams graphically display internal shear force \(V\) and bending moment \(M\) along the beam length.
- Positive shear: left portion moves upward relative to right
- Negative shear: left portion moves downward relative to right
- Positive moment: bends beam concave upward (sagging)
- Negative moment: bends beam concave downward (hogging)
Resisting Bending Moment
12.5 Relationship Between Load, Shear and Moment
Relationship between load, shear, and moment defines calculus derivatives connecting applied loads to shear force and bending moment curves.
Shear as Slope of Moment Diagram
Load as Slope of Shear Diagram
Where: \(w\) = distributed load intensity
12.6 Properties of Shear and Moment Diagrams
Properties of shear and moment diagrams dictate geometric features and critical stress points along a beam.
- Moment change: area under shear diagram equals moment change
- Shear change: area under load diagram equals shear change
- Maximum moment: occurs where shear force equals zero
- Point of inflection: location where bending moment equals zero
- Concentrated load: causes abrupt step in shear diagram
- Uniform load: produces linear shear and parabolic moment
13. Stresses in Beams
Stresses in beams represent internal flexural normal stresses and transverse shearing stresses developed across beam cross sections.
13.1 Radius of Curvature
Radius of curvature measures the radius of the bent neutral axis of a beam under flexure.
Radius of Curvature of Beam
Where: \(E\) = Young’s modulus, \(I\) = moment of inertia, \(M\) = bending moment
13.2 Flexure Formula
Flexure formula calculates normal bending stress developed at any distance from the neutral axis.
- Plane sections before bending remain plane after bending
- Material obeys Hooke’s law within elastic limit
- Modulus of elasticity equal in tension and compression
- Beam is initially straight with uniform cross section
Flexure Formula
Where: \(M\) = bending moment, \(y\) = distance from neutral axis, \(I\) = centroidal moment of inertia
Maximum Flexural Stress
Where: \(c\) = distance to extreme fiber, \(S = \dfrac{I}{c}\) = section modulus
CE Board May 2003
Example. A simple beam \(10\ \text{m}\) long carries a concentrated load of \(500\ \text{kN}\) at midspan. Compute the maximum bending stress if the beam is rectangular with width \(b = 300\ \text{mm}\) and depth \(d = 600\ \text{mm}\).
Solution. Calculate maximum bending moment and apply flexure formula:
Answer: \(69.44\ \text{MPa}\)
13.3 Shear Flow
Shear flow is the longitudinal shearing force per unit length transmitted across longitudinal planes of a beam.
Shear Flow
Where: \(V\) = vertical shear force, \(Q = A \bar{y}\) = first moment of area, \(I\) = moment of inertia
Beam Shearing Stress Formula
Where: \(b\) = width of cross section at cut level
Formula Summary
| Formula Name | Formula |
|---|---|
| Simple Stress | \(\sigma = \dfrac{P}{A}\) |
| Normal Stress | \(\sigma = \dfrac{P}{A}\) |
| Shearing Stress | \(\tau = \dfrac{V}{A}\) |
| Punching Shear Force | \(P = \tau (\pi d t)\) |
| Bearing Stress | \(\sigma_b = \dfrac{P_b}{A_b}\) |
| Tangential Stress | \(\sigma_T = \dfrac{p D}{2t}\) |
| Longitudinal Stress | \(\sigma_L = \dfrac{p D}{4t}\) |
| Spherical Wall Stress | \(\sigma = \dfrac{p D}{4t}\) |
| Simple Strain | \(\varepsilon = \dfrac{\delta}{L}\) |
| Hooke’s Law | \(\sigma = E \varepsilon\) |
| Axial Deformation | \(\delta = \dfrac{P L}{A E} = \dfrac{\sigma L}{E}\) |
| Self-Weight Deformation | \(\delta = \dfrac{\rho g L^2}{2E} = \dfrac{M g L}{2 A E}\) |
| Axial Stiffness | \(k = \dfrac{P}{\delta} = \dfrac{A E}{L}\) |
| Modulus of Rigidity | \(G = \dfrac{\tau}{\gamma}\) |
| Shearing Deformation | \(\delta_s = \dfrac{V L}{A_s G} = \dfrac{\tau L}{G}\) |
| Poisson’s Ratio | \(\nu = -\dfrac{\varepsilon_y}{\varepsilon_x}\) |
| Relation Between Elastic Constants | \(G = \dfrac{E}{2(1 + \nu)}\) |
| Thermal Deformation | \(\delta_T = \alpha L \Delta T\) |
| Thermal Stress | \(\sigma_T = \alpha E \Delta T\) |
| Torsional Shearing Stress | \(\tau = \dfrac{T \rho}{J}\) |
| Solid Shaft Maximum Shear Stress | \(\tau_{\max} = \dfrac{16 T}{\pi d^3}\) |
| Hollow Shaft Maximum Shear Stress | \(\tau_{\max} = \dfrac{16 T D}{\pi (D^4 - d^4)}\) |
| Angle of Twist | \(\theta = \dfrac{T L}{J G}\) |
| Power Transmitted in Watts | \(P = 2 \pi f T\) |
| Light Helical Spring Shear Stress | \(\tau = \dfrac{16 P R}{\pi d^3} \left( 1 + \dfrac{d}{4R} \right)\) |
| Heavy Helical Spring Shear Stress (Wahl Formula) | \(\tau = \dfrac{16 P R}{\pi d^3} \left( \dfrac{4m - 1}{4m - 4} + \dfrac{0.615}{m} \right)\) |
| Helical Spring Deflection | \(\delta = \dfrac{64 P R^3 n}{G d^4}\) |
| Spring Constant | \(k = \dfrac{P}{\delta} = \dfrac{G d^4}{64 R^3 n}\) |
| Springs in Series Stiffness | \(\dfrac{1}{k_{eq}} = \dfrac{1}{k_1} + \dfrac{1}{k_2} + \dots + \dfrac{1}{k_n}\) |
| Springs in Parallel Stiffness | \(k_{eq} = k_1 + k_2 + \dots + k_n\) |
| Static Equilibrium Equations | \(\sum F_x = 0 \quad \sum F_y = 0 \quad \sum M = 0\) |
| Degree of Indeterminacy | \(D_i = R - N_e\) |
| Resisting Shear Force | \(V = \sum F_y\) |
| Resisting Bending Moment | \(M = \sum M_c\) |
| Shear as Slope of Moment Diagram | \(\dfrac{dM}{dx} = V\) |
| Load as Slope of Shear Diagram | \(\dfrac{dV}{dx} = -w\) |
| Radius of Curvature of Beam | \(\rho = \dfrac{E I}{M}\) |
| Flexural Strain Relation | \(\varepsilon = \dfrac{y}{\rho}\) |
| Flexure Formula | \(\sigma = \dfrac{M y}{I}\) |
| Maximum Flexural Stress | \(\sigma_{\max} = \dfrac{M c}{I} = \dfrac{M}{S}\) |
| Rectangular Beam Maximum Bending Stress | \(\sigma_{\max} = \dfrac{6 M}{b d^2}\) |
| Shear Flow | \(q = \tau b = \dfrac{V Q}{I}\) |
| Beam Shearing Stress Formula | \(\tau = \dfrac{V Q}{I b}\) |
| Rectangular Beam Maximum Shear Stress | \(\tau_{\max} = \dfrac{3 V}{2 b d} = \dfrac{3 V}{2 A}\) |
| Fastener Spacing in Built-Up Beams | \(s = \dfrac{R I}{V Q}\) |