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1. Simple Stress and Strain

This section covers idealized simple stress and strain. In more complex cases, some of these simplifications may not apply.

1.1. Loading

Components of engineering structures or mechanisms experience various forces. For example, the cylindrical bar below is used in a lifting mechanism and so it is subjected to tension force along the axis.

A member of a mechanism may experience loading in one of the following ways.

(a) Static, dead, or non-fluctuating loads e.g. force due to gravity
(b) Live loads e.g. train on the railway
(c) Impact or shock loads e.g. hammering
(d) Fatigue, fluctuating, or altering loads e.g. repeated loaded swivel, crack propagation

1.2. Direct or Normal Stress (\(\sigma\))

This is when a uniform-structured member or a bar is subjected to uniform tension or compression.

\[\begin{align}\tag{1} \sigma\ (\text{stress}) = \frac{F\ (\text{load})}{A\ (\text{area})}\ \frac{N}{m^2} \end{align}\]

1.3. Direct or Normal Strain (\(\epsilon\))

This is a dimensionless measure of a change in length of a member, after it has been subjected to tension or compression.

\[\begin{align}\tag{2} \epsilon\ (\text{strain}) = \frac{\Delta L\ (\text{change in length})}{L\ (\text{original length})} \end{align}\]

The strain in Equation 2 may be expressed as a percentage, \(\epsilon = \frac{\Delta L}{L} \times 100 \%\). Tensile stress and strain are considered positive, while compressive ones are considered negative.

1.4. Elastic Materials (Hooke’s Law, \(F = ke\))

A member is said to be elastic or obey Hooke’s law if it returns to its original length when the load is withdrawn. In this case, the stress is proportional to strain.

\[\frac{\sigma}{\epsilon} = \text{constant}\]

Common classification of materials:

Homogeneous material — uniform structure without any flaws or discontinuities e.g. steel
Inhomogeneous material — the structure varies from point to point e.g. concrete, some cast iron
Isotropic — uniform properties throughout in all directions
Anisotropic — varying properties
Orthotropic — different properties in different planes e.g. wood, some composites

Here, we assume homogeneous materials with isotropic properties.

1.5. Modulus of Elasticity (Young’s Modulus, \(E\))

The constant derived above, within which Hooke’s law applies, is referred to as the modulus of elasticity or the Young’s modulus, E.

\[\begin{align}\tag{3} \frac{\sigma}{\epsilon} = E \end{align}\]

\(E = \frac{F}{A}/ \frac{\Delta L}{L} = \frac{FL}{A \Delta L}\)

The modulus of elasticity in Equation 3 is assumed to be the same, both in tension and in compression. Since its value is large for most engineering materials, \(~ \times 10^9 N/m^2\), the strain will be small, \(~ 0.3\%\). Usually, the deformations are typically small with respect to the original dimension.

The modulus of elasticity of a material is a measure of its stiffness, or its resistance to elastic deformation under load. For an actuator in a mechatronics system, the modulus of elasticity of the material from which it’s made will significantly influence its performance.

A material with a high modulus of elasticity (like steel) will deform less under the same load compared to a material with a lower modulus of elasticity (like rubber).

Therefore, in an actuator, a higher modulus of elasticity would result in less deformation when force or torque is applied, which means the actuator would be more efficient at converting the electrical energy into mechanical movement.

However, materials with a high modulus of elasticity tend to be less ductile and may fail more abruptly than materials with a lower modulus of elasticity.

1.6. Tensile Test

The modulus of elasticity is determined by carrying out a standard test as shown in the figure below. In this test, a circular bar of a uniform cross-section is subjected to a gradually increasing tensile load until failure occurs. The change in the gauge length is recorded as the loading operation continues.

tensileTest
Figure 1. Tensile test.

Origin to point A: Hooke’s law is obeyed. Stress is proportional to strain.
At point A: limit of proportionality
Between points A and B: the material may still be elastic but Hooke’s law is not obeyed.
At point B: elastic limit
In most practical cases, points A and B are coincident.
Beyond point B: plastic deformation
Strains are not totally recoverable and permanent deformations are recorded when load is removed.
At point C: upper yield point
At point D: lower yield point
A rapid increase in strain occurs without a corresponding increase in stress.
Between points E and F: necking occurs
The stress reaches the maximum value, the ultimate tensile strength.
At point F: the specimen breaks

  • The ductility is the capacity of a material to allow large plastic deformations.

  • For some materials, say, non-ferrous and high-carbon steel, the difference between points C and D cannot be detected. In such cases, proof stress is used to show the onset of plastic strain.

