ISIS Educational Module 1:

Mechanics Examples Incorporating FRP Materials

Prepared by ISIS Canada A Canadian Network of Centres of Excellence www.isiscanada.com Principal Contributor: L.A. Bisby, Ph.D., P.Eng.Department of Civil Engineering, Queen’s UniversityContributor: M. RangerMarch 2006

ISIS Education Committee:

N. Banthia, University of British ColumbiaL. Bisby, Queen’s UniversityR. Cheng, University of AlbertaR. El-Hacha, University of CalgaryG. Fallis, Vector Construction GroupR. Hutchinson, Red River CollegeA. Mufti, University of ManitobaK.W. Neale, Université de SherbrookeJ. Newhook, Dalhousie UniversityK. Soudki, University of WaterlooL. Wegner, University of Saskatchewan

ISIS Canada Educational Module No. 1: Mechanics Examples Incorporating FRP Materials

Objectives of This ModuleThe objective of this module is to provide civil engineering instructors and students with a series of worked examples which are demonstrative of field applications and which can be used to incorporate fibre reinforced polymer (FRP) materials into first and second-year engineering curricula across North-America. This document is one of a series of modules on innovative FRP technologies available from ISIS Canada. Further information on FRP materials and on the use of FRPS in a variety of innovative applications can be found on the internet at www.isiscanada.com.

An overall knowledge of currently available FRP materials is essential for the new generation of civil engineers. Experience has shown that the problems of the future cannot generally be solved with the materials and

methodologies of the past. Thus, the primary objectives of this manual can be summarized as follows:1. to provide civil engineering students with a general

awareness of the properties and behaviour of FRP materials;

2. to provide information on some of the potential uses of FRPs in civil engineering applications;

3. to facilitate the use of FRP reinforcing materials in the construction industry; and

4. to provide guidance for students seeking additional information on this topic.Both students and instructors are encouraged to consult

ISIS Educational Modules 2 through 4 for additional information on the properties, behaviour, and uses of FRP materials in civil engineering.

1

Additional ISIS Educational ModulesAvailable from ISIS Canada (www.isiscanada.com)

Module 2 – Introduction to FRP Composites for Construction

FRP materials are discussed in detail at the introductory level. This module seeks to expose undergraduate students to FRP materials such that they have a basic understanding of the components, manufacture, properties, mechanics, durability, and application of FRP materials in civil infrastructure applications. A suggested laboratory is included which outlines an experimental procedure for comparing the stress-strain responses of steel versus FRPs in tension, and a sample assignment is provided.

Module 3 – Introduction to FRP-Reinforced Concrete

The use of FRP bars, rods, and tendons as internal tensile reinforcement for new concrete structures is presented and discussed in detail. Included are discussions of FRP materials relevant to these applications, flexural design guidelines, serviceability criteria, deformability, bar spacing, and various additional considerations. A number of case studies are also discussed. A series of worked example problems, a suggested assignment with solutions, and a suggested laboratory incorporating FRP-reinforced concrete beams are all included.

Module 4 – Introduction to FRP-Strengthening of Concrete Structures

The use of externally-bonded FRP reinforcement for strengthening concrete structures is discussed in detail. FRP materials relevant to these applications are first presented, followed by detailed discussions of FRP-strengthening of concrete structures in flexure, shear, and axial compression. A series of worked examples are presented, case studies are outlined, and additional, more specialized, applications are introduced. A suggested assignment is provided with worked solutions, and a potential laboratory for strengthening concrete beams in flexure with externally-bonded FRP sheets is outlined.

Module 5 – Introduction to Structural Health Monitoring

The overall motivation behind, and the benefits, design, application, and use of, structural health monitoring (SHM) systems for infrastructure are presented and discussed at the introductory level. The motivation and goals of SHM are first presented and discussed, followed by descriptions of the various components, categories, and classifications of SHM systems. Typical SHM methodologies are outlined, innovative fibre optic sensor technology is briefly covered, and types of tests which can be carried out using SHM are explained. Finally, a series of SHM case studies is provided to demonstrate four field applications of SHM systems in Canada.

Module 6 – Application & Handling of FRP Reinforcements for Concrete

Important considerations in the handling and application of FRP materials for both reinforcement and strengthening of reinforced concrete structures are presented in detail. Introductory information on FRP materials, their mechanical properties, and their applications in civil engineering applications is provided. Handling and application of FRP materials as internal reinforcement for concrete structures is treated in detail, including discussions on: grades, sizes, and bar identification, handling and storage, placement and assembly, quality control (QC) and quality assurance (QA), and safety precautions. This is followed by information on handling and application of FRP repair materials for concrete structures, including: handling and storage, installation, QC, QA, safety, and maintenance and repair of FRP systems.

Module 7 – Introduction to Life Cycle Engineering & Costing for Innovative Infrastructure

Life cycle costing (LCC) is a well-recognized means of guiding design, rehabilitation and on-going management decisions involving infrastructure systems. LCC can be employed to enable and encourage the use of fibre reinforced polymers (FRPs) and fibre optic sensor (FOS) technologies across a broad range of infrastructure applications and circumstances, even where the initial costs of innovations exceed those of conventional alternatives. The objective of this module is to provide undergraduate engineering students with a general awareness of the principles of LCC, particularly as it applies to the use of fibre reinforced polymers (FRPs) and structural health monitoring (SHM) in civil engineering applications.

Module 8 – Durability of FRP Composites for Construction

Fibre reinforced polymers (FRPs), like all engineering materials, are potentially susceptible to a variety of environmental factors that may influence their long-term durability. It is thus important, when contemplating the use of FRP materials in a specific application, that allowance be made for potentially harmful environments and conditions. It is shown in this module that modern FRP materials are extremely durable and that they have tremendous promise in infrastructure applications. The objective of this module is to provide engineering students with an overall awareness and understanding of the various environmental factors that are currently considered significant with respect to the durability of fibre reinforced polymer (FRP) materials in civil engineering applications.

