5 Axial Loading

Learning Objectives

Explain stress concentrations and use geometry to calculate maximum stress
Calculate axial deformation in a bar subjected to axial load
Determine deformation in a series of parallel bars
Define statically indeterminate problems and solve them using knowledge of deformation
Calculate thermal deformation and solve statically indeterminate problems involving changes in temperature

This chapter takes a closer look at the effects of axial loading. Axial loads are those applied perpendicular to the cross-section of an object (Figure 5.1). Such loads affect the object in several ways.

Horizontally oriented rectangular solid with its longer dimension labeled A to indicate the axial direction. Equal and opposite forces labeled F are applied along this axial direction: F acts leftward on the hidden left face and rightward on the visible right face, both centered and perpendicular to the faces. The square cross-sectional face on the right has side length labeled T, indicating the transverse direction. — Figure 5.1: Axial loads act perpendicular to the cross-section. They cause normal stress on the cross-section and normal strains in both the axial (A) and transverse (T) directions.

Axial loads create axial stresses, as seen in Section 2.1 and revisited briefly in Section 5.1. We previously considered the average normal stress in a cross-section, but the geometry of the cross-section can cause large localized stresses much higher than the average. These stress concentrations are discussed in Section 5.2.

This text is concerned with both stress and deformation, so Section 5.3 explores the deformation caused by axial loads. This extends to a special case of axial deformation involving parallel bars in Section 5.4, before we apply our knowledge of axial deformation to solve statically indeterminate problems in Section 5.5. These are problems where the equilibrium equations are insufficient to determine the reaction and internal forces.

Finally we’ll investigate the effects of temperature in Section 5.6. As the temperature of an object changes, the object expands or contracts. Thus deformation can occur even in the absence of any applied forces.

5.1 Axial Stress

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Axial loads create axial stresses, also known as normal stresses. As discussed in Section 2.1, the average normal stress is calculated from

\[ \boxed {\sigma=\frac{N}{A}}\text{ ,} \]

where
𝜎 = Average normal stress [Pa, psi]
N = Internal normal force [N, lb]
A = Cross-sectional area [m², in.²]

5.2 Stress Concentrations

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5.2.1 Saint-Venant’s Principle

The normal stress mentioned in the introductory section is averaged over the cross-section and assumes that stresses (and therefore strains) at the cross-section are uniform. In reality stresses and strains can vary across a cross-section, especially if the cross-section is close to an applied load, a support, or a change in geometry. At these points localized stress concentrations occur, leading to large, localized stresses and strains. The intensity of these concentrations depends on the type of loading, support, or geometry (Figure 5.2).

Two plates modeled in finite element software. Two rectangular plates are shown for comparison under uniaxial tension, both displayed using color stress contours labeled in psi. In the top image, the solid plate has no holes. The left edge is fixed, and uniform tensile forces are applied on the right edge. The stress distribution is nearly uniform across the plate, showing an average normal stress of about 333 psi, indicated mostly by green coloring on the legend scale. There are no significant stress concentrations. In the bottom image, the same plate includes a circular hole near the center. The left edge is fixed, and the right edge is again pulled in tension. The stress contours show a clear concentration around the hole, highest along the horizontal sides of the hole and lowest along the vertical sides. Far from the hole, the stresses return to about 333 psi, matching the uniform plate. Both plots include a color legend from blue (0 psi) to red (1,000 psi), showing the range of normal stress magnitudes. — Figure 5.2: Each bar has a fixed support on the left, a cross-section of 30 in.², and is subjected to a force of 10,000 lb. The top bar is uniform and experiences a uniform average normal stress of \(\sigma=\frac{10,000}{30}=333\) psi at all points, except those close to the support and load. The bottom bar experiences the same stress at most points, but significantly higher stress concentrations close to the hole.

However, these effects disappear a certain distance away from the force, support, or local geometry, so it is acceptable to assume uniform stress and strain (as we have done so far) provided that we also assume our cross-section is a sufficient distance away from these points. This is known as Saint-Venant’s principle, which can be formally stated thus: “The stresses and strains created at a point in a body by two statically equivalent loads are equivalent at points sufficiently far removed from the applied load” (Figure 5.3).

Four diagrams demonstrating static equivalence and internal force distribution in axial systems. In the top-left diagram, a fixed-end horizontal bar is subjected to three 20 kip forces acting to the right. A vertical dashed line indicates a cut at the midpoint of the bar. The corresponding bottom-left diagram shows the free-body diagram of the right segment after the cut: the three 20 kip forces continue to act to the right on the right end, and an internal force of 60 kips acts to the left on the cut face at the left end of the segment. The top-right diagram shows the same bar, but the three 20 kip forces are replaced with a single equivalent 60 kip force acting to the right. A dashed line again indicates a cut at the midpoint. The bottom-right diagram shows the free-body diagram of the right segment: a 60 kip force acts to the right on the right end, and an internal force of 60 kips acts to the left on the cut face at the left end. — Figure 5.3: Two statically equivalent loads applied to a bar. Close to the point of application there will be localized stresses and deformations, but at a cross-section sufficiently far away from the applied loads the internal effects are equivalent.

5.2.2 Stress Concentrations Around Changes in Geometry

More advanced courses use the theory of elasticity to study the areas of variable stress and strain, but for now we’ll continue to apply Saint-Venant’s principle and assume that our cross-sections are sufficiently far away from the applied loads to eliminate the need to consider these stress concentrations. Here, however, the focus is on the question of stress concentrations around changes in geometry. We’ll study two specific examples; holes and fillets. Figure 5.2 shows an example of the stress concentrations around a hole. Figure 5.4 shows an example of the stress concentrations around a fillet, which is a rounded corner used to help transition between two geometries.

Finite element model of a 3D stepped bar under uniaxial tension, shown with color stress contours labeled in psi. The bar transitions smoothly from a larger rectangular cross-section on the left to a smaller rectangular cross-section on the right, connected by a rounded fillet at the step. The left face of the larger section is fixed, and the right face of the smaller section is loaded in tension through evenly spaced arrows. A color legend from blue (0 psi) to red (1,000 psi) represents the normal stress in the X-direction. Most of the larger section appears darker green, indicating lower stress. The smaller section is shown in lighter green, representing higher stress due to its reduced cross-sectional area. The curved fillet region between the two sections displays red and blue contours, highlighting the local stress concentration caused by the change in geometry. — Figure 5.4: The fillet helps prevent a sharp corner but still causes stress concentrations. The bar has a thickness of 3 in. The height of the smaller section is 6 in. and the larger section is 10 in. The applied load is 10,000 kips, so the average normal stress in the smaller section is 556 psi and in the larger section is 333 psi. The maximum stress at the fillet, however, is 973 psi.

