Behavior of Fibrous Reinforced Concrete Splices

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RESEARCH ARTICLE

Behavior of Fibrous Reinforced Concrete Splices

The Open Civil Engineering Journal 31 Dec 2021 RESEARCH ARTICLE DOI: 10.2174/1874149502115010441

Abstract

Background:

The tangent of the relationship between bond stress and displacement (slip) is called the modulus of displacement and gives the basis for the theory. This theory is used to determine the stress distribution along the spliced reinforcement bars.

Objective:

This research presents a modification on the theory of the modulus of displacement to determine the stress distribution along the spliced reinforcement bond for fibrous reinforced concrete.

Methods:

1- General differential equations are derived for concrete stress, stress in reinforcement bars and bond stress between reinforcement bars and surrounding concrete.

2-The general solutions of these D.E. are determined and Excel data sheets are prepared to apply these solutions and determine the concrete, steel and bond stresses.

Results:

Excel data sheets are prepared to determine the concrete, steel and bond stresses. The stresses are determined along the bar splice length considering the effect of steel fiber content.

Conclusion:

The maximum concrete stress is obtained at center x=0 and minimum at . Maximum bond stress obtained at and minimum at the center. The maximum steel stress at and minimum at . The value of (σcmax) increased linearly with increasing of (ρ). The concrete stress increased nonlinearly with (ρ%) and linearly with ( fy) and (fc’). Also increasing of (k) and bar diameter have small effects. The value of bond stress decreased linearly with (Qf) and (ρ%).

Keywords: Bond stress, Fibrous concrete, Concrete, Splices, Volume fraction, Modulus of displacement.

1. INTRODUCTION

Modulus of displacement theory is used to determine the stress distribution along the spliced reinforcement bars. The tangent of the relationship between bond stress and displacement (slip) is called the modulus of displacement and gives the basis for the theory. Fig. (1) shows the relation between bond stress and displacement (slip).

Losberg investigated the bond properties of long concrete specimens with pulled, axially embedded reinforcement bars [1]. He used the modulus of displacement theory for the calculation of the stresses and the crack widths. Also, he extended the investigations to the curtailment of reinforcement in accordance with the moment diagram, and the influence of cracks was also studied.

Feldman and Bartlett showed that bond stress magnitude varies along the length of plain reinforcing bars in pullout specimens by using analytical methods and experimental tests [2]. They established analytical relationships between bond stress and slip at the unloaded end of the bar and along the length of the bar. The analytical relationship showed that the bond stress is a function of bar slip, slip is a function of bar force, and bar force is a function of bond stress. Also, they concluded that the maximum load occurs just before slip occurs at the unloaded end of the bar, and the location of the peak bond stress shifts from the loaded end towards the unloaded end of the specimen with increasing applied load.

Thompson et al. studied the anchorage behavior of headed reinforcement in lap splices by experimental tests [3]. They concluded that the stress is transferred between opposing bars in non-contact lap splices through struts acting at an angle to the direction of the bar. Anchorage length in non-contact lap splices could be determined by drawing the struts between opposing bars propagate at an angle of 55° with respect to the bar axis.

Coogler et al. tested two commercially available offset mechanical splice systems in direct tension with the splice both restrained and unrestrained from rotation [4]. They concluded that pullout failure was the most common failure mode observed. This mode of failure results in a decrease in apparent ultimate stress for the system because of the inability to develop the full strength of the cross-section. A 345 MPa stress range for fatigue testing results in fatigue-induced reinforcing bar rupture at a very low number of cycles. A more reasonable stress range of 138 MPa is suggested for assessing the performance of this type of splice. Also, for all in-place testing, concrete was unable to properly confine the offset splice near ultimate load levels.

Feldman and Bartlett concluded that there is a complicated interaction between flexural and shear behavior, band stress, and cracking two full-scale T-beams, 4.2m long with a/d=7.5 were tested in four-point loading [5]. Both were reinforced with plain hollow steel bars that had roughened surfaces to simulate bars found in historic structures. The flexural reinforcement ratios were 0.33% for the HSS bar and 0.98% for plane bars. Arch action initiated in the beam reinforced with plane bars due to loss of bond in the constant shear region near midspan when the applied load reached 60% of its maximum value. The beam reinforced with the HSS bars had lesser bond demand, and arch action due to bond loss did not initiate in the beam until the maximum load was achieved. Also, when shear is carried by beam action, bond demand is greatest within the elastic-uncracked region adjacent to the first flexural crack.