Contents

This educational module contains 19 worked examples dealing with mechanics of materials and incorporating FRP materials. In many cases, additional information has been included in the problem definition to inform students as to

the real-world significance of a particular example. The table below provides a list of the primary and secondary topics covered in the various examples:

No. Application Topics Covered Page #

1 FRP Cable-Supported Walkway Equilibrium, Average Normal Stress 3

2 FRP Repair of Metallic Structures Average Shear Stress, Allowable Stress 4

3 FRP Dowel Bars for Concrete Average Shear Stress 5

4 Stress-Strain Behaviour: FRP vs. Steel Stress-Strain, Hooke’s Law, Axial Elongation 6

5 FRP Repair of Metallic Structures Thermal Stress and Strain, Indeterminacy 8

6 FRP Poles Shear Stress and Deformation due to Torsion 10

7 FRP Reinforcing Bars Axial Deformation, Poisson’s Ratio 11

8 FRP Repair of Metallic Structures Average Shear Stress 12

9 External FRP Reinforcement for Concrete Average Shear Stress and Strain/Deformation 13

10 FRP Stay Cables for Bridges Normal Stress, Strain, and Elongation 15

11 FRP versus Steel and Aluminum Thermal Stress and Strain 16

12 Pultruded FRP Flexural Members Inertia, SFD, BMD, Bending Stress 17

13 Timber Flexural Strengthening with FRP Composite Beams, Bending Stress 18

14 FRP-Reinforced Concrete Beam Composite Beams, Bending Stress, Concrete 20

15 FRP-Strengthened Concrete Beam Composite Beams, Bending Stress, Concrete 22

16 Hybrid Aluminum/FRP Sections Composite Beams, Bending Stress 25

17 FRP-Strengthened Timber Beams Composite Beams, Bending Stress, Deflection 27

18 FRP Bridge Deck Panels Composite Beams, Deflection 29

19 Near Surface Mounted FRP Reinforcement Composite Beams, Bending Stress 31

Example 1: A section of a stainless steel overhead walkway, section BD, is supported by two carbon fibre reinforced polymer (CFRP) cables, AB and CD, in a corrosive industrial setting. CFRP cables have been used because they are extremely strong, light, and non-corrosive, unlike conventional steel cables. The cross-sectional area of CFRP cable AB is 100 mm2, and that of CD is 120 mm2. Determine the distance, x, from point B that the 4 kN load must be applied such that the average normal stress in each CFRP bar is equal.

Solution:

Draw a free body diagram of the walkway.

Consider equilibrium in the vertical direction. Sum the forces in the y-direction:

Thus, [1]

Consider moment equilibrium. Sum the moments about point B:

Thus, [2]

The average normal stress in each of the rods can be found using:

In this case, since we know that the stresses in AB and CD are equal, we have:

Thus, [3]

Substituting [3] into [1] gives:

which gives, [4]

Substituting [4] into [3]:

[5]

Finally, substituting [5] into [2] gives:

Therefore, the average axial stress in the CFRP cables will be equal when x = 2.73 m.

A

B

C

D

4 kN

x

5 m

B D

4 kN

x

5 m

FAB FCD

x

y

M

Example 2:

Occasionally, metal bars subjected to repeated cycles of load can fail due to a phenomenon called fatigue. One method that can be used to repair a fatigue fractured metal bar is to join the two ends of the bar using adhesively bonded fibre reinforced polymer (FRP) plates. These plates are extremely light and strong, and will not corrode. In one specific application of this repair technique, shown on the right, FRP plates are to be bonded to both sides of a fractured flat aluminum bar. The fractured ends of the bar have been shaped to form a uniform space of 8 mm. Determine the length L required for a factor of safety (F.O.S.) of 3.0, given that the ultimate (failure) average shear stress of the adhesive used to bond the FRP to the aluminum plates is 4.5 MPa.

30 kN

30 kN

L

120 mm

8 mm

Fracture

Aluminum

FRP Plate

Solution:

Given the known allowable average shear stress in the bond and the required factor of safety, we can determine the allowable shear stress on the bonded section:

Thus, MPa [1]

The average shear stress in the bonded section can be found using the average shear stress equation:

[2]

The shear force, V, that must be transferred through the bonded area is 30 kN. The area resisting the shear force is

equal to twice the area of each bonded surface (double shear in this case). The area of each bonded surface can be expressed as its width, b, times its length, d, thus:

Thus, [3]

Now we can determine L by substituting [1] and [3] into [2]:

Gives mm

Therefore the overall length of the FRP plates must be greater than 175 mm to ensure a factor of safety of 3.0.

Example 3:

Concrete pavements (slabs) are often used in lieu of asphalt in many highways and airport runways across North America. To prevent uncontrolled cracking in these pavements resulting from shrinkage and thermal cycling, construction joints are normally placed in the concrete slabs at various desired locations. To prevent differential settlement of adjacent slabs, dowel bars are placed across the construction joints to ensure that adjacent slabs stay level with each other. As a result, these dowel bars sometimes carry shear stresses.

Traditionally, concrete dowels have been fabricated from steel rods. However, when subjected to moisture and deicing salts, these steel dowels corrode (rust) and expand, causing concrete cracking and often leading to failure. Recently, fibre reinforced polymer (FRP) bars have been used as an alternative to steel bars in this application, as shown in the figures to the right in which glass FRP dowels are being placed prior to pouring a concrete pavement.

For the dowel bar layout shown, determine the number of 25 mm diameter FRP dowels required to transfer a vertical force of 50 kN across the construction joint if the dowel bars are capable of carrying an average shear stress of 25 MPa. How much stronger is the joint than it needs to be?

50 kN

Construction Joint FRP Dowel

Concrete Slabs

50 kN

Solution:

The dowels transfer shear force across a circular area represented by their cross-section.

Given the known ultimate average shear strength and diameter of the FRP dowels, the average vertical force that can be transferred per dowel can be obtained by multiplying the shear strength by the cross-sectional area:

N

Thus, kN.