Fully modeling these stress concentrations is very complicated, but for our purposes it is sufficient to find only the maximum stress that occurs around these stress concentrations. These can be significantly larger than the average stress we have been calculating and can cause localized failure even if the average stress is below the yield stress for the material. The maximum stress can be found simply by multiplying the average stress by a stress concentration factor, K.

\[ \boxed{\sigma_{max}=K \sigma_{avg}}\text{ ,} \tag{5.1}\]

where
𝜎_max = Maximum normal stress [Pa, psi]
K = Stress concentration factor
𝜎_avg = Average normal stress [Pa, psi]

This stress concentration factor depends on the geometry at hand and can be found from curves in design handbooks. Figure 5.5 shows curves for holes and fillets. The average normal stress is calculated at whichever plane gives the largest value. For the case with a hole, this is the plane passing through the center of the hole. For the case with a fillet, this is a plane passing through the smaller segment of the member. See Example 5.1 and Example 5.2 to see how these curves can be used.

Two charts that compare stress concentration factors for tension members with geometric changes. On the left, a flat plate with a central circular hole is shown in the inset. The plate has a total width D, a hole of diameter d, and thickness t, with equal tensile loads P applied at both ends. The inset also includes the equations: K = sigma sub max / sigma sub avg and sigma sub average = P / [t × (D − d)]. The vertical axis shows the stress concentration factor K, and the horizontal axis shows the ratio d / D. The single curve starts near K = 3 when d / D = 0 and gradually decreases toward about K = 2 as d / D increases, showing that larger holes relative to plate width lead to lower stress concentration factors. On the right, the inset shows a stepped bar with a fillet radius r connecting a large diameter D to a smaller diameter d. The inset also includes the equations: K = sigma sub max / sigma sub average and sigma sub average = P / (t × d). The vertical axis again shows K, and the horizontal axis shows the ratio r / d. Several colored curves represent different diameter ratios (D/d = 1.10, 1.20, 1.30, 1.50, and 2.00). Each curve decreases as r / d increases, indicating that larger fillet radii reduce stress concentration. For a given r / d, higher D/d ratios correspond to higher K values, meaning that a sharper diameter change produces greater stress concentration. — Figure 5.5: Graphs showing how the stress concentration factor, K, changes according to the geometry of the object.

Example 5.1

Example 5.1

A hole is drilled through a steel plate to allow two components to be bolted together. The plate has the dimensions shown and is 15 mm thick. The plate is subjected to an axial load of P = 50 kN.

Determine the maximum stress in the plate if the hole diameter is d = 20 mm.

Horizontally oriented rectangular plate measuring 1000 mm in length and 100 mm in height is shown under axial tension, with force P applied outward at both ends. A small circular hole is located at the center of the plate, with its diameter labeled as d.

Solution

Start by calculating the average normal stress in the plate at the location of the hole. The cross-section is a rectangle with a base of 15 mm (0.015 m) and a height of 100 mm (0.1 m), with a 20 mm (0.02 m) diameter hole cut out.

Vertical cross-section of a rectangular plate measuring 100 mm in height and 15 mm in width. A shaded region in the center, bounded by horizontal dashed lines at the top and bottom, represents the vertical projection of a centered circular hole. This shaded area measures 20 mm in height and corresponds to the hole shown in the original plate from Example 5.1.

\[ \sigma_{a v g}=\frac{N}{A}=\frac{50,000{~N}}{0.015{~m} *(0.1-0.02){~m}}=41.7{~MPa} \]

Determine the ratio \(\frac{d}{D}\), where d = hole diameter and D = height of the cross-section. Use the appropriate stress concentration curve to read off the stress concentration factor K for this geometry.

Graph of stress concentration factor K versus hole ratio d/D. The x-axis (0 to 0.8) is d/D (hole diameter to height of the cross section). The y-axis (2 to 3) is K = sigma sub max / sigma sub nom. A blue curve decreases as d/D increases. A red marker at d/D ≈ 0.2 indicates K ≈ 2.5. An inset shows a rectangular plate with a central hole under tension, labeled D (height of the cross section), d (hole diameter), and t (thickness), with K = sigma sub max / sigma sub nom and sigma sub nom = P / [ t (D − d) ].

\[ \begin{aligned} &\frac{d}{D}=\frac{20{~mm}}{100{~mm}}=0.2\\ &K=2.525 \end{aligned} \]

Calculate the maximum stress.

\[ \sigma_{max }=K \sigma_{avg}=2.525*41.7{~MPa}=105{~MPa} \]

Example 5.2

Example 5.2

A 1 in. thick connecting rod in an engine assembly has the dimensions shown. The maximum allowable stress in the rod is 40 ksi.

Determine the maximum axial load that may be applied to the rod.

Two-dimensional top view of a key-shaped mechanical part. The part has a narrow arm on the left and a wider, rounded head on the right. The narrow arm contains a small circular hole with a 2-inch diameter, centered within its 4-inch width. The arm gradually widens into the head, which contains a larger central circular hole with a 6-inch diameter. The full width of the head is labeled as 8 inches. A fillet with a 1-inch radius is indicated at the transition between the narrow arm and the wider head.

Solution

The connecting rod has two holes and a fillet. We can start by determining which of these geometries experiences the largest maximum stress, considering applied load P. For each geometry, calculate the average normal stress and then use the appropriate stress concentration curve to determine the maximum normal stress.

For the smaller circle:

Graph with the y-axis labeled "Stress-concentration factor K" (ranging from 2 to 3) and the x-axis labeled "Ratio d over D" (ranging from 0 to 0.8). A red vertical line intersects the x-axis at approximately d/D = 0.5, and a red horizontal line intersects the curve at K ≈ 2.15. The blue curve shows that K decreases as the d/D ratio increases. An inset in the top corner shows a rectangular plate with a central hole under tension, labeled with D (plate height), d (hole diameter), and t (thickness). The equation K = sigma sub max / sigma sub nom is shown, with sigma sub nom = P / [ t × (D − d) ].

\[ \begin{aligned} & \sigma_{avg}=\frac{N}{A}=\frac{P}{1 *(4-2)}=\frac{P}{2}=0.5 P \\ & \frac{d}{D}=\frac{2{~in.}}{4{~in.}}=0.5 \\ & K=2.15 \\ & \sigma_{max }=K \sigma_{avg}=2.15 * 0.5 P=1.075 P \end{aligned} \]

For the larger circle:

Graph similar to Image 2, with the same axes and decreasing blue curve. A red horizontal line is drawn just above K = 2.0, at K ≈ 2.05, and a red vertical line intersects the x-axis at d/D = 0.75. The same inset diagram appears in the top-right corner, showing the rectangular plate with hole and formula as described in Image 2.

\[ \begin{aligned} & \sigma_{avg}=\frac{N}{A}=\frac{P}{1 *(8-6)}=\frac{P}{2}=0.5 P \\ & \frac{d}{D}=\frac{6{~in.}}{8{~in.}}=0.75 \\ & K=2.05 \\ & \sigma_{max}=K \sigma_{avg}=2.05 * 0.5 P=1.025 P \end{aligned} \]

For the fillet:

Graph with the y-axis labeled "Stress-concentration factor K" and the x-axis labeled "Ratio r over d," ranging from 0 to 0.30. Several curves are plotted, each representing a different D/d ratio (1.10, 1.20, 1.30, 1.50, and 2.00). Higher D/d curves are positioned higher, indicating that for a given r/d value, a higher D/d ratio results in a greater stress concentration factor K. A red vertical line intersects r/d ≈ 0.25, and a red horizontal line intersects K ≈ 1.82 for a D over d ratio equal to 2. An inset in the top-right corner shows a key-shaped part with a central notch and fillet radius r, along with labels D (cross-section height), d (hole diameter), and force arrows labeled P pointing outward on both ends. The formula K = sigma sub max / sigma sub nom is shown, with sigma sub nom = P / (t × d).

\[ \begin{aligned} & \sigma_{avg}=\frac{N}{A}=\frac{P}{1 * 4}=\frac{P}{4}=0.25 P \\ & \frac{D}{d}=\frac{8{~in.}}{4{~in.}}=2 \\ & \frac{r}{d}=\frac{1{~in.}}{4{~in.}}=0.25 \\ & K=1.82 \\ & \sigma_{max}=K \sigma_{avg}=1.82 * 0.25 P=0.455 P \end{aligned} \]

The largest stress is at the small hole, where σ_max = 1.075P.

Since the maximum allowable stress is 40 ksi, \(40{~ksi}=1.075 P \quad\rightarrow\quad P=\frac{40{~ksi}}{1.075}=37.2{~kips}\).

Apparent from these examples is that sharp corners and tight radii lead to high stress concentrations. When designing parts, it is best to use smooth, flowing transitions where possible and avoid sharp corners to minimize stress concentrations.

Step-by-Step: Stress Concentrations

Calculate the average normal stress using \(\sigma_{avg}=\frac{N}{A}\). Do this at the hole or at the thinner section of the fillet.
Use the appropriate stress concentration curve to determine the stress concentration factor, K, for the given geometry.
Calculate the maximum stress from \(\sigma_{max}=K \sigma_{avg}\).

5.3 Axial Deformation

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Consider a simple bar of uniform cross-section subjected to an axial force (Figure 5.6).

Horizontal rectangular bar under axial tension. Equal and opposite horizontal forces labeled F are applied at both ends, indicating a pulling force. Below the bar, a horizontal dimension line labeled L represents the original length of the bar. — Figure 5.6: A bar of uniform cross-section, A, and length, L, subjected to axial load, F.

Assuming elastic behavior, we have equations for stress and strain, as well as Hooke’s law, which relates the two equations.

\[ \begin{aligned} & \sigma=\frac{F}{A} \\ & \varepsilon_{long}=\frac{\Delta L}{L} \\ & \sigma=E \varepsilon_{long}\end{aligned} \]

Replace the stress and strain terms in Hooke’s law.

\[ \frac{F}{A}=E \frac{\Delta L}{L} \]

Rearrange the equation.

\[ \Delta L=\frac{F L}{A E} \]

This last equation can be used to directly find the change in length of an object subjected to an axial load.

In practice, multiple axial loads may be applied to the bar. The cross-sectional area may change at different points along the bar, or perhaps different materials with different elastic moduli are connected in series to form the bar. In any of these cases we can split the bar into sections where each section has a constant F, A, and E. We can calculate the change in length of each section separately and sum them to find the total. See Example 5.3 for deformation of a simple bar subjected to a single load and Example 5.4 for a problem involving multiple segments of a bar.

\[ \boxed{\Delta L=\sum \frac{F L}{A E}}\text{ ,} \tag{5.2}\]

where
ΔL = Change in length [m, in.]
F = Internal axial load [N, lb]
L = Original length [m, in.]
A = Cross-sectional area [m², in.²]
E = Elastic modulus [Pa, psi]

Example 5.3

Example 5.3

A wooden beam (E = 12 GPa) of length L = 6 m has a rectangular cross-section of base b = 50 mm and height h = 80 mm. The beam is subjected to a tensile load P = 30 kN as shown.

Determine the change in length of the beam.

Horizontal rectangular bar is fixed to a wall on the left side and extends 6 meters to the right. A red arrow labeled 30 kN points rightward at the free end, indicating an external axial force.

Solution

We can find the change in length of the beam using \(\Delta L = \Sigma\frac{FL}{AE}\).

Here only one load is applied and the beam is made of a single material with a uniform cross-section, so F, A, and E are constant.

We next find the internal load in the beam by cutting a cross-section and drawing a free body diagram (FBD).

The same bar is shown with a vertical dashed line on the left side, representing a cut made anywhere before the 30 kN force. This is the free-body diagram of the right segment. A 30 kN force points rightward at the right end, and a red arrow labeled F points leftward at the cut face on the left, representing the internal axial force.

\[ \begin{aligned} \Sigma F_x &= 30 - F = 0\\ F &=30~kN \end{aligned} \]

The beam has a rectangular cross-section so the area is given by

\[ \begin{aligned} A&=bh\\ &=0.05~m*0.08~m\\ &=0.004~m^2 \end{aligned} \]

We are given L = 6 m and E = 12 GPa, so the change in length of the beam is given by

\[ \begin{aligned} \Delta L &= \frac{FL}{AE}\\ &=\frac{30,000~N*6~m}{0.004~m^2*12*10^9~\frac{N}{m^2}}\\ &=0.00375~m\\ &=3.75~mm \end{aligned} \]

The beam lengthens by 3.75 mm.

Example 5.4

Example 5.4

A component is made by welding together two circular rods. Rod 1 is made of steel (E = 29 x 10⁶ psi) and is hollow, with an outer diameter of 4 in. and an inner diameter of 2 in. Rod 2 is made of copper (E = 17 x 10⁶ psi) and is solid, with a diameter of 5 in.

If the component is subjected to the axial loads shown, determine the total deformation of the component.

A stepped axial bar is shown horizontally, with a narrow segment on the left and a thicker segment on the right. The narrow segment, labeled segment (1), is 9 inches long. The thicker segment, labeled segment (2), is 14 inches long and ends in a fixed support on the far right. A leftward force of 60 kips is applied at the far left end of segment (1). A rightward force of 80 kips is applied on segment (2), located 4 inches to the right of the junction with segment (1) and 10 inches to the left of the fixed end.

Solution

We can find the change in length of the component using \(\Delta L=\sum \frac{F L}{A E}\).

First, break the component into three segments and determine the internal load in each segment. The first segment covers the 9 in. steel rod. After the first 9 in. of the component, both the cross-sectional area and elastic modulus change, so a second segment is needed. This segment covers the next 4 in. At this point the loading changes, so a third segment is required to cover the final 10 in. of the component.