1.7. Poisson’s Ratio (\(v\))

When a bar is subjected to a tension load, it experiences an increase in length, \(\Delta L\), giving a longitudinal strain in the bar, \(\epsilon_{Long}\).

\(\epsilon_{Long} = \frac{\Delta L}{L}\)

The breath, \(b\), and the depth, \(d\), of the bar will also reduce, thereby altering the lateral dimensions of the bar. The lateral strain, \(\epsilon_{Lat}\), can be given by the following equation.

\(\epsilon_{Lat} = - \frac{\Delta b}{b} = - \frac{\Delta d}{d}\)

\[\text{Poisson's ratio, } v = \frac{\epsilon_{Lat}}{\epsilon_{Long}}\]
\[\begin{align}\tag{4} \text{Poisson's ratio, } v = - \frac{\Delta d}{d} \times \frac{L}{\Delta L} \end{align}\]

\(\epsilon_{Long} = \frac{\sigma_{Long}}{E} = \frac{\sigma}{E}\)

\[\epsilon_{Lat} = v \times \frac{\sigma}{E}\]

1.8. Modulus of Rigidity (\(G\))

1.8.1. Shear Stress (\(\tau\))

\[\begin{align}\tag{5} \tau\ (\text{shear stress}) = \frac{F\ (\text{shear load})}{A\ (\text{area resisting shear})} \end{align}\]
\[\begin{align}\tag{6} \tau\ (\text{in double shear stress}) = \frac{F}{2 \times A } \end{align}\]

1.8.2. Shear Strain (\(\gamma\))

\[\begin{align}\tag{7} \gamma = \frac{\Delta L}{L} \end{align}\]
  • The modulus of rigidity, \(G\), or shear modulus is comparable to modulus of elasticity and it is calculated as follows.

\[\begin{align}\tag{8} G = \frac{\tau}{\gamma} = \text{constant} \end{align}\]

The shear modulus of a material, also known as the modulus of rigidity, is a measure of the material’s ability to resist deformation under shear stress or torsional loading. In a torsionally-loaded component, like a torsion bar in an actuator system, the shear modulus plays a vital role in determining the component’s performance and efficiency.

A material with a high shear modulus will be more resistant to deformation under torsional loading, meaning that it will twist less for a given torque.

Therefore, in a mechatronic system like an actuator, using a material with a high shear modulus for components subjected to torsion will provide more precise and efficient torsional movements, as less of the applied torque will be "lost" to deformation of the component.

However, it’s important to note that materials with a high shear modulus will generally be less ductile and may be more prone to brittle fracture under high torsional loads, as they are less capable of absorbing energy through deformation. Therefore, the selection of material for such components should consider not only the shear modulus but also other properties such as toughness and fatigue strength.

  • Before attempting a question, make sure you understand the fundamental concept. Draw sketches where necessary. Make sure you derive your equations correctly, then substitute the values. This is useful in preparing a general algorithm and reducing approximation errors.

  • Your answers should be in SI units and scientific notation. Use the following form \(p.qrs \times 10^t\) units. Where \(p,q,r,s\) is a number between \(0-9\), \(t\) is a signed integer, and units is the relevant SI units. This is useful in establishing a quick objective meaning of size/scale.

1.9. A bar with various cross-sections

Idea

Example 1

A sectional bar of \(E = 210\ GN/m^2\) is subjected to an axial tensile load of \(F_1 = F_2 = 25\ kN\). The circular section \(C_{s1}\) has a diameter of \(25\ mm\). A square cross-section of dimension \(C_{s2} = 40\ mm\), and circular section \(C_{s3} = 20\ mm\) in diameter. \(L_{1} = 200\ mm\), \(L_{2} = 110\ mm\), \(L_{3} = 450\ mm\)

What is the stress in each section and the total extension of the sectional bar?

sectionalBar
Figure 2. Sectional bar.
Solution:

stress in each section

\[\begin{align*} \sigma &= \frac{F}{A} \\ \sigma_{s1} &= \frac{F}{\pi \times \big( \frac{D_1}{2} \big)^2} \\ \sigma_{s2} &= \frac{F}{L_2^2} \\ \sigma_{s3} &= \frac{F}{\pi \times \big( \frac{D_3}{2} \big)^2} \end{align*}\]

total extension (\(\Delta L\))

\[\begin{align*} \epsilon &= \frac{\sigma}{E} \\ \epsilon &= \frac{\Delta L}{L} \\ \Delta L &= \Delta L_{s1} + \Delta L_{s2} + \Delta L_{s3} \\ \Delta L &= \frac{ \sigma_{s1} L_1 + \sigma_{s2} L_2 + \sigma_{s3} L_3}{E} \end{align*}\]

1.10. A bored circular cross-section

Idea

Example 2

A \(30\ mm\) diameter bar is subjected to an axial tensile load of \(115\ kN\). Under the action of this load a \(200\ mm\) gauge length is found to extend \(0.15 \times 10^{-3}\ mm\). Determine the modulus of elasticity for the bar material.

