Fiber Reinforced Concrete

     Fiber reinforced concrete (FRC) is concrete containing fibrous material which increases its structural integrity.
Factors controlling performance of composites
•  Physical properties of reinforcement and matrix
•  Strength of bond between reinforcement and matrix

Characteristics of fiber reinforced systems
•  Fibers distributed through given cross section, reinforcement bars only placed where required
•  Fibers are short and closely spaced, reinforcing bars are continuous
•  Small reinforcement ratio when compared to reinforcement bars

Reinforcement Ratio =Area of Reinforcement/Area of Concrete

Advantages
•  Easily placed
         –Cast
         –Sprayed
         –Less labor intensive than placing rebar
•  Can be made into thin sheets or irregular shapes
•  Used when placing rebar is difficult
Disadvantages
• Efficiency factors as low as .4 2-D(spray placement method), or .25 3-D placement (casting method)
• Not highly effective in improving compressive strength
Types of Natural Fibers
• Sisal (Castro, 1981)
– From AgaveSisalana in Mexico
– Durability problems caused by chemical decomposition
in alkaline environment
• Coir (Balaguru, 1985)
– Coconut husks
– Very durable to natural weathering
– Increases modulus of rupture of concrete (MOR)
• Bamboo (Ghavami 2005; Rodrigues, 2006)
– E is very similar to that of concrete
– Susceptible to volume changes in water
– Increases ultimate tensile strength and MOR
• Jute (Balaguru, 1985)
– Grow in India, Bangladesh, China, and Thailand
– Increases tensile, flexural and compressive strengths,
as well as flexural toughness
• Akwara (Balaguru, 1985)
– Abundant in Nigeria
– No dimensional changes due to variations in water
– Alkali resistant
– No changes in flexural or compressive strengths
– Impact strength 5 to 16 times greater than unreinforced
cement matrix
• Elephant Grass (Balaguru, 1985)
– Very durable – good rot and alkali resistance as well as
small dimensional changes
– Increases flexural and impact strength
History of Natural Fiber Reinforcement
• Egyptians used straw in making mud bricks 1200-1400BC (Exodus 5:6)
• 2500 BC asbestos fibers used in Finland to make clay pots (Active Asbestos )
• Hornero bird native to South America builds nests out of straw and clay (Mehta, 2006)
• Replacement for asbestos Current Uses
• Fiber Cement Board
     – Siding
     – Backer board
     – Roofing materials
     – Non pressure pipe
• Buckeye Technologies
     – UltraFiber 500
     – Slab on grade concrete
     – Precast concrete
     – Decorative concrete
     – 85% crack reduction
     – Improved hydration
     – Improved freeze/thaw resistance

Wood Pulp Fiber Composition
• Cellulose –(C₆H₁₂O₅)_n
– n degree of polymerization
– 600-1500 for commercial wood pulps
– Determines the character of the fiber
– As cellulose increases fiber tensile strength and E increase linearly
• Hemicelluloses – polysaccharides of five different sugars
– Easier to degrade than cellulose
– Highly variable with fiber type
• Lignin – complex polymer
composition
– Binds wood together
– Found in the middle lamella
– Used in concrete as a set retarder
• Extractives
– No physical structure
– Give properties such as color, odor, taste
– Some can be incompatible with concrete
 

Fresh Properties
• Workability
• Setting time
• Cement hydration
• Fiber clumping/consolidation
• Shrinkage
– Plastic
– Free
– Drying
• Internal curing and autogenous shrinkage
Durability Improvement
• Pressure treatment on fiber cement board
• Reduction in w/c ratio to decrease porosity
• Addition of SCMs eliminated degradation
due to wet/dry cycles (Mohr, 2005)
– 30%, 50% Silica Fume
– 90% Slag
– 30% Metakaolin 235
– 10% SF / 70% SL
– 10% MK235 / 70% SL
– 10% MK235 / 10% SF / 70% SL
• Chemically coated fibers
Conclusions
• Natural fibers offer many benefits for
reinforcement
– Low cost and abundant
– Renewable
– Non hazardous – replacement of asbestos
• Can improve characteristics of concrete
– Increase flexural strength and toughness
– Increase impact resistance
– Reduce shrinkage and cracking
– Improve durability by stabilization of
microcracks and decrease in permeability
Future Research
• Sources of pulp fibers
– Thermomechanical fibers
– Paper mill residual solids
• Optimal fiber ratios for specific uses
• Durability
– Additional research on freeze thaw
– Wet/dry cycle effects
• SCM addition