Three free-body diagrams of bar segments are shown stacked vertically. The top diagram shows a cut from segment (1), representing the left portion of the bar. It includes a leftward external force of 60 kips on the far left end and an internal rightward force labeled N sub 1 at the cut face, marked by a vertical dashed line. The middle diagram shows a cut from segment (2), just before the applied 80 kip force, representing the left portion of segment (2) (i.e., after the junction). It includes a leftward force of 60 kips on the left end and an internal rightward force labeled N sub 2 at the dashed cut face. The bottom diagram shows the full bar without the fixed support, representing the left segment of the complete free-body diagram (cut just before the fixed end). It includes a leftward force of 60 kips at the far left end, a rightward force of 80 kips applied 4 inches to the right of the junction, and an internal rightward force labeled N sub 3 at the far right cut face, marked by a vertical dashed line.

\[ \begin{gathered} N_1=N_2=60{~kips} \\ N_3=-20{~kips} \end{gathered} \]

Now calculate the deformation of each segment. Be sure to use the appropriate dimensions and material properties for each segment.

Cross-sectional areas:

\[ \begin{gathered} A_1=\pi*(2^2-1^2){~in.}^2=9.42{~in.}^2 \\ A_2=\pi*(2.5^2){~in.}^2=19.6{~in.}^2 \end{gathered} \]

Deformation of segment 1:

\[ \Delta L_1=\frac{60,000{~lb} * 9{~in.}}{9.42{~in.}^2 * 29 \times 10^6\frac{lb}{in^2}}=0.00198{~in.} \]

Deformation of segment 2:

\[ \Delta L_2=\frac{60,000{~lb} * 4{~in.}}{19.6{~in.}^2 * 17 \times 10^6\frac{lb}{in.^2}}=0.000719{~in} . \]

Deformation of segment 3:

\[ \Delta L_3=\frac{-20,000{~lb} * 10{~in.}}{19.6{~in.}^2 * 17 \times 10^6\frac{lb}{in.^2}}=-0.000599{~in.} \]

Note that segment 3 is in compression and therefore becomes shorter, while segments 1 and 2 are in tension and therefore become longer. The total deformation in the component is simply the sum of these three deformations.

\[ \Delta L=0.00198{~in.}+0.000719{~in.}-0.000599{~in.}=0.002{~in.} \]

Step-by-Step: Axial Deformation

Split the bar into sections of constant F, A, and E. Any time any term changes, begin a new section.
Determine the length of each section.
Calculate the change in length of each section using \(\Delta L=\frac{F L}{A E}\).
Determine the total change in length by summing the change in length of each section. Remember that some sections may elongate while others contract.

5.4 Deformation in Series of Bars

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Sometimes a structure has more than one axially loaded member. In such cases multiple bars will experience a change in length and the amount of deformation won’t necessarily be the same in each member. The deformable members are generally connected by a rigid (nondeformable) beam. These problems generally involve a little geometry alongside the deformation equation (Figure 5.7).

Two diagrams comparing the original configuration of a structural frame with its potential deformed shape. The top diagram shows a horizontal beam extending from point A to point B, with point C located between them, closer to point A. The beam is supported from above by two vertical bars labeled (1) and (2), which are fixed at the top and extend downward to the beam. The original lengths of the bars are labeled L sub 1 and L sub 2. A uniformly distributed load W is applied downward from point C to point B. The bottom diagram shows the same frame with possible deformation under the load. The beam is shown detached from supports at points A and B to highlight the displacements of the vertical bars. The beam tilts slightly downward to the southeast, with displacements labeled ΔL sub 1 for bar (1) and ΔL sub 2 for bar (2), marked by dashed vertical lines extending downward from the original bar lengths. The distributed load W remains applied from point C to point B, as in the top diagram. — Figure 5.7: Rigid beam AB is attached to deformable poles 1 and 2. The poles may deform different amounts, so point A is displaced by amount ΔL₁, point B by amount ΔL₂, and point C by an amount in between these two values.

By identifying the force in each member through equilibrium, we may calculate the deformation of each member separately by using

\[ \Delta L=\frac{F L}{A E}\text{.} \]

Once each member’s change in length is known, use simple geometry to find the displacement at different points on the rigid beam (Figure 5.8). This process is demonstrated in Example 5.5.

The top diagram shows a horizontal beam supported by two vertical bars labeled (1) and (2), both originally the same length. Under loading, bar (2) undergoes vertical displacement, illustrated by a dashed vertical line extending downward from its original position. This causes the beam to tilt diagonally in the southeast direction, shown with a dashed outline of the displaced beam. The vertical displacement of bar (2) is labeled ΔL sub 2, indicating the downward movement of point B. The bottom diagram represents the same deformation using geometry and similar triangles. Points A, C, and B are connected by a diagonal dashed line sloping downward to the southeast, representing the tilted beam. A horizontal line is drawn beneath it to represent the original, undeformed beam length, labeled L, with point A at the left end. The distance from point A to point C is labeled L sub AC. The vertical displacement between the original and deformed positions of point C is labeled ΔL sub C, and the vertical displacement at point B is labeled ΔL sub 2. — Figure 5.8: Beam deflection in the case where only bar 2 deforms. Point B will deflect downward by amount ΔL₂. The deflection at point C can be determined through geometry.

The deflection of point C will be somewhere between the deflection at A and the deflection at B. Assuming that there is no deflection at A and that the deflections (and thus the angle at A) are small, use similar triangles to find

\[ \frac{\Delta L_C}{L_{A C}}=\frac{\Delta L_2}{L} \rightarrow \Delta L_C=\Delta L_2 \frac{L_{A C}}{L}\text{.}\]

If a deflection exists at point A (Figure 5.9), this simply becomes

\[ \Delta L_C=\Delta L_1+\left(\Delta L_2-\Delta L_1\right) \frac{L_{A C}}{L}\text{.}\]

Horizontal line segment labeled A–C–B representing the original position of a structural element. Below it, the same A–C–B line is shown rotated clockwise (south east direction) in black dashed lines, representing its displaced configuration. A green dashed horizontal line extends from point A to the original position of point B align vertically with its displaced position. The vertical distance between the original and displaced position of point A is labeled ΔL sub 1. The vertical distance from the original position of point B to its displaced position is divided into ΔL sub 1 and ΔL sub 2 - ΔL sub 1, indicating additional downward displacement. — Figure 5.9: Similar triangles for calculating deflection at a point when both bars experience a change in length.

Example 5.5

Example 5.5

A rigid beam is supported by two nonrigid poles and subjected to a distributed load w = 75 kN/m. Pole 1 is made of steel (E = 200 GPa) and has a diameter of 30 mm. Pole 2 is made of cast iron (E = 70 GPa) and has a diameter of 50 mm.

Determine the deflection at point C of the rigid beam.

Horizontal beam AB is supported by two vertical columns. Column (1), on the left, is 5 m tall; Column (2), on the right, is 3 m tall. The total beam length is 12 m, divided into AC (4 m) and CB (8 m), with point C between A and B. A uniformly distributed vertical load W is applied downward along CB.