To reduce weight whilst keeping the external diameter constant, the bar is bored axially to produce a cylinder of uniform thickness, what is the maximum diameter of bore possible given that the maximum allowable stress is \(245\ MN/m^2\)? The load can be assumed to remain constant at \(115\ kN\).

What will be the change in the outside diameter of the bar under the limiting stress quoted above? (\(E = 210\ GN/m^2\) and \(v =\ 0.3\))

Solution:

modulus of elasticity

\[\begin{align*} \frac{\sigma}{\epsilon} &= E \\ \epsilon &= \frac{\Delta L}{L} \\ E &= \sigma \frac{L}{\Delta L} \\ E &= \frac{F}{A} \times \frac{L}{\Delta L} \\ E &= \frac{F}{\pi \times \big(\frac{D}{2}\big)^2} \times \frac{L}{\Delta L} \end{align*}\]

maximum diameter of bore

\[\begin{align*} \sigma &= \frac{F}{A} \\ A &= \pi \times \big( \frac{D}{2} \big)^2 - \pi \times \big( \frac{d}{2} \big)^2 \\ A &= \frac{\pi}{4} \times (D^2 - d^2) \\ (D^2 - d^2) &= \frac{F}{\sigma} \times \frac{4}{\pi} \\ d &= \sqrt{D^2 - \big( \frac{F}{\sigma} \times \frac{4}{\pi} \big)} \end{align*}\]

change in the outside diameter

\[\begin{align*} \epsilon &= v \times \frac{\sigma}{E} \\ \frac{\Delta d}{d} &= v \times \frac{\sigma}{E} \\ \Delta d &= v \times \frac{\sigma}{E} \times d \end{align*}\]

1.11. A stressed coupling

Idea

Example 3

The coupling shown below is constructed from steel of a rectangular cross-section and is designed to transmit a tensile force of \(50\ kN\). If the bolt is of \(14.5\ mm\) diameter calculate:

\(F = F_1 = F_2 = 50 kN\)
\(F_3 = F_4 = 25 kN\)
\(W_1 = 55 mm\)
\(T_1 = T_2 = T_3 = 7 mm\)

(a) the shear stress in the bolt;
(b) the direct stress in the plate;
(c) the direct stress in the forked end of the coupling.

coupling
Figure 3. Coupling plate.
Solution:

shear stress in the bolt

\[\begin{align*} \tau &= \frac{F}{A} \\ \tau &= \frac{F}{2 \times A} \\ \tau &= \frac{4 \times F}{2 \times \pi D^2} \end{align*}\]

direct stress in the plate

\[\begin{align*} \sigma = \frac{F}{A} \end{align*}\]

direct stress in the forked end

\[\begin{align*} \sigma = \frac{F}{A} \end{align*}\]

1.12. Extension of tapered bar

Idea

Example 4

Derive an expression for the total extension of the tapered bar of the circular cross-section shown below when it is subjected to an axial tensile load \(W\).

taperedBar
Figure 4. Tapered bar.
Solution:

Consider a very small length \(dx\) with diameter \(P\), experiencing the tensile force of \(W\), and then integrate over the whole length, \(L\).

\[\begin{align*} \Delta L &= \frac{\sigma}{E} \times L \\ \Delta L &= \frac{\sigma}{E} \times dx \\ \Delta L &= \frac{W}{AE} \times dx \\ \Delta L &= \frac{W}{\frac{\pi}{4} \times P^2 \times E} \times dx \end{align*}\]
taperedBarSec
Figure 5. Tapered bar section.
\[\begin{align*} P &= (2 q)+d \\ \frac{q}{x} &= \frac{D-d}{2}/L \\ q &= \frac{(D-d)}{2L} \times x \\ P &= \frac{(D-d)}{L} x + d \\ \text{Letting } k &= \frac{(D-d)}{L} \text{; } P = d + kx \\ \Delta L &= \frac{W}{\frac{\pi}{4} \times (d + kx)^2 \times E} dx \\ \Delta L &= \int_{0}^{L} \frac{4W}{\pi \times (d + kx)^2 \times E} \,dx \\ \Delta L &= \frac{4W}{\pi E} \int_{0}^{L} (d + kx)^{-2} \,dx \\ \text{Let } u &= d + kx \text{; } \frac{du}{dx} = k \text{; } dx = \frac{du}{k} \\ \Delta L &= \frac{4W}{k \pi E} \int_{0}^{L} (u)^{-2} \,du \\ \Delta L &= \frac{4W}{k \pi E} \begin{bmatrix} -(u)^{-1} \end{bmatrix} _{0}^{L} \\ \Delta L &= \frac{4W}{k \pi E} \begin{bmatrix} -(d + kx)^{-1} \end{bmatrix} _{0}^{L} \\ \Delta L &= \frac{4W}{k \pi E} \biggr( (\frac{-1}{d + kL}) - (\frac{-1}{d})\biggr) \\ \Delta L &= \frac{4W}{k \pi E} \biggr( \frac{1}{d} - \frac{1}{d + kL} \biggr) \\ \Delta L &= \frac{4W}{k \pi E} \biggr( \frac{kL}{d(d + kL)} \biggr) \\ \Delta L &= \frac{4W}{\pi E} \times \frac{L}{d(d + \Big( \frac{(D-d)}{L} \Big) \times L)} \\ \Delta L &= \frac{4WL}{\pi E d D} \end{align*}\]