Yang and Ashour developed a mechanical analysis based on an upper-bound theorem to predict the optimum failure surface and concrete breakout capacity of single anchors under tensile loads [6]. The predicted results obtained from ACI 318-05 are compared with the experimental test results. They concluded that the shape of the failure surface predicted by the mechanism analysis is significantly influenced by the ratio of effective tensile and compressive strengths of concrete. The concrete breakout capacity of anchors predicted from mechanism analysis responds sensitively to the variation of the ratio between effective tensile and compressive strengths of concrete. Conservation of ACI 318-05 sharply increases in specimens having concrete strength above 50 MPa, whereas the mechanism analysis shows good agreement with tests results, regardless of concrete strength.

Sezen and Setzler showed that the contribution of bar slip deformations to total member lateral displacement could be significant [7]. In addition to flexural deformations, bar slip deformations should be considered in the modeling and analysis of reinforced concrete members. They present a method for computing slip for bars stressed, its unloaded end, and hooked bars. The proposed model is compared with five models found in the literature and three independent sets of experimental data. Also, the proposed model is used to calculate the lateral load-slip displacement relations for seven columns from two different studies, and the computed relations compare well with the measured test results.

Howell and Higgins studied the bond performance of square and round deformed reinforcing bars [8]. They concluded that the application of the simplified ACI development length equations to characterize the reinforcing bar stress provided a reasonable lower bound for both square and round bars across all test types. Also, the ACI approach was conservative for partial reinforcing bar embedment of round and square results and indicated that linear interpolation of available reinforcing bar stress for embedment lengths less than the computed development length also appears reasonable for the square reinforcing bar. Computation of development length using the ACI formula with an equivalent round diameter for square reinforcing bar results in lengths (13%) larger than when the side dimension is used. This is conservative and available for large square reinforcing bar sizes and alternative test configurations.

Eligehausen et al. proposed a model to predict the average failure load of anchorage using adhesive banded anchors based on numerical and experimental investigations [9]. The model is similarly cast in place, and post-installed mechanical anchors are incorporated in ACI 318-05, but with the following modifications:

(1) The basic strength of a single adhesive anchor predicts the pullout capacity and not the concrete breakout capacity.

(2) The critical spacing and critical edge distance of adhesive anchorages depend on the anchor diameter and the bond strength and not on the anchor embedment depth.

(3) The proposed model results agree very well with the results of (415) group tests contained in a worldwide database.

This research presents a modification on the theory of the modulus of displacement that has been presented by Tepfors [10] to determine the stress distribution along the spliced reinforcement bond for fibrous reinforced concrete.

2. MATERIALS AND METHODS

2.1. Theoretical Analysis

2.1.1. Basic Equations

  • 1- The splice area is located in the region of the constant moment and has no shear.
  • 2- The stress in the reinforcement at both ends of the splice is equal.
  • 3- Moment cracks in the concrete are located at the ends of the splice, and the stress in the reinforcement is (σs). Fig. (2) shows the tension reinforcement splice.
Fig. (1). Bond stress-slip relationship.
Fig. (2). Tension reinforcement splice.

From Fig. (1); the bond stress between reinforcement and concrete:

(1)

Where τ= bond stress. (MPa) N/mm2

k = modulus of displacement (N/mm3)

= displacement or slip (mm)

The connection between bond stress (τ) and normal stress of concrete (σc) and steel stress (σs) shown in Fig. (3):

Fig. (3). Equilibrium of splice bar.

The horizontal equilibrium of splice bar at gives the following equations:

(2)

Also by the same way:

(3)

Where: d = diameter of the bar (mm).

As = Cross-sectional area of the bar (mm2)

u = perimeter of the bar (mm)

τ1 = band stress at x=- (MPa)

σs1 = tension steel stress at (MPa)

τ2 = band stress at x= (MPa)

σs2 = tension stress at x=- (mpa)

Fig. (4). Equilibrium of steel concrete connection within the splice.

Concrete stress is determined from the horizontal equilibrium of the concrete steel connection shown in Fig. (4);

(4)

Where σc = tensile stress in the effective concrete (MPa).

Ac = effective area of the concrete around the steel (mm2).

The equilibrium condition for the splice is shown in Fig. (5)

(5)

Where: fs = tension stress of the steel reinforcement (MPa).

Fig. (5). Equilibrium of the splices.

From Eq.(1); derivate on both sides :

(6)

The strain is the difference in displacement between the two materials, concrete, and steel.

(7)

The change in shear stress for element length (dx) due to the displacement between the reinforcement bar and the concrete for both parts of the splice is:

(8)

Or

(9)

And

(10)

Or

(11)

Take the derivative of both sides of Eq. (4);

(12)

Substitute Eq. (9 and 11) into Eq. (12) to get;

(13)
(14)

From Eq. (5);

(15)

the bar diameter is taken the same at both side of the splice, thus:

Subacute Eq. (15) into Eq. (14);

(16)

taking ρ = reinforcement index

and n = modular ratio =

(17)
(18)

but

re-arrange Eq. (18) to get:

(19)

Let

(20)

The general solution of Eq. (20) is:

(21)

The constants A and B are determined by application of the boundary conditions:

at for fibrous concrete

σc=ο for concrete without fibers.