Given a total force of 50 kN to be transferred across the joint, the number of dowels required is calculated as follows:

However, we must use an integer number of bars, so we require 5 dowels across the joint. Hence, the actual force that can be transferred across the joint is:

kN

And thus the joint is:

stronger than it actually needs to be.

Therefore use five dowel bars, which results in a construction joint that is 23% stronger than required.

Example 4:

The relationship between stress and strain is an extremely important characteristic for all engineering materials. The figure to the right shows typical stress-strain curves for two different materials that are currently used for reinforcement of concrete. Steel, which has traditionally been the material used for concrete reinforcement, displays an elastic-plastic response, while glass FRP, an innovative material which has been gaining acceptance as a reinforcing material for concrete, is linear-elastic to failure. Given the stress-strain curves shown in the figure:

(a) Using Hooke’s Law, determine the modulus of elasticity (Young’s modulus) of both steel and glass FRP.

(b) If both materials were subjected to a strain of 1.8%, what would be the stress in each material?

(c) If a 920 mm long circular steel rod with a diameter of D = 10 mm is subjected to an axial load of 25 kN,

what would be its axial elongation? What if the same rod were made from glass FRP?

Steel

Strain (%)

Stress (MPa)Rupture

Glass FRP

2.00.2

1400

400

Yield Point

Solution (a):

The modulus of elasticity is determined using Hooke’s law. For the reinforcing steel, the modulus of elasticity is determined in the pre-yield zone only, where the steel is still behaving elastically. For glass FRP, which is a linear elastic material, the modulus can be determined at any point on the stress-strain curve. Thus, for steel:

GPa

For the glass FRP:

GPa

Therefore, the moduli of elasticity for the steel and glass FRP are 200 GPa and 70 GPa respectively.

Solution (b):

Since steel behaves in an elastic-plastic manner, at any strain larger than the yield strain, 0.2%, the stress remains constant at approximately 400 MPa. Thus, at a strain of 1.8 % the stress in the steel is 400 MPa.

For the glass FRP, we obtain the stress at a strain of 1.8% from Hooke’s Law:

MPa

Therefore, the stress in the steel and glass FRP at a strain of 1.8% are 400 MPa and 1260 MPa respectively.

Solution (c):

The change in length of an axially loaded rod of an elastic material can be determined using the following equation:

Thus, for the steel, we have:

mm

And, for the glass FRP:

mm

Therefore, bars of the same diameter and length made from steel or glass FRP would elongate by 1.46 mm and 4.18 mm, respectively, under a 25 kN applied load.

We should also check to ensure that the steel has not yielded under the applied load and that the FRP has not failed in tension. The stress in either bar is given by:

MPa

Which is less than the yield stress of 400 MPa for the steel, or the failure stress of the FRP, and hence our calculations are correct.

Example 5:

Spurred on by the widespread and severe deterioration from which the North American infrastructure is currently suffering, structural engineers are turning to new and innovative materials, such as non-corrosive ultra-high-modulus (UHM) carbon fibre reinforced polymers (FRP) to repair and strengthen steel bridge girders. In these applications, FRP plates are bonded to the tension flanges of steel bridge girders to increase the girders’ flexural capacity using an industrial grade epoxy adhesive.

In one specific application of this technique, a simply-supported steel I-girder is plated on both its top and bottom flanges with UHM carbon FRP plates, as shown in the figure below. In practice, a thin layer of glass fibres is

placed between the carbon plates and the steel to prevent galvanic corrosion.

Neglecting self-weight and assuming that the steel girder has an elastic modulus of 200 GPa and a coefficient of thermal expansion of 1310-6/˚C, and that the carbon FRP has an elastic modulus of 340 GPa and a zero coefficient of thermal expansion, determine (assuming carbon FRP is fully effective in compression, which is not strictly true):

(a) The stress in each material resulting from a change in temperature of 40˚C.

(b) The overall change in length of the beam when subjected to the temperature change in (a).

Solution (a):

100 mm

150 mm

12 mm

10 mm

10 mm

12 mm

10 mm

2000 mm

Steel FRP

The overall strain in both materials must be equal, since they are bonded to each other. In addition, the stresses in each material must be opposite and have equal stress resultant magnitudes, such that equilibrium of internal forces is maintained. The stress in each material is due to a combination of thermal strain and strain due to the restraint provided by the other material. For the steel, the free thermal strain is calculated as:

The strain in the steel due to restraint is:

Similarly, the stress in the carbon FRP due to restraint is:

There is no thermal strain in the carbon FRP since its coefficient of thermal expansion is zero.

By superposition, the strain in the steel is equal to the sum of the thermal and mechanical strains, and this must be equal to the strain in the FRP. Thus:

And the stress in each material can now be calculated using the average stress equation:

MPa (i.e. compression)

MPa (i.e. tension)

Therefore, the stresses in the steel and carbon FRP are 56 MPa compression and 82 MPa tension, respectively.

Solution (b):

The overall change in length can be obtained by examining either of the two materials in the cross-section. Taking the carbon FRP, we have:

Which gives:

mm

Therefore, the overall change in length of the strengthened carbon FRP-strengthened steel bridge girder under the temperature change of 40˚C is 0.48 mm.

Alternatively, we could have solved the problem by examining the steel, which would give:

And, as before:

mm

Example 6: Fibre reinforced polymer (FRP) materials are light, strong, and have outstanding environmental durability. Thus, FRPs are now being used in a variety of applications where metals such as steel and aluminum have traditionally been used. One example of this technology is the use of FRP tubes for highway sign standards, where a single FRP ‘monopole’ can be used to support a large billboard or highway sign.

In one specific application of this technology, an FRP tube has been used, as shown to the right, to support a highway sign which is subject to wind loading. As a result of the worst-case wind loading, the FRP tube must resist an applied torque of 9 kNm. If the FRP tube has an outside diameter of 200 mm, a wall thickness of 5 mm, and a shear modulus (modulus of rigidity) of 30 GPa, determine:

(a) The maximum shear stress in the FRP tube due to the applied torque.