1.13. Error in using the mean diameter

Idea

Example 5

The mean diameter is used in calculating the modulus of elasticity, \(E\), of a bar tapered from \((D + a)\) diameter to \((D-a)\) diameter. Show that the error involved in calculating \(E\) is \(\big( \frac{10a}{D} \big)^2 \%\).

Solution:

\(E\) of a tapered circular bar should be derived similar to Extension of tapered bar :

\[\begin{align*} \Delta L &= \frac{4WL}{\pi E d D} \\ E &= \frac{4WL}{\pi \Delta L d D} \\ E &= \frac{4WL}{\pi \Delta L (D-a) (D+a)} \\ E &= \frac{4WL}{\pi \Delta L (D^2-a^2)} \end{align*}\]

but when mean diameter is used instead,

\[\begin{align*} D_m &= \frac{(D-a)+(D+a)}{2} = D \\ \Delta L &= \frac{WL}{AE_m} \\ \Delta L &= \frac{WL}{(\frac{\pi}{4}D^2)E_m} \\ E_m &= \frac{4WL}{\pi \Delta L (D^2)} \\ \% \text{error} &= \frac{E-E_m}{E} \times 100 \\ \% \text{error} &= \frac{\big( \frac{4WL}{\pi \Delta L (D^2-a^2)} \big) - \big( \frac{4WL}{\pi \Delta L (D^2)} \big)} {\frac{4WL}{\pi \Delta L (D^2-a^2)}} \times 100 \\ \% \text{error} &= \frac{\big( \frac{1}{ (D^2-a^2)} \big) - \big( \frac{1}{D^2} \big)} {\frac{1}{(D^2-a^2)}} \times 100 \\ \% \text{error} &= \big( \frac{D^2 - (D^2 - a^2)}{ D^2(D^2-a^2)} \times (D^2-a^2) \big) \times 100 \\ \% \text{error} &= \Big( \frac{10 a}{D} \Big)^2 \% \end{align*}\]

1.14. Rectangular tapered bar

Idea

Assignment 1

A brass plate of uniform thickness \(7\ mm\) and length \(550\ mm\), varies in width from \(100\ mm\) to \(185\ mm\) and is subjected to a tension load of \(5\ kN\). Find the elongation of the bar. \(E\) for brass \(= 82\ GPa\).

Solution:

Please attempt this assignment.

1.15. Load-extension graph

Idea

Assignment 2

During a tensile test on a specimen the following results were obtained:

Load (kN)

15

30

40

50

55

60

65

Extension (mm)

0.05

0.094

0.127

0.157

1.778

2.79

3.81

Load (kN)

70

75

80

82

80

70

Extension (mm)

5.08

7.62

12.7

16.0

19.05

22.9

Diameter of gauge length = 19 mm
Gauge length = 100mm
Diameter at fracture = 16.49 mm
Gauge length at fracture = 121 mm

Plot the complete load extension graph and the straight line portion to an enlarged scale. Hence determine:

(a) the modulus of elasticity;
(b) the percentage elongation;
(c) the percentage reduction in the area;
(d) the nominal stress at fracture;
(e) the actual stress at fracture;
(f) the tensile strength.

Solution:

Please attempt this assignment.



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Comments (5)

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Brenda

Ae we doing chapters 6, 7, 8 & 9?

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Dr. Sam

No. That's for Solid II.

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Elsie

Kindly confirm for us the formula for volumetric strain

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Brenda

Hello, are the notes here all we will have for SSM2 or is there a possibility for addition?

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Dr. Sam

Solid Mechanics II starts from chapter 6.