Strain Energy

     Consider an elastic spring as shown in the Fig.1.0 When the spring is slowly pulled, it deflects by a small amount u₁. When the load is removed from the spring, it goes back to the original position. When the spring is pulled by a force, it does some work and this can be calculated once the load-displacement relationship is known. It may be noted that, the spring is a mathematical idealization of the rod being pulled by a force P axially. It is assumed here that the force is applied gradually so that it slowly increases from zero to a maximum value P. Such a load is called static loading, as there are no inertial effects due to motion. Let the load-displacement relationship be as shown in Fig. 2.0. Now, work done by the external force may be calculated as,

 

      The area enclosed by force-displacement curve gives the total work done by the externally applied load. Here it is assumed that the energy is conserved i.e. the work done by gradually applied loads is equal to energy stored in the structure. This internal energy is known as strain energy. Now strain energy stored in a spring is,

Work and energy are expressed in the same units. In SI system, the unit of work and energy is the joule (J), which is equal to one Newton metre (N.m). The strain energy may also be defined as the internal work done by the stress resultants in moving through the corresponding deformations. Consider an infinitesimal element within a three dimensional homogeneous and isotropic material. In the most general case, the state of stress acting on such an element may be as shown in Fig. 3.0. There are normal stresses (σ₁,σ₂ and σ₃), and and shear stresses (τ₁,τ₂ and τ₃), and acting on the element. Corresponding to normal and shear stresses we have normal and shear strains. Now strain energy may be written as,

in which σ^T is the transpose of the stress column vector i.e.,

    The strain energy may be further classified as elastic strain energy and inelastic strain energy as shown in Fig. 4.0 If the force P is removed then the spring shortens. When the elastic limit of the spring is not exceeded, then on removal of load, the spring regains its original shape. If the elastic limit of the material is exceeded, a permanent set will remain on removal of load. In the present case, load the spring beyond its elastic limit. Then we obtain the load-displacement curve OABCDO as shown in Fig. 4.0. Now if at B, the load is removed, the spring gradually shortens. However, a permanent set of OD is till retained. The shaded area BCD is known as the elastic strain energy. This can be recovered upon removing the load. The area OABDO represents the inelastic portion of strain energy.

     The area corresponds to strain energy stored in the structure. The area is defined as the complementary strain energy. For the linearly elastic structure it may be seen that OABCDOOABEO
Area OBC = Area OBE
i.e. Strain energy = Complementary strain energy
    This is not the case always as observed from Fig. 4.0. The complementary energy has no physical meaning. The definition is being used for its convenience in structural analysis as will be clear from the subsequent chapters.

     Usually structural member is subjected to any one or the combination of bending moment; shear force, axial force and twisting moment. The member resists these external actions by internal stresses. In this section, the internal stresses induced in the structure due to external forces and the associated displacements are calculated for different actions. Knowing internal stresses due to individual forces, one could calculate the resulting stress distribution due to combination of external forces by the method of superposition. After knowing internal stresses and deformations, one could easily evaluate strain energy stored in a simple beam due to axial, bending, shear and torsional deformations.

Strain energy can be described in 4 types of cases.

    1.Strain energy under axial load
    2.Strain energy due to bending
    3.Strain energy due to transverse shear
    4.Strain energy due to torsion

Strain energy due to torsion

Consider a circular shaft of length L radius R, subjected to a torque T at one end (see Fig. 1.0). Under the action of torque one end of the shaft rotates with respect to the fixed end by an angle dφ. Hence the strain energy stored in the shaft is,

Consider an elemental length ds of the shaft. Let the one end rotates by a small amount dφ  with respect to another end. Now the strain energy stored in the elemental length is,

We know that

where, G is the shear modulus of the shaft material and J is the polar moment of area. Substituting for  dφ from (3.0) in equation (2.0), we obtain

Now, the total strain energy stored in the beam may be obtained by integrating the above equation.