Solution

Although the beam is rigid, poles 1 and 2 will both elongate. We can find the force in each pole by drawing an FBD and applying equilibrium equations.

The beam AB is shown isolated in a horizontal position. Upward support reactions are labeled F sub 1 at point A and F sub 2 at point B. A concentrated downward force of 600 kilonewtons is applied at a point 8 meters to the right of A (or 4 meters to the left of B).

\[ \begin{gathered} \sum M_A=-(600{~kN}* 8{~m})+(F_2 * 12{~m})=0 \quad\rightarrow\quad F_2=400{~kN} \\ \sum F_y=F_1-600{~kN}+400{~kN}=0 \quad\rightarrow\quad F_1=200{~kN} \end{gathered} \]

We can then calculate each pole’s change in length.

\[ \begin{aligned} \Delta L_1 & =\frac{F_1 L_1}{A_1 E_1}=\frac{200,000{~N} * 5{~m}}{\pi * (0.015{~in.})^2 * 200 * 10^9\frac{N}{m^2}}=0.00707{~m}=7.07{~mm} \\ \Delta L_2 & =\frac{F_2 L_2}{A_2 E_2}=\frac{400,000{~N} * 3{~m}}{\pi * (0.025{~in.})^2 * 70 * 10^9\frac{N}{m^2}}=0.00873{~m}=8.73{~mm} \end{aligned} \]

To determine the deflection at point C, first determine that point B has deflected: (8.73 mm – 7.07 mm) = 1.66 mm more than point A.

Two lines are shown: a dashed horizontal line for the original, undeformed beam AB (A-C-B) and a solid black line for the deformed beam A′-C′-B′ sloping downward from left to right (southeast direction). Point A has moved downward 7.07 mm; point B, 8.73 mm. The vertical distance between C and C′ is labeled ΔL sub C, indicating the relative vertical displacement due to the difference in movement between A and B.

We can find how much more point C has deflected than point A by using similar triangles.

The deformed beam A′B′ is shown as a solid black line connecting points A′, C′, and B′. A horizontal dashed line extends from point A′ to the original (undeformed) position of point B, highlighting the beam’s rotation. The beam has rotated downward in a clockwise direction, as shown by the slanted orientation of line A′B′. The vertical distance from C′ to the dashed reference line is labeled ΔL sub C slash A, representing the displacement of point C relative to the straight line connecting A′ to B′. The horizontal spans are: A′ to undeformed C is 4 meters, C to B is 8 meters, and the total vertical drop from A′ to B′ is 1.66 millimeters.

\[ \frac{1.66{~mm}}{12{~m}}=\frac{\Delta L_{C / A}}{4{~m}} \quad\rightarrow\quad \Delta L_{C / A}=0.553{~mm} \]

So point C deflects 0.553 mm more than point A, which is a total deflection at point C of

\[ \Delta L_C=7.07{~mm}+0.553{~mm}=7.62{~mm}\text{.} \]

Step-by-Step: Deformation in Series of Bars

Use equilibrium to determine the internal force in each bar.
Calculate the change in length of each bar using \(\Delta L=\frac{F L}{A E}\).
Use similar triangles to find the deflection at any point between the parallel bars.

5.5 Statically Indeterminate Problems

Click to expand

A statically indeterminate problem is one that has more unknowns than we have equilibrium equations to solve for them. This issue prevents us from finding the internal loads, and so we cannot calculate stress or deformation.

There are two types of statically indeterminate problems. The first type has additional supports beyond those needed to maintain equilibrium. These are known as redundant supports and are quite common in practice (Figure 5.10).

A long cable-stayed bridge spans across a wide valley with multiple tall piers supporting the deck. The bridge features vertical pylons from which multiple diagonal cables fan out symmetrically to support the roadway. — Figure 5.10: An example of redundancy on a suspension bridge. There are multiple supports and multiple cables at each support, such that if one cable fails the entire structure does not collapse.

In such problems, determining all the reaction forces using equilibrium alone isn’t possible. However, we can use our knowledge of deformation to help. If a member is held between two supports, then its total deformation must be zero. Since the change in length depends on the internal force in the member, this constraint introduces an additional equation to use alongside the equilibrium equations. Two approaches are possible here.

Approach 1: To determine the reaction force at the redundant support, begin by removing the redundant support from the problem and determining the deformation that would occur without the support. Then replace the reaction force at the support, which will cause the member to deform in the opposite direction (Figure 5.11).

Four subfigures labeled (a) through (d) show a horizontal bar composed of a larger left segment and a smaller right segment. In subfigure (a) the bar is fixed between two rigid supports at both ends. In (b) the same bar is shown as a free body diagram: a rightward force F acts at the interface between segments, with leftward reaction forces F sub A at the left end and F sub B at the right end. In subfigure (c) the right support is removed, leaving the smaller segment’s end free so the bar can elongate under F. In (d) The right support is replaced with an equivalent leftward force F sub B, causing the bar to contract. — Figure 5.11: (a) A bar held between two rigid supports. (b) A free body diagram of the beam reveals that there are two unknowns but only one equilibrium equation. (c) Removing one of the supports allows the bar to elongate. (d) Replacing the support force causes the bar to contract.

The sum of these two deformations must equal the member’s actual deformation. If the member is held between two rigid supports, the total deformation will be zero. If there is a small gap or the supports allow a certain amount of movement, the total deformation will be equal to the size of this gap. Example 5.6 shows this process applied to a bar made of two materials.

Example 5.6

Example 5.6

A 15 ft tall concrete (E = 4,000 ksi) column has a square cross-section of 6 in. x 6 in. A 10-foot-tall-copper (E = 17,000 ksi) cylinder with an outer diameter of 4 in. and inner diameter of 3 in. is attached to the top of the concrete. The structure is fixed between two supports. A force of 25 kips is applied as shown.

Determine the average normal stress in each material.

A vertically oriented stepped column with two segments of different widths. The upper segment is 10 feet tall and 4 inches wide, and the lower segment is 15 feet tall and 6 inches wide. The column is fixed at both the top and bottom ends. A downward vertical force of 25 kips is applied at the junction between the two segments.

Solution

To determine the internal force in each material, first find the reaction forces at the supports. Usually we draw an FBD and apply equilibrium equations to find these forces. However, this won’t work here because the problem is statically indeterminate.

A free-body diagram of the same column with both supports replaced by reaction forces. An upward force labeled F sub 1 is shown at the top of the upper segment, and an upward force labeled F sub 2 is shown at the base of the lower segment. The 25-kip downward force is applied at the same junction.

\[ \sum F_y=F_1+F_2-25{~kips}=0 \quad\rightarrow\quad F_1+F_2=25{~kips} \]

To solve this statically indeterminate problem, remove one of the supports and allow the structure to deform. Either support may be removed.

Remove the top support. In this scenario there is no load in the copper cylinder and there is a compressive load of 25 kips in the concrete column.

The column fixed only at the bottom, with the top support removed. The 25-kip downward force is applied at the junction, allowing the structure to compress.