The tensile strength of plain concrete with steel fiber can be calculated from the following equations (11, 12).

σfu =0.82τF

where: τ = Interfacial bond strength between steel fiber and concrete matrix.

f = Fiber factor (13, 14) =

Q f = volume fraction of steel fiber (%).

d f = Bond factor depends on the type of steel fiber.

L f = Length of steel fiber (mm).

D f = Diameter of steel fiber (mm).

These give the constants as:

(22)
(23)

The final form of the concrete normal stress is:

(24)

for concrete without steel fiber: σ fu=0;

(25)

The normal stress (σs1) is determined by derivation of Eq. (3) and combined with Eq. (9) to get:

(26)
(27)

rearrangement of Eq. (27);

(28)

Let

(29)

The homogeneous solution of Eq. (29) is:

(30)

because of σc is also a function of (x), thus the particular solution is taken as the following:

(31)

Substitute σs1ρ & its 2nd derivative into eq. (29) to find the constants C, D, & E.

(32)

substitute Eqs. (31 and 32) into Eq. (29);

C=0;

finally:

or

(33)

for concrete without steel fiber; σ fu=0;

(34)

Applying B.C.:

and x=- ; σs1 = σs

(35)
(36)

adding the above two equations:

(37)

for σ fu=0; A=0

substitute in Eq. (35);

(38)

Finally:

or

(39)

Check: @

By the same way. The derivation of Eq. (2) is combined with Eq. (11) to get;

(40)

By the same previous procedures, the general solution is:

(41)

The B.C. are:

@x=- ; σs2 = σs

@x=- ;σs2 = 0

(42)
(43)

adding the above two equations:

(44)

for σ fu=0; A=0

sub. in Eq. (42);

(45)

Finally:

(46)

Check: @

σs2=σs

@

The bond stresses (τ1 & τ2) at x= & x=- are determined from Eqs. (2 & 3):

(47)
(48)

from Eq. (39);

(49)

from Eq. (46);

(50)

For the special case:

when ρ = ∞ ; at x = ∓ ; concrete cracks and Ac=0, then ρ=

from Eq. (24);

(51)

for concrete without fiber, σ fu=0;

from Eq. (39);

(52)

and from Eq. (49);

(53)
(54)

from Eq. (50);

(55)

3. RESULTS AND DISCUSSION

Effect of steel fiber content is considered on the bond stress between steel bars and surrounding fibrous concrete and concrete stress at the lap splice, taking into account the following variables:

  • 1- Steel fiber content (Qf%).
  • 2- Value of modulus of displacement k N/mm3.
  • 3- Reinforcement index (ρ%).
  • 4- Bar diameter (db) mm.
  • 5- Steel bar yield strength (fy) N/mm2.
  • 6- Compressive strength of concrete  fc' N/mm2.

Excel data sheets are prepared to apply equations of concrete stress (σc), steel stress (σs), and bond stress (τ). The value of these stresses is determined along the bar splice length considering the variables mentioned above.