Strain energy due to transverse shear

The shearing stress on a cross section of beam of rectangular cross section may be found out by the relation

where is the first moment of the portion of the cross-sectional area above the point where shear stress is required about neutral axis, Vis the transverse shear force,b is the width of the rectangular cross-section and I_zz is the moment of inertia of the cross-sectional area about the neutral axis. Due to shear stress, the angle between the lines which are originally at right angle will change. The shear stress varies across the height in a parabolic manner in the case of a rectangular cross-section. Also, the shear stress distribution is different for different shape of the cross section. However, to simplify the computation shear stress is assumed to be uniform (which is strictly not correct) across the cross section. Consider a segment of length ds subjected to shear stress τ. The shear stress across the cross section may be taken as

in which A is area of the cross-section and k is the form factor which is dependent on the shape of the cross section. One could write, the deformation du as

where Δy is the shear strain and is given by

Hence, the total deformation of the beam due to the action of shear force is

Now the strain energy stored in the beam due to the action of transverse shear force is given by,

The strain energy due to transverse shear stress is very low compared to strain energy due to bending and hence is usually neglected. Thus the error induced in assuming a uniform shear stress across the cross section is very small.

Strain energy due to bending

    Consider a prismatic beam subjected to loads as shown in the Fig. 1.0. The loads are assumed to act on the beam in a plane containing the axis of symmetry of the cross section and the beam axis. It is assumed that the transverse cross sections (such as AB and CD), which are perpendicular to centroidal axis, remain plane and perpendicular to the centroidal axis of beam (as shown in Fig 1.0).

    Consider a small segment of beam of length ds subjected to bending moment as shown in the Fig. 1.0. Now one cross section rotates about another cross section by a small amount dθ. From the figure,

where R is the radius of curvature of the bent beam and EI is the flexural rigidity of the beam. Now the work done by the moment M while rotating through angle dθ will be stored in the segment of beam as strain energy dU. Hence, 

Substituting for dθ in equation (2.0), we get,

Now, the energy stored in the complete beam of span L may be obtained by integrating equation (3.0). Thus,

 

Strain energy under axial load

     Consider a member of constant cross sectional area A, subjected to axial force Pas shown in Fig. 2.8. Let E be the Young’s modulus of the material. Let the member be under equilibrium under the action of this force, which is applied through the centroid of the cross section. Now, the applied force P is resisted by uniformly distributed internal stresses given by average stress σ =P/A as shown by the free body diagram (vide Fig. 2.8). Under the action of axial load P applied at one end gradually, the beam gets elongated by (say) . This may be calculated as follows. The incremental elongation of du small element of length  dx of beam is given by,

Now the work done by external loads   W= 1/2XPu                                    (3.0)
  

    In a conservative system, the external work is stored as the internal strain energy. Hence, the strain energy stored in the bar in axial deformation is,

Substituting equation (2.0) in (4.0) we get,

Rock Fibre Slabs

Description
    Rock Fibre Slabs are manufactured from non-combustible rock fibre bonded with a high performance binder.
    They are designed for the thermal and acoustic insulation of flat and curved surfaces.
    Rock Fibre Slabs are available in semi-rigid and rigid slabs and in a range of densities to suit almost any thermal or acoustic insulation application. They can be supplied cut to size and shape, and can be faced or totally enclosed in a wide range of materials including aluminium foil, tissue and glass cloth.