(b) The angle of rotation of the sign at A under this worst-case wind load with respect to the fixed support at B.

Solution (a):

The maximum shear stress in the tube will be found at the outside surface. We use the torsional shear stress formula:

The polar moment of inertia, J, can be calculated for a hollow tube as follows:

The maximum shear stress can now be calculated:

MPa

Therefore, the maximum shear stress in the FRP pole is at the outside surface of the tube and has a magnitude of 30.9 MPa.

Solution (b):

The angle of rotation of the sign under this worst-case wind loading scenario is determined using the torque-twist relationship:

In this case, we have the following:

radians

Therefore, the angle of rotation of the sign under the worst-case wind loading is 0.0309 radians, or 1.77˚.

Exit 2009

ISIS CanadaInnovation Ahead!A

B

300

0 m

m

Example 7: Glass fibre reinforced polymer (GFRP) materials are becoming increasingly popular in structural engineering applications due to their high strength, light weight, and non-corrosive nature. These materials are composed of long and thin microscopic glass fibres embedded in a polymer matrix to create a composite material.

A uniform circular rod is composed of GFRP with all of the glass fibres running along the axis of the rod (the rod is thus called unidirectional). If the rod is 250 mm long and 20 mm in diameter, determine the changes in its length and diameter if it is subjected to a 100 kN axial load. Assume

that the modulus of elasticity of the rod is Efrp = 40 GPa and that its Poisson’s ratio is frp = 0.38 (note that GFRP is not isotropic in reality).

Solution:

The average axial stress in the GFRP bar can be determined from the following:

Thus,

GFRP is a linear-elastic material. We can thus use Hooke’s law, , to determine the longitudinal strain given the stress and elastic modulus:

Or, in units of microstrain,

The axial elongation of the bar is determined using the definition of axial strain:

Thus,

Now, the lateral strain can be obtained using Poisson’s ratio:

Thus,

The change in the diameter of the bar is found, again using the definition of strain (lateral strain in this case):

Thus,

Therefore, the changes in the bar’s length and diameter are 1.99 mm (elongation) and 0.060 mm (contraction), respectively.

250 mm

100 kN

20 mm

Example 8: Many steel cranes and aluminum sign standards are fabricated from metal tubular hollow structural sections (HSS). In some cases, these members are subjected to repeated load cycles (fatigue cycles) which can eventually lead to cracking, and occasionally can cause failure. Fibre reinforced polymer (FRP) sheets can be used to patch these metal structures by adhesively bonding FRP to the exterior of the structure to bridge cracks. This is shown in the photo

below, where an aluminum overhead sign is being repaired with glass FRP sheets.

In one specific application of this technology, a 70 mm diameter cracked steel tube, shown below, has been rehabilitated using a carbon FRP wrap. Determine the average shear stress over the bonded area when a tensile axial load of 200 kN is applied to the ends of the tube.

Solution:

The average shear stress can be determined using the following equation, in which V is the shear force to be transferred through the bond and A is its area parallel to the direction of the applied load:

On each side of the crack, the bonded area through which the force must be transferred is:

Since the total force to be transferred through the bond is 200 kN, the average shear stress can be calculated as:

Therefore, the average shear stress over the bonded area is 12.1 MPa when a tensile axial load of 200 kN is applied to the ends of the fractured tube.

CrackFRP

200 kN

D = 70 mm

200 kN

L = 150 mm

Example 9: One of the most widely implemented applications of fibre reinforced polymers (FRPs) in civil engineering applications is for externally-bonded reinforcement of concrete structures. In this application, FRP plates or sheets are bonded to the exterior surface of a reinforced concrete structural member to provide tensile reinforcement which supplements that provided by the internal reinforcing steel.

In order to rehabilitate a particular concrete beam using a carbon FRP plate, we will assume that the bond between the concrete and the FRP must be able to withstand an allowable bond shear stress of 10 MPa to maintain an

overall factor of safety of 2.5. A common method of testing the bond between the FRP and the concrete is the single lap pull-off test, which is illustrated below.

(a) If the FRP is expected to carry a tensile load 30 kN, what length of bond, x, is required to obtain the desired factor of safety? Also, what is the average shear stress in the bond at failure?

(b) If the concrete is assumed to be an elastic material with a Young’s modulus of E = 28 GPa and a shear modulus of G = 9 GPa, what is the horizontal displacement of point A when the load is applied?

Solution (a):

The average shear stress (bond stress in this case) can be calculated by:

where V is the force to be transferred and A is the bond area. In this case we want the bond area, so rearranging gives:

Now, since,

Thus,

Therefore the required bond length to maintain the average bond stress below 10 MPa is 60 mm.

Now, if the factor of safety is 2.5 with an allowable bond stress of 10 MPa, the average bond stress at failure can be obtained as follows:

Thus,

The average shear stress in the bond at failure is 25 MPa.

It is interesting to note that if we actually calculate the load at failure we obtain the following:

which is equal to the factor of safety of 2.5 times the actual load on the joint, as expected.

w = 50 mm

x

Concrete Block 30 kN

Carbon FRP Sheet

30 kN

x

ELEVATION

PLAN

A

60 mm

60 mm Concrete

200 mm

Solution (b):

To solve this portion of the problem, we must recognize that a shear force of 30 kN is being transferred from the top surface of the concrete to its bottom surface (the testing machine), and that this will cause a shear stress and a corresponding shear strain in the concrete. The average shear stress on any horizontal plane in the concrete is given by:

MPa

The relationship between shear stress and shear strain is given by Hooke’s Law for shear stress:

Thus, in this case, we can calculate the shear strain as follows:

radians

To find the horizontal displacement of point A, we must recognize that we will have an angular displacement of 0.27810-3 radians over a vertical distance of 60 mm. Thus:

And so:

Therefore, the horizontal displacement of point A under the applied loading is 0.017 mm.