The total deformation of the structure in this scenario is

\[ \Delta L=\sum \frac{F L}{A E}=0-\frac{(25{~kips}) *(15 * 12){~in.}}{(6 * 6){~in.}^2 *(4,000)\frac{kips}{in.^2}}=-0.03125{~in}\text{.} \]

Then replace force F₁ and calculate the deformation caused by this force. Because force is applied at the top of the structure, both the concrete and the copper elongate.

The bottom of the column remains fixed, and the top support is replaced by an upward force labeled F sub 1 applied at the top of the 10-foot upper segment. This force causes both segments to elongate under tension.

\[ \Delta L=\sum \frac{F L}{A E}\\ \Delta L=\frac{F_1 *(10 * 12){~in.}}{\pi(2^2-1.5^2){~in.}^2 *(17,000)\frac{kips}{in.^2}}+\frac{\left(F_1\right) *(15 * 12){~in.}}{(6 * 6){~in.}^2 *(4,000)\frac{kips}{in.^2}} \\ \Delta L=0.001284F_1+0.00125 F_1 \\ \Delta L=0.002534 F_1 \]

Since the structure is fixed at both ends, the actual total deformation must be zero.

\[ -0.03125{~in.}+0.002534 F_1=0 \quad\rightarrow\quad F_1=12.3{~kips} \]

Returning to the equilibrium equation, also find force F₂.

\[ F_1+F_2=25{~kips} \\ 12.3{~kips}+F_2=25{~kips} \\ F_2=12.7{~kips} \]

The internal force in the copper cylinder is 12.3 kips, and the internal force in the concrete column is 12.7 kips. Finally, calculate the stress in each material. Note from your original FBD that the copper cylinder is in tension and the concrete column is in compression. Introduce a negative sign here to indicate compressive stress in the concrete.

\[ \begin{aligned} & \sigma_{copper}=\frac{N_1}{A_1}=\frac{12.3{~kips}}{\pi(2^2-1.5^2){~in.}^2}=2.24 {~ksi} \\ & \sigma_{concrete}=-\frac{N_2}{A_2}=-\frac{12.7{~kips}}{(6 * 6){~in.}^2}=-0.353 {~ksi} \end{aligned} \]

Approach 2: An alternative approach is to start by noting that the total deformation of the bar must be zero. In the case shown in Figure 5.12, the deformation of segment 1 plus the deformation of segment 2 must add up to zero.

The image contains two subfigures labeled (a) and (b), both illustrating the second approach where the total deformation of the bar must be zero. In diagram (a), a two-segment axial bar is shown fixed at both ends. A rightward force labeled F is applied at the interface between the two segments. In diagram (b), the same two-part bar is shown isolated (with supports removed). The same rightward force F is applied at the junction between the segments. Additionally, a leftward force labeled F sub A is applied at the left end, representing the reaction at the left support, and a rightward force labeled F sub B is applied at the right end, representing the reaction at the right support. — Figure 5.12: (a) A bar held between two rigid supports. (b) A free body diagram of the bar.

The internal load in segment 1 is F_A and the internal load in segment 2 is F_B, so there are currently two unknowns. We may use an equilibrium equation to solve for these two unknowns simultaneously.

\[ \begin{aligned} & \frac{F_A L_1}{A_1 E_1}+\frac{F_B L_2}{A_2 E_2}=0 \\ \\ & F-F_A-F_B=0\end{aligned} \]

Example 5.7 solves Example 5.6 using this method instead.

Example 5.7

Example 5.7

A 15 ft tall concrete (E = 4,000 ksi) column has a square cross-section of 6 in. x 6 in. A 10 ft tall copper (E = 17,000 ksi) cylinder with an outer diameter of 4 in. and inner diameter of 3 in. is attached to the top of the concrete. The structure is fixed between two supports. A force of 25 kips is applied as shown.

Determine the average normal stress in each material.

Solution

As before, begin with an equilibrium equation that relates the support loads to the applied load. In this equation we use the convention that forces pointing upward are positive and those pointing downward are negative.

\[ \begin{aligned} & \sum F_y=F_1+F_2-25{~kips}=0 \\ & F_1+F_2=25{~kips} \end{aligned} \]

This bar consists of two segments: the copper cylinder (segment 1) and the concrete column (segment 2). Since the bar is held between two rigid supports, the total deformation of these two segments must sum to zero. Note that in this diagram segment 1 is in tension and segment 2 is in compression. Remember to be consistent with the sign convention that tension is positive and compression is negative.

\[ \frac{F_1 L_1}{A_1 E_1}-\frac{F_2 L_2}{A_2 E_2}=0 \]

Here F₁ and F₂ are the internal forces in segments 1 and 2 of the bar. These will be the same as the reaction loads at the supports that we are trying to solve for.

\[ \frac{F_1 *(10 * 12){~in.}}{\pi(2^2-1.5^2){~in.}^2 *(17,000)\frac{kips}{in.^2}}-\frac{F_2 *(15 * 12){~in.}}{(6 * 6){~in.}^2 *(4,000)\frac{kips}{in.^2}}=0 \]

Rearrange the equilibrium equation and substitute into the deformation equation.

\[ \begin{gathered} F_1=25{~kips}-F_2 \\ \frac{(25-F_2){~kips} *(10 * 12){~in.}}{\pi(2^2-1.5^2){~in.}^2 *(17,000)\frac{kips}{in.^2}}-\frac{F_2 *(15 * 12){in.}}{(6 * 6){in.}^2 *(4,000)\frac{kips}{in.^2}}=0 \end{gathered} \]

Simplify and solve for force F₂.

\[ \begin{gathered} 0.0321-0.001284 F_2-0.00125 F_2=0 \\ 0.0321=0.002534 F_2 \\ F_2=12.7 {~kips} \end{gathered} \]

Then use the equilibrium equation again to find F₁.

\[ F_1=25{~kips}-12.7{~kips}=12.3{~kips} \]

These are the same reactions found in Example 5.6 when we solved this problem using the other approach. From here find the stress in each material as before, noting again from the diagram that segment 2 is in compression.

The second type of indeterminate problem involves two materials bonded together in parallel. In these problems it is possible to find the reactions at the supports, but not possible to find the internal force in each material using only equilibrium (Figure 5.13).

The top diagram shows a rectangular block divided horizontally into three equal sections by two dashed blue lines. The top layer is labeled (1), and the middle (thicker shaded) section is labeled (2). The top and bottom layers are equal in size. Equal horizontal forces labeled F are applied at both ends of the block, indicating axial loading. The bottom diagram shows the block cut vertically in half, with the left segment isolated in a free-body diagram. A leftward force F is applied on the left face, while two rightward forces labeled F sub 1 and F sub 2 are applied separately along the same horizontal line, representing internal force distribution across the layers. — Figure 5.13: When two materials in parallel are subjected to force F, the force is split between the two materials (F₁ and F₂). By equilibrium F₁ + F₂ = F, but this is not sufficient to find the force in each material.