Colour - Greyish Brown
Application
   Rock Fibre Slabs are used extensively for thermal and acoustic insulation in domestic, commercial and industrial buildings; in original equipment such as cookers, boilers, tanks, vessels, generators,industrial plant and equipment, ceiling tiles and marine, road and rail transport.
Operating Temperature
   Rock Fibre Slabs can be used at continuous operating temperatures up to 230ºC and up to 850ºC in selected applications. Further details available on request.
Fire Performance
   Rock Fibre Slabs achieve a Euroclass A1 fire rating (non-combustible) when tested in accordance with BS EN 13501-1.
They are non-combustible, when tested in accordance with BS476: Part 4: 1970 (1984).
They comply with the Class 'O' requirements of the Building Regulations, when tested to BS476: Part 6: 1981 and Part 7: 1987.

Thermal Conductivity

Acoustic Performance
   Typical test data is given below for comparative purposes. The letter (S)denotes results obtained with a solid backing and the letter (C) denotes results obtained with a 300mmcavity behind the insulation.

†NRC (Noise Reduction Coefficient) is the average coefficient over the frequency range250-2000Hz inclusive.

Dimensions and Density
Rock Fibre Slabs can be die-cut to size and shape.
Note: Other slab sizes, density and thicknesses may be available subject to minimum order quantities. Further details available on request.

Permanence
    Rock Fibre Slabs are rot-proof, odourless, non-hygroscopic, will not sustain vermin and will not encourage the growth of fungi, mould or bacteria. They do not settle under vibration, when tested in accordance with BS2972: 1975 and are dimensionally stable under varying conditions of temperature and humidity.

PRINCIPLE OF SUPERPOSITION

        The principle of superposition is a central concept in the analysis of structures. This is applicable when there exists a linear relationship between external forces and corresponding structural displacements. The principle of superposition may be stated as the deflection at a given point in a structure produced by several loads acting simultaneously on the structure can be found by superposing deflections at the same point produced by loads acting individually. This is illustrated with the help of a simple beam problem. Now consider a cantilever beam of length L and having constant flexural rigidity EI subjected to two externally applied forces P₁ and P₂ as shown in Fig.1. From moment-area theorem we can evaluate deflection below , which states that the tangential deviation of point from the tangent at point A is equal to the first moment of the area of the M/EI diagram between A and C about . Hence, the deflection below due to loads P₁ and P₂ acting simultaneously is (by moment-area theorem),

u=A₁x₁ +A₂x₂ +A₃x₃

where u is the tangential deviation of point C with respect to a tangent at A. Since, in this case the tangent at A is horizontal, the tangential deviation of point C is nothing but the vertical deflection at C.  x₁,x₂ and x₃are the distances from point C to the cancroids of respective areas respectively.

Hence

After simplification one can write,

Now consider the forces being applied separately and evaluate deflection at in each of the case.

 

where u₂₂ is deflection at C(2) when load P₁ is applied at (2) itself. And,

where u₂₁ is the deflection at C(2) when load is applied at  B(1). Now the total deflection at C when both the loads are applied simultaneously is obtained by adding u₂₁ and u₂₂

Hence it is seen from equations (2.3) and (2.6) that when the structure behaves linearly, the total deflection caused by forces P₁,P₂,P₃,....P_n at any point in the structure is the sum of deflection caused by forces acting P₁,P₂,P₃,....P_n independently on the structure at the same point. This is known as the Principle of Superposition.
The method of superposition is not valid when the material stress-strain relationship is non-linear. Also, it is not valid in cases where the geometry of structure changes on application of load. For example, consider a hinged-hinged beam-column subjected to only compressive force as shown in Fig. 2.3(a). Let the compressive force P be less than the Euler’s buckling load of the structure. Then deflection at an arbitrary point C u_c¹  is zero. Next, the same beam-column be subjected to lateral load Q with no axial load as shown in Fig. 2.3(b). Let the deflection of the beam-column at C u_c² be . Now consider the case when the same beam-column is subjected to both axial load and lateral load. As per the principle of superposition, the deflection at the centre u_c³ must be the sum of deflections caused by P and Q when applied individually. However this is not so in the present case. Because of lateral deflection caused by Q, there will be additional bending moment due to at C.Hence, the net deflection u_c³ will be more than the sum of deflections u_c¹ and u_c².