Example 10: Since the Second World War, a new class of cable-supported bridges, called cable-stayed bridges have seen increasing popularity. These structures are constructed by first building a tower, and then suspending the deck from the tower one section at a time, each deck portion being supported by a series of stay cables. One such bridge during construction is shown schematically in the figure to the right.

Traditionally, cable-stayed bridges have been constructed from steel or concrete deck sections with steel stay cables. Recently however, fibre reinforced polymers (FRP) have been used, both for the cables and for the prefabricated deck sections, due to their light weight and outstanding resistance to corrosion.

In a new design of this type, shown schematically to the right, aramid (KevlarTM) FRP cables and a glass FRP deck have been employed. What would be the axial strain and increase in length of cable AB during construction with the addition of the bridge deck section BC? Assume that the cable has a 50 mm diameter, and that each bridge deck

section is 3 m long and has a total mass of md = 9600 kg. The bridge deck sections are assumed to be pin-connected. Young’s modulus for the FRP cables is 125 GPa.

Solution:

A

B C

45˚

12 m

First, we draw a free body diagram of the deck section BC. Since there is a pin connection at C, both horizontal and vertical reactions are present:

The weight of the bridge deck section BC is:

The force in the cable is determined using the equations of equilibrium in 2-dimensions:

[1]

[2]

[3]

Solving [1], [2], and [3], gives:

FCY = 47088 N, FCX = 47088 N, and FBA = 66592 N.

The stress in the cable can now be calculated:

And Hooke’s law can be used to determine the strain:

The elongation of the cable is given by:

Therefore, the strain in the cable is 271 microstrain, and its total elongation is 4.6 mm.

FBA FCY

FCX

Wd

45˚

B

C

1.5 m 1.5 m

Example 11: Different materials behave differently when subjected to a change in temperature. For instance, most materials, like steel, aluminum, or concrete, expand on heating. Some materials, like aramid fibre reinforced polymer (AFRP) actually contract on heating. What would be the respective stresses and strains induced in the W350 steel, aluminum, and AFRP bars shown below when their temperatures rise from 0C to 40C. The coefficient of thermal expansion of A-36 steel can be taken as αs = 12x10-6/C, that of aluminum as αa = 24x10-6/C, and that of AFRP (in the longitudinal direction) can be taken as αAFRP = -4x10-6/C. The Young’s modulus can be assumed as 200 GPa for steel, 70 GPa for aluminum, and 60 GPa for AFRP.

Solution:

2 m

W350 steel

Aramid FRP

Aluminum

The thermal strains that would result from free thermal expansion under these conditions can be determined using:

Thus:

Since the bars are all fully restrained, the thermal stress can be determined from the free thermal strain using:

Thus:

Therefore, the strain in the bars is zero, since they are fully restrained. However, the stresses due to temperature change are -96 MPa (i.e. compression), -67 MPa (i.e. compression), and +9.6 MPa (i.e. tension) in the steel, aluminum, and AFRP bars, respectively.

This problem illustrates a special case of loading where stress is induced in the absence of strain.

Example 12: One of the major problems facing existing steel structures is corrosion of their component members, which can lead to reduced flexural capacity and eventually to failure. In an effort to prevent corrosion in structures, civil engineers are looking to non-corrosive materials, such as strong and light weight fibre reinforced polymers (FRPs) to replace steel in many construction applications. One example of an FRP structural member that is currently available for use is the pultruded glass FRP I-beam. These members can be

manufactured in almost any size, and perform well in a variety of environments and applications.

In one specific application, a pultruded glass FRP I-beam is subjected to a point load of 25 kN, applied 5 m from the left-hand end, over a simply-supported span of 8 m. The beam has cross-sectional dimensions as shown below. Determine the maximum bending (normal) stress in the beam under the load condition shown.

Solution:

To find the bending stress at any location in the cross-section we can use the flexure formula:

The maximum bending stress will occur furthest from the neutral axis, or at y = 175 mm in the current example, at the location of the maximum moment along the span. The maximum moment, M, must be determined from the bending moment diagram. Thus:

From the bending moment diagram we note that the maximum moment on the beam is 46.9 kNm.

The moment of inertia of the cross-section must be determined. In this case:

Now, the maximum normal stress due to bending can be determined as (with all units in N, mm, and MPa):

25 kN

5 m 3 m

25 mm

25 mm

200 mm

C.G.150 mm

150 mm

25 mm

00 SFD(kN)

BMD(kNm)

9.4 9.4

15.6 15.6

46.9

+

-

25

FBD(kN)

+00

9.4

0

15.6

5 m 3 m

y

Example 13: Timber has been used as a construction material for thousands of years throughout human history. However, timber structures deteriorate over time due to various factors, and modern timber bridges are no exception. One technique that has been successfully used to strengthen and rehabilitate aging timber bridges in Canada is externally-bonded fibre reinforced polymer (FRP) materials. In this application, high strength FRP plates, sheets, or bars are bonded to the tension faces of timber beams to increase their flexural capacity, as illustrated in the following example.

A simply-supported timber bridge beam with a 6 m span and a depth of 300 mm is subjected to a uniformly distributed load of 12 kN/m.

a) Determine the required width of the beam such that the maximum bending stress in the wood does not exceed ±18 MPa.

b) At some later stage in the life of the bridge, the loads on the bridge are increased to 15 kN/m and the decision is made to increase the flexural capacity of the structure by bonding carbon FRP sheets to the tension face of the beam. If the elastic moduli of the timber and CFRP are taken as Et =13.1 GPa and ECFRP = 155 GPa, respectively, determine the width of a 4 mm thick CFRP plate that is required to maintain the stress in the timber below the original allowable stress of 18 MPa.

Solution (a):

First, we construct shear force and bending moment diagrams:

Bending stress can be calculated using the flexure formula:

The maximum bending stress will occur at the top or bottom of the section, where y = ±150 mm, at midspan, where M is 54 kNm. The moment of inertia of the cross-section can be obtained from:

Thus, the required width of the cross-section can be determined as:

which gives a required width, b = 200 mm.