We have one equilibrium equation but two unknown internal forces. However, since the materials are bonded together, they must deform by the same amount. By setting the deformation the same for each material, we can define a second equation that involves the two internal forces and, combined with the equilibrium equation, now solve for both internal forces. See Example 5.8 for a demonstration.

Example 5.8

Example 5.8

A 20 ft tall concrete (E = 4,000 ksi) column has a square cross-section 6 in. on each side. It is reinforced by six pieces of steel (E = 30,000 ksi) rebar that extend through the column’s length. Each has a diameter of 0.5 in. The column is subjected to a compressive load of 70 kips.

Determine the stress in each material.

A 3D rectangular concrete column is shown with a height of 20 feet and a square base of 6 inches by 6 inches. The top face of the column displays six small black circles, labeled as 6 x 0.5-inch diameter, representing reinforcing steel rods embedded within the concrete.

Solution

The 70 kips force will be split between the two materials. We need to find the force in each material before we can calculate the stress. Start by cutting a cross-section through the column, drawing an FBD, and writing an equilibrium equation.

A square cross section of the column is shown with a downward vertical force of 70 kips applied at the top center. Two upward vertical forces, labeled F sub C and F sub S, act from below the column, representing the resisting forces from concrete and steel respectively.

\[ \sum F_y= F_C+F_S=70{~kips} \]

This problem is statically indeterminate, since we have two unknowns and only one equilibrium equation. However, because the two materials are bonded together, they must both deform the same amount.

\[ \Delta L_C=\Delta L_S \]

Since we know \(\Delta L=\frac{F L}{A E}\),

\[ \frac{F_C L_C}{A_C E_C}=\frac{F_S L_S}{A_S E_S}\text{.} \]

The length of both materials is 20 ft, so this term will cancel.

Although the steel comprises six individual pieces, it is fine to determine the total area of the steel and the total force in the steel. The total area of the steel is \(A_S=6 * \pi * (0.25{~in.})^2=1.178{~in.}^2\).

For the area of the concrete, calculate the area of the square and then remove the area of the six rebar rods.

\[ A_C=(6 * 6){~in.}^2-1.178{~in.^2}=34.82{~in.}^2 \]

Substitute these into the deformation equation and rearrange.

\[ \begin{gathered} \frac{F_C}{34.82{~in.}^2 * 4,000{~ksi}}=\frac{F_S}{1.178{~in.}^2 * 30,000{~ksi}} \\ F_C=F_S\left[\frac{34.82{~in.}^2 * 4,000{~ksi}}{1.178{~in.}^2 * 30,000{~ksi}}\right] \\ F_C=3.941 F_S \end{gathered} \]

Substitute this into the equilibrium equation.

\[ \begin{aligned} &F_C+F_S=70{~kips} \\ &3.941 F_S+F_S=70{~kips} \\ &4.941 F_S=70{~kips} \\ &F_S=14.2{~kips} \end{aligned} \]

Then since \(F_C=3.941 F_S \quad\rightarrow\quad F_C=3.941 * 14.2{~kips}=55.8{~kips}\).

Now that we know the force in each material, calculate the stress in each material.

\[ \begin{aligned} & \sigma_S=\frac{14.2{~kips}}{1.178{~in.}^2}=12.0{~ksi} \\ & \sigma_C=\frac{55.8{~kips}}{34.82{~in.}^2}=1.60{~ksi} \end{aligned} \]

Step-by-Step: Statically Indeterminate Axial Problems

Problems with redundant supports

Approach 1:

Write out equilibrium equations. There will be too many unknowns, so put these to one side for now.
Remove one of the supports and calculate the amount of deformation that would occur if the support were not there, using \(\Delta L_1=\sum \frac{F L}{A E}\).
Replace the force from the removed support and determine the deformation, ΔL₂, caused by this force in terms of the force itself.
Set ΔL₁ + ΔL₂ equal to the total allowed deformation for the bar and use this to solve for the unknown force at the redundant support.
Now that one force is known, use the equilibrium equations to determine the other unknown forces.

Approach 2:

Draw an FBD of the entire structure and write out the relevant equilibrium equations. For axial loads this will be just one sum of force equation, either horizontal or vertical depending on the applied loading.
Write an equation in which the total deformation in each segment of the bar must sum to the total deformation of the bar (zero if the bar is held between two rigid supports). A new segment must be made any time the internal load, cross-sectional area, or material changes.
This deformation equation involves the same two unknowns as the equilibrium equation. Solve these equations simultaneously to find the reaction loads at the supports.

Problems with two materials in parallel

Cut a cross-section through the member and set up an equilibrium equation for the internal forces where the internal forces will sum to equal the external applied force.
Set the deformation of the two materials equal.
Solve the deformation equation and equilibrium equation simultaneously to determine the internal force in each material.

5.6 Thermal Deformation and Thermal Stresses

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So far we’ve studied the effects of axial forces on an object and how they create stresses and deformations. Temperature changes will also cause an object to deform. As seen in Section 4.6, the strain due to temperature can be predicted by

\[ \varepsilon_T=\alpha \Delta T\text{.} \]

As before, strain is dimensionless. The coefficient of thermal expansion is a material constant that can be looked up in handbooks or in Appendix C. Note that for a given change in temperature, the thermal strain will be the same in the axial and transverse directions.

Since strain is also defined as \(\varepsilon=\frac{\Delta L}{L}\), we can also predict the deformation resulting from a change in temperature.

\[ \varepsilon_T=\alpha \Delta T=\frac{\Delta L}{L} \quad\rightarrow\quad \Delta L=\alpha \Delta T L \]

That the object is free to expand or contract doesn’t cause any issues and is easy to account for. Many real applications include a small gap to allow for changes in length due to temperature changes (Figure 5.14). It is even possible that an object is subjected to both a physical force and a temperature change and that the total change in length is simply the sum of these effects.

\[ \boxed{\Delta L=\Delta L_F+\Delta L_T=\frac{F L}{A E}+\alpha \Delta T L}\text{ ,} \tag{5.3}\]

where
∆L = Change in length [m, in.]
∆L_F = Change in length due to applied load [m, in.]
∆L_T = Change in length due to temperature change [m, in.]
F = Internal force [N, lb]
L = Original length [m, in.]
A = Original cross-sectional area [m², in.²]
E = Elastic modulus [Pa, psi]
𝛼 = Coefficient of thermal expansion \(\left[\frac{1}{^\circ C}, \frac{1}{^\circ F}\right]\)
∆T = Change in temperature \([^\circ C, ^\circ F]\)

A close-up view of an expansion joint on a concrete bridge deck. The image shows a series of interlocking metal plates with evenly spaced gaps, embedded in the surface of the road to accommodate thermal expansion and contraction. — Figure 5.14: Expansion joint on a bridge, allowing for thermal expansion and contraction of the bridge as temperature changes through the year.