Therefore, the section must be 200 mm wide in order for the bending stress in the wood to remain less than 18 MPa.

Solution (b):

-

12 kN/m

0 SFD(kN)

BMD(kNm)

FBD(kN)

+

+

0

0

0

36

36

54

36 36

0

6 m

If the load on the beam is increased to 15 kN/m, the maximum moment at midspan increases from 54 kNm to 67.5 kNm. The carbon FRP strengthened beam will have the cross-sectional dimensions shown below.

The problem is to determine the width of FRP plate that maintains the stress in the wood below 18 MPa. First, we must determine the transformed section moment of inertia. In this case we will transform the FRP to timber. The modular ratio is described by:

Thus, the transformed width of the FRP is 11.83w. We first determine the neutral axis depth:

[1]

The transformed moment of inertia is calculated using the parallel axis theorem:

Therefore:

[2]

Now, again using the flexure formula and recognizing that the maximum bending stress in the timber will occur at the top fibre of the cross-section:

[3]

Equations [1], [2], and [3] can be solved simultaneously to give w = 202 mm.

Therefore, the CFRP plate must be 202 mm wide to ensure that the maximum bending stress in the timber remains less than 18 MPa under the increased load condition.

NA

200 mm

300 m

m

4 mm

FRP

Timber

w y

Example 14: One of the most pressing problems with conventional steel-reinforced concrete construction is corrosion of reinforcing steel, which can cause the steel reinforcement to expand, subsequently cracking the concrete. This cracking can result in increased corrosion, spalling of the concrete cover, and can eventually lead to failure of the member. In an effort to combat this problem, structural engineers are now using non-corrosive fibre reinforced polymer (FRP) reinforcing materials, in some cases, in place of steel reinforcement for concrete.

The concrete beam shown to the right is reinforced with three 12 mm diameter carbon FRP (CFRP) reinforcing bars, and has the cross-sectional properties shown. The beam is subjected to a maximum in-service bending moment of 35 kN·m. The modulus of elasticity of the concrete is taken as Ec = 25 GPa, and that of CFRP is taken as ECFRP = 147 GPa. Assuming all materials behave linear-elastically (which is not strictly true for concrete at higher stress levels!), and that concrete is ineffective in tension:

(a) What is the maximum bending stress in the concrete?

(b) Is the FRP reinforcement capable of withstanding this moment if its ultimate tensile strength is 1500 MPa?

Solution (a):

First, we will transform the area of carbon FRP to concrete. The area of CFRP is:

The modular ratio, n, can be found from the ratio of elastic moduli:

The area of FRP, Afrp, can be transformed into an equivalent area of concrete, Afrpt, by multiplying by the modular ratio:

We must account for the fact that concrete is ineffective in tension (since it’s cracked). This means that the transformed section is actually as shown below:

Equilibrium of forces at any cross-section dictates that the centroid of the effective area must lie on the neutral axis, thus:

and solving for the positive root, the neutral axis depth, h’, is 61.3 mm. The moment of inertia can now be calculated using the parallel axis theorem:

Now, the maximum bending stress in the concrete can be determined using the flexural formula:

Therefore, the maximum bending stress in the concrete is 24.8 MPa in compression.

Concrete

12 mm diameter CFRP bars

300 m

m

200 mm

50 mm

NA

Afrpt = 1995 mm2

Concrete

250 m

m

200 mm

h’

Solution (b):

The maximum normal stress in the transformed area of FRP is determined from the flexure formula:

However, in reality we have less area of FRP, so the stress is higher by a factor of the modular ratio, n, thus:

Therefore, since 449.3 MPa < 1500 MPa, the FRP rods are sufficiently strong to resist the applied moment.

Example 15: Many existing concrete beams in buildings, bridges, and parking garages have suffered severe deterioration as a consequence of corrosion of reinforcing steel within the concrete, which can lead to cracking, spalling, and eventually failure. In an effort to repair deteriorated concrete structures, engineers are now turning to fibre reinforced polymers (FRP) as alternatives to steel. These lightweight, strong, and durable materials can be bonded to the exterior of concrete members to increase their flexural and/or shear strength.

The concrete beam shown to the right was originally built using three 10 mm diameter steel reinforcing bars and with cross-sectional properties as shown in the figure.

Since its original construction, the reinforcing steel has corroded and the diameter of the steel reinforcing bars has been reduced to only 7.5 mm. It is proposed to strengthen the beam by bonding carbon FRP sheets to its bottom face. The carbon FRP sheets are 5 mm thick and have an elastic modulus of Efrp = 120 GPa. The modulus of elasticity of the concrete is taken as Ec = 25 GPa and that of steel is taken as Es = 200 GPa. If the beam was originally designed to carry an applied moment of 17.5 kN·m, determine, assuming all materials behave linear-elastically (which is not strictly true for concrete at higher stress levels!) and that concrete is ineffective in tension:

(a) the bending stress in the steel under the original design; and

(b) if the stress in the steel in the repaired beam is less than in the original design, as required by the engineers implementing the repair strategy.

Solution (a):

First, for the original beam we must transform the area of steel to concrete. The area of steel is:

The modular ratio, n, can be found from the ratio of elastic moduli:

The area of steel, As, can be transformed into an equivalent area of concrete, Ast, by multiplying by the modular ratio:

We must account for the fact that concrete is ineffective in tension (since it’s cracked). This means that the effective area of the cross-section is as shown in the following figure:

Equilibrium of forces at any section dictates that the centroid of the effective area lies on the neutral axis, thus:

Original Construction

Concrete

10 mm diameter steel bars

300 m

m

200 mm

50 mm

Concrete

7.5 mm diameter steel bars

200 mm

5 mm thick carbon FRP sheet

Repaired Beam

NA

Ast = 1885 mm2

Concrete

250 m

m

200 mm

h’

Solving for the positive root, the neutral axis depth, h’, is 59.9 mm. The moment of inertia can now be calculated using the parallel axis theorem:

Now, the maximum bending stress in the transformed steel reinforcement can be determined using the flexural formula:

To get the actual maximum bending stress in the steel reinforcement we must multiply by the modular ratio:

Therefore, the bending stress in the steel in the original construction is 240 MPa (tension).

Solution (b):

For the strengthened beam, we have a reduced steel area and an additional contribution from the FRP sheet. First we must transform the steel to concrete:

And, for the deteriorated beam:

Thus:

For the FRP sheet:

And, so the transformed area of FRP is:

Thus, the transformed cross-section of the beam can be illustrated as follows:

The centroid can be found using the same procedure as in part (a) above:

and solving for the positive root, the neutral axis depth, h’, is 105.0 mm. The moment of inertia can now be calculated using the parallel axis theorem, and adding an additional term to account for the presence of the FRP:

Ast = 1060.3 mm2

Concrete25

0 m

m

200 mm

h’

300

mm

Afrp = 4800 mm2

Now, the maximum bending stress in the transformed steel reinforcement can be determined using the flexural formula:

To get the actual maximum bending stress in the steel reinforcement we must multiply by the modular ratio:

Therefore, the bending stress in the steel in the repaired construction is 75.9 MPa (tension), which is substantially less than in the original construction.

Example 16: Aluminum has a number of significant advantages as a structural material. It is non-corrosive in most environments and is much lighter than steel. However, aluminum has much lower yield strength and elastic modulus than steel, and this has restricted its use in structural engineering applications. Recently, it has been suggested that hybrid structural sections could be created by combining aluminum with high-modulus, non-corrosive, and light weight carbon FRP. As an example, an aluminum beam with the cross-sectional properties shown below is subjected to a moment of 60 kN·m. What would be the percentage decrease in the maximum stress in the aluminum with the addition of a 100 mm 1.4 mm carbon FRP plate to:

(a) the bottom face of the beam only?

(b) both the top face and the bottom face of the beam (assuming carbon FRP is fully effective in compression, which is not strictly true)?

Note that material properties for aluminum and carbon FRP can be assumed as follows:

Elastic modulus of aluminum, EAl = 70 GPa Elastic modulus of carbon FRP, Efrp = 300 GPa

Solution (a):

The moment of inertia of the cross-section must be determined. In this case:

The maximum bending stress in the original aluminum beam without carbon FRP plates can be calculated using the flexure formula:

For the beam with carbon FRP attached to the bottom face only, the beam looks as follows:

The modular ratio, n, can be found by:

and the equivalent width of transformed FRP can be determined:

Thus, the transformed beam looks as follows:

The neutral axis depth can be determined by:

25 mm

25 mm

200 mm

C.G.150 mm

150 mm

25 mm

1.4 mm

200 mm

Aluminum

Carbon FRP

100 mm

y1.4 mm

200 mm

Aluminum

Transformed FRP

429 mm

y

The transformed moment of inertia can be taken as:

The maximum stress in the aluminum after FRP has been added to the bottom face of the beam will occur at the top fibre of the section (farthest from the neutral axis), and can be calculated as:

Now, the decrease in maximum stress in the aluminum is:

Therefore, by adding a single 100 mm 1.4 mm carbon FRP strip to the bottom face of the beam, the stress in the aluminum is decreased by 0.7 MPa.

Solution (b):

If carbon FRP strips are added to both the top and bottom faces of the beam, the transformed section looks as follows:

The beam is now symmetric about the neutral axis, so the neutral axis depth can be taken as:

The transformed moment of inertia can be taken as:

mm 4.176y1.4 mm

200 mm

429 mm

1.4 mm

The modified maximum stress in the aluminum is now:

The decrease in stress as compared with plain aluminum can be calculated as:

Therefore, by adding 100 mm 1.4 mm carbon FRP strips to both the top and the bottom faces of the beam, the stress in the aluminum is decreased by 3.4 MPa.

Example 17: Many timber bridges constructed in the past are no longer strong enough to carry the larger, heavier trucks that are now using our roads. As a consequence, many timber bridges in Canada require strengthening to enable them to carry these increased loads. One way to increase the bending strength and stiffness of a timber bridge is to bond FRP plates or sheets to the bottom face of the timber bridge

beams. The following example demonstrates the basic application of FRP strengthening of a timber bridge beam.

At the time of its initial construction, the design load on a simply supported timber bridge was approximated as shown below, with an 8 kN/m uniformly distributed load (UDL) over a span, L, of 6 metres:

Problem (a):

The beam has cross-sectional dimensions b and h as shown. If the depth of the beam is limited to 300 mm, what is the required cross-sectional width such that the maximum

deflection is less than L/240? The elastic modulus of timber can be taken as 13.1 GPa.

Solution (a):

The limit on deflection has been specified as a ratio of the span length, L, thus:

The deflection of the timber beam can be calculated by any appropriate method. Here we will use integration. The bending moment at any section at a distance, x, from the left hand end of the beam can be expressed as:

Now, integrating twice, dividing by the flexural rigidity, EI, and subbing in x = L/2, gives an expression for the maximum deflection at midspan:

[1]

where:

w = 8 kN/m = 8 N/mm

L = 6000 mm

E = 13.1 GPa = 13.1 103 MPa

Equation 1, above, can be rearranged to determine the required width of cross-section by substituting in the above values:

h <

300

mm

b

8 kN/m

6 m

x

w = 8kN/m

Problem (b):

Later in the life of the bridge, the average weight of vehicles crossing the bridge has increased and the loads are estimated to be greater by 10% than those considered in the bridge’s original design. It is decided that the bridge should be strengthened by bonding carbon FRP plates to the bottom faces of the timber beams. Determine the

number of carbon FRP plates required to maintain the same deflection limit of L/240 under the increased loads. The properties of the carbon FRP plates are as follows:

Elastic modulus, Efrp = 165 GPa = 165 103 MPa Thickness, t = 1.2 mm Width, w = 50 mm

Solution (b):

The load has increased 10%, thus:

With a single layer of FRP applied to the bottom face of the timber beam, the beam would look as illustrated below:

We will transform the timber section into FRP. The transform factor, n, can be calculated as follows:

Hence, the equivalent transformed width of the wood, bwt, can be calculated as:

Since each FRP strip is 50 mm wide, and because we cannot stack the FRP strips, we will try using 3 strips

bonded to the bottom face of the beam. Thus, the total width of the FRP is 150 mm. The neutral axis depth can be calculated as follows:

The transformed moment of inertia can now be determined:

The deflection of the beam with FRP bonded to its bottom face can be calculated as:

Therefore, since 24.6 mm < 25 mm, three side-by-side carbon FRP plates will allow the timber beam to resist the deflection caused by a 10% increase in load.