However if we do not design a gap, or if the gap isn’t large enough, then the object is not free to expand or contract. As the member pushes or pulls on its supports, a physical force is created. This in turn creates a stress in the object, and these stresses can be very large. Such problems are statically indeterminate because the force the support applies on the member is unknown and cannot be found using only equilibrium.

Solving these problems is very similar to solving the first type of statically indeterminate problem. First, remove a support and determine the amount of deformation that would occur as a result of the change in temperature if the object were free to deform. Then replace the force from the removed support, which will cause the object to deform in the other direction. The sum of these two deformations will equal the total deformation of the member as before. Example 5.9 works through a statically indeterminate thermal expansion problem.

Example 5.9

Example 5.9 Steel (E = 200 GPa, α = 11.7 x 10^-6 /°C) train rails are laid end-to-end. Each rail is 20 m long. A small section of this track is shown below. The rails are laid in winter when the temperature is 0 °C. In summer, the maximum temperature is 40 °C.

Determine the compressive stress in the rail as a result of the temperature change if

a. In a particular section of track, the rails are laid with no gap between them.

b. In another section, the rails have a 5 mm gap between them.

Three railway track segments, each 20 meters long, are shown laid horizontally on a brown platform labeled "ground." The segments are separated by two small labeled gaps, indicating space left to accommodate thermal expansion.

Solution

a. First determine the deformation of each rail that would occur if it were free to expand.

\[ \Delta L_T=\alpha \Delta T L=11.7 \times 10^{-6}/^\circ{C} * 40^\circ{C} * 20{~m}=0.00936{~m} \]

Since each rail contacts the rail next to it as it tries to expand, each rail will experience a compressive force. This force causes compression in the rail.

\[ \Delta L=-\frac{F L}{A E} \]

The actual deformation of each rail must be zero, given that the rails are in contact end-to-end. Thus these deformations must sum to zero.

\[ 0.00936{~m}-\frac{F * 20{~m}}{A * 200 * 10^9{~Pa}}=0 \]

Although we don’t know the cross-sectional area of the rail, we can replace \(\sigma=\frac{F}{A}\).

\[ \begin{gathered} 0.00936{~m}-\sigma * \frac{20{~m}}{200 * 10^9{~Pa}}=0 \\ \sigma=\frac{0.00936{~m} * 200 * 10^9{~Pa}}{20{~m}}=93.6{~MPa} \end{gathered} \]

b. In this case again begin by determining the deformation of each rail that would occur if it were free to expand.

\[ \Delta L_T=\alpha \Delta T L=11.7 \times 10^{-6}/^\circ{C} * 40^\circ{C} * 20{~m}=0.00936{~m} \]

Each rail would expand 9.36 mm, but the gap between rails is only 5 mm, so the rails close this gap and make contact, which means there will again be some compressive stress. Set up the deformation equation as in part (a), but this time the deformation is 5 mm (0.005 m) instead of zero.

\[ 0.00936{~m}-\frac{F * 20{~m}}{A * 200 * 10^9{~Pa}}=0.005{~m} \]

Replace \(\sigma=\frac{F}{A}\) as before and determine the stress.

\[ \begin{gathered} 0.00936{~m}-\sigma * \frac{20{~m}}{200 * 10^9{~Pa}}=0.005{~m} \\ 0.00936{~m}-0.005{~m}=\sigma * \frac{20{~m}}{200 * 10^9{~Pa}} \\ \sigma=\frac{0.00436{~m} * 200 * 10^9{~Pa}}{20{~m}}=43.6{~MPa} \end{gathered} \]

We can see that installing the rails with a small gap has significantly reduced the thermal stress. Accurately predicting the amount of deformation and leaving a suitably sized gap would reduce the stress even further.

Step-by-Step: Thermal Deformation

Remove one of the supports and calculate the total deformation that would occur if the support were not there, using \(\Delta L_T=\alpha \Delta T L\) for thermal deformation and \(\Delta L=\frac{F L}{A E}\) for any applied axial loads.
If the total deformation causes the member to contact a support that prevents some of the deformation, replace the force from the removed support and determine the deformation caused by this force in terms of the force itself, using \(\Delta L=\frac{F L}{A E}\).
Sum these deformations and set them equal to the total allowed deformation. Use this equation to solve for the force at the redundant support.
Use equilibrium to determine the force at any other supports.

Summary

Click to expand

Key Takeaways

Axial loads cause normal stresses. While we generally calculate the average normal stress, stress concentrations occur close to applied loads, supports, and changes in geometry.

These stress concentrations can be large and cause localized failure even if the average stress is within acceptable limits. Modeling stress concentration is beyond the scope of this course, but we can determine maximum stresses for different geometries using stress concentration curves.

It is often desirable to limit the total change in length of a bar subjected to axial loads. Change in length can be predicted directly.

Leaving a gap to allow for change in length is useful in practice. If a change in length is prevented from happening, this creates a stress in the bar.

Changes in temperature also cause axial deformation of a bar, known as thermal deformation. Thermal deformation can lead to a physical stress in the bar if a support prevents the deformation from occurring.

Key Equations

Stress concentrations:

\[ \sigma_{avg}=\frac{N}{A} \]

\[ \sigma_{max}=K \sigma_{avg} \]

Axial deformation:

\[ \Delta L=\frac{F L}{A E} \]

Thermal strain:

\[ \varepsilon_T=\alpha \Delta T \]

Thermal deformation:

\[ \Delta L=\alpha \Delta T L \]

References

Click to expand

Figures

All figures in this chapter were created by Kindred Grey in 2025 and released under a CC BY license, except for

Figure 5.2: Each bar has a fixed support on the left, a cross-section of 30 in.², and is subjected to a force of 10,000 lb. James Lord. 2024. CC BY-NC-SA.
Figure 5.4: The fillet helps prevent a sharp corner but still causes stress concentrations. James Lord. 2024. CC BY-NC-SA.
Figure 5.5: Graphs showing how the stress concentration factor, K, changes according to the geometry of the object. Adapted under fair use by Kindred Grey from Figures 5.14 and 5.15 in Philpot, T.A. (2020) Mechanics of Materials, 4th Edition, Wiley.
Example 5.1: First and second image: Kindred Grey. 2024. CC BY. Third image: Adapted under fair use by Kindred Grey from Figures 5.14 and 5.15 in Philpot, T.A. (2020) Mechanics of Materials, 4th Edition, Wiley.
Example 5.2: First image: Kindred Grey. 2024. CC BY. Second, third, and fourth image: Adapted under fair use by Kindred Grey from Figures 5.14 and 5.15 in Philpot, T.A. (2020) Mechanics of Materials, 4th Edition, Wiley.
Figure 5.10: An example of redundancy on a suspension bridge. Unknown author. 2017. Public domain. https://pxhere.com/en/photo/759379.
Figure 5.14: Expansion joint on a bridge, allowing for thermal expansion and contraction of the bridge as temperature changes through the year. James Lord. 2024. CC BY-NC-SA.