1.2 mm

3

00 m

m

183 mm

bfrp = ?

Example 18: Both steel and concrete structures are susceptible to the damaging effects of corrosion. This is particularly true for bridge decks, where the presence of de-icing salts on the road’s surface can drastically increase the risk of corrosion. Engineers are now looking to non-corrosive fibre reinforced polymers (FRPs) as alternative materials for bridge deck construction. Because FRP materials are extremely strong and lightweight, FRP bridge deck panels can also drastically reduce the dead weight of the deck, enhancing the overall efficiency of the structure and allowing increased live loads.

In one specific FRP bridge deck system, glass FRP sections are individually pultruded (a manufacturing process for FRP structural elements) and then bonded

together to form a rigid deck structure. This is shown schematically in the figure below.

If the FRP bridge deck panel is simply-supported, spans 5m (into the page), and the deflection is limited to L/360:

(a) Can the bridge carry a uniformly distributed load (UDL) of 52 kN/m2?

(b) What is the maximum deflection if a 1.4 mm thick layer of carbon FRP is added to the top and bottom faces of the panel? Is this satisfactory?

Note: Assume that the modulus of elasticity of glass FRP, EGFRP, is 75 GPa and that the elastic modulus of carbon FRP, ECFRP, is 300 GPa.

Solution (a):

The deflection at midspan of a simply-supported beam subjected to a UDL can be found by the following equation:

The only unknown quantity in this expression is the moment of inertia, I, which can be found as follows:

The deflection can now be calculated as follows:

The limiting deflection of this deck is L/360 = 13.9 mm. Since 14.9 mm > 13.9 mm, the glass FRP bridge deck cannot carry the 13 kN/m UDL.

Solution (b):

Section through a bridge deck panel

Two individual pultruded glass FRP sections

Bond between sections

250 mm

20 mm200

mm

If a 1.4 mm thick layer of carbon FRP is added to the top and bottom of the bridge deck panel, each individual section will look as shown below:

We will transform the carbon FRP into glass FRP. The modular ratio, n, can be found by:

Therefore, the equivalent width of the transformed CFRP, bCFRPt, is:

The moment of inertia of the transformed section is simply the moment of inertia for the glass FRP section without carbon, plus the contributions from the CFRP layers at the top and bottom of the section. This is calculated as follows:

The deflection can now be calculated:

Since L/360 = 13.89 mm > 11.4 mm, the addition of the carbon FRP to the top and bottom of the section gives the bridge deck panel the required flexural rigidity to adhere to the deflection limit.

1.4 mm

250 mm

20 mm200

mm

Example 19: Due to decades of neglect and overuse, increased traffic loads, and environmental degradation, many timber bridges in Canada are in need of rehabilitation and strengthening. In some cases, these timber structures have historic value, and so it is important to maintain the appearance of the structure for aesthetic reasons. One strengthening technique for these historically-sensitive timber structures is called near surface mounting (NSM) of fibre-reinforced polymer (FRP) materials. In this application, small grooves are cut into the bottom face of the timber bridge girders and FRP plates or bars are bonded inside the grooves using an epoxy adhesive. This increases the flexural rigidity and strength of the timber beams.

For a specific timber bridge beam, a maximum allowable bending stress of 10 MPa is permitted in the wood. If the cross-sectional properties of the beam are as shown below:

(a) What is the maximum bending moment the existing bridge girders are capable of resisting?

(b) What is this maximum allowable bending moment if the girders are strengthened using the NSM technique, as shown, with two carbon FRP plates (60 mm 1.4 mm)?

Assume that the modulus of elasticity of the wood, Ew, is 13.1 GPa, that the elastic modulus of the carbon FRP, Efrp, is 210 GPa, and that the allowable stress in the FRP is 1500 MPa.

Solution (a):

The maximum allowable moment can be obtained by rearranging the flexure formula:

For the original (unstrengthened) beam, the moment of inertia for the unstrengthened beam, Ii, can be calculated as:

The maximum moment the beam can resist, such that the stress in the wood does not exceed 10 MPa, is thus:

Therefore, the original timber bridge beams can each carry a moment of ±22.5 kNm.

iy

150 mm

300

mm

Originalbeam

150 mm

60 mm

150 mm

Beam with grooves

Beam with NSMFRP strips

NANA

y

y

Solution (b):

To solve the second part of this question, we must determine the transformed section moment of inertia for the NSM strengthened beam. The modular ratio, n, to transform the FRP plates into wood, can be found by:

The equivalent width of wood for the carbon FRP plates, bfrpt, is:

And so the transformed section looks as follows:

The neutral axis for the transformed section can be found by the following:

Now, the moment of inertia for the transformed section can be determined as:

And again using the rearranged version of the flexure formula we find the moment that the NSM-strengthened timber beams can carry, again ensuring that the stress in the wood remains less than 10 MPa:

Therefore, the maximum moment that can be carried by the NSM strengthened timber bridge beams is 23.8 kNm. This represents an increase of about 8%.

Note: strictly speaking, we should also check that the stress in the FRP does not exceed 1500 MPa.

192.1 mm

60 mm

NA

y

240 mm

150 mm