Table 1.
Sample data sheet for calculating concrete stress (σc).
K(N/mm^3) = 50 100 150 50 100 150 50 100 150 50 100 150
L (mm) = 500 500 500 500 500 500 500 500 500 500 500 500
Fy(N/mm^2)= 420 420 420 420 420 420 420 420 420 420 420 420
Rho % = 1 1 1 1 1 1 1 1 1 1 1 1
Qf (fiber) %= 0 0 0 0.5 0.5 0.5 1 1 1 2 2 2
L/D Fiber = 100 100 100 100 100 100 100 100 100 100 100 100
df fiber = 1 1 1 1 1 1 1 1 1 1 1 1
fc'(N/mm^2)= 28 28 28 28 28 28 28 28 28 28 28 28
Es(N/mm^2)= 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000
bar diameter= 10 10 10 10 10 10 10 10 10 10 10 10
F= 0 0 0 0.5 0.5 0.5 1 1 1 2 2 2
Sigma fu= 0 0 0 1.7015 1.7015 1.7015 3.403 3.403 3.403 6.806 6.806 6.806
u= 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159
As= 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975
Ec= 25034.099 25034.099 25034.099 25034.099 25034.099 25034.099 25034.1 25034.1 25034.099 25034.099 25034.1 25034.099
n= 7.9891032 7.9891032 7.9891032 7.9891032 7.9891032 7.9891032 7.989103 7.989103 7.9891032 7.9891032 7.989103 7.9891032
K2^2= 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030
K1^2= 0.000116 0.000232 0.0003479 0.000116 0.000232 0.0003479 0.000116 0.000232 0.0003479 0.000116 0.000232 0.0003479
K1*L/2= 2.6923295 3.8075288 4.6632514 2.6923295 3.8075288 4.6632514 2.692329 3.807529 4.6632514 2.6923295 3.807529 4.6632514
K2*L/2= 2.5 3.5355339 4.330127 2.5 3.5355339 4.330127 2.5 3.535534 4.330127 2.5 3.535534 4.330127
Fy*Rho/(1+2nRho)= 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137
Sfu - above = -3.62137 -3.62137 -3.62137 -1.91987 -1.91987 -1.91987 -0.21837 -0.21837 -0.21837 3.18463 3.18463 3.18463
K2= 0.01 0.0141421 0.0173205 0.01 0.0141421 0.0173205 0.01 0.014142 0.0173205 0.01 0.014142 0.0173205
K1= 0.0107693 0.0152301 0.018653 0.0107693 0.0152301 0.018653 0.010769 0.01523 0.018653 0.0107693 0.01523 0.018653
Cosh(K1L/2)= 7.4168779 22.530603 52.994771 7.4168779 22.530603 52.994771 7.416878 22.5306 52.994771 7.4168779 22.5306 52.994771
Cosh(K2L/2)= 6.1322895 17.171237 37.98355 6.1322895 17.171237 37.98355 6.132289 17.17124 37.98355 6.1322895 17.17124 37.98355
- - - - - - - - - - - - -
X Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete) Sc(Concrete)
-250 0 0 0 1.7015 1.7015 1.7015 3.403 3.403 3.403 6.806 6.806 6.806
-200 1.4891048 1.9273371 2.1956344 2.4909492 2.7232781 2.865516 3.492794 3.519219 3.5353976 5.4964824 5.111101 4.8751608
-150 2.3448858 2.823921 3.0585701 2.9446417 3.1986023 3.3230015 3.544398 3.573284 3.587433 4.7439093 4.322646 4.1162959
-100 2.8215263 3.2352882 3.3954314 3.1973327 3.4166887 3.5015886 3.573139 3.598089 3.6077458 4.3247519 3.96089 3.8200602
-50 3.0605979 3.4117395 3.5210979 3.3240766 3.5102344 3.5682107 3.587555 3.608729 3.6153236 4.1145126 3.805719 3.7095492
0 3.1331093 3.4606389 3.5530355 3.3625185 3.5361584 3.5851425 3.591928 3.611678 3.6172494 4.0507461 3.762717 3.6814633
50 3.0605979 3.4117395 3.5210979 3.3240766 3.5102344 3.5682107 3.587555 3.608729 3.6153236 4.1145126 3.805719 3.7095492
100 2.8215263 3.2352882 3.3954314 3.1973327 3.4166887 3.5015886 3.573139 3.598089 3.6077458 4.3247519 3.96089 3.8200602
150 2.3448858 2.823921 3.0585701 2.9446417 3.1986023 3.3230015 3.544398 3.573284 3.587433 4.7439093 4.322646 4.1162959
200 1.4891048 1.9273371 2.1956344 2.4909492 2.7232781 2.865516 3.492794 3.519219 3.5353976 5.4964824 5.111101 4.8751608
250 0 0 0 1.7015 1.7015 1.7015 3.403 3.403 3.403 6.806 6.806 6.806
Table 2.
Sample data sheet for calculating bond stress (τ).
K(N/mm^3) = 50 100 150 50 100 150 50 100 150 50 100 150
L (mm) = 500 500 500 500 500 500 500 500 500 500 500 500
Fy(N/mm^2)= 420 420 420 420 420 420 420 420 420 420 420 420
Rho % = 1 1 1 1 1 1 1 1 1 1 1 1
Qf (fiber) %= 0 0 0 0.5 0.5 0.5 1 1 1 2 2 2
L/D Fiber = 100 100 100 100 100 100 100 100 100 100 100 100
df fiber = 1 1 1 1 1 1 1 1 1 1 1 1
fc'(N/mm^2)= 28 28 28 28 28 28 28 28 28 28 28 28
Es(N/mm^2)= 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000
bar diameter= 10 10 10 10 10 10 10 10 10 10 10 10
F= 0 0 0 0.5 0.5 0.5 1 1 1 2 2 2
Sigma fu= 0 0 0 1.7015 1.7015 1.7015 3.403 3.403 3.403 6.806 6.806 6.806
u= 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159
As= 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975
As/u= 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5 2.5
Ec= 25034.09891 25034.099 25034.099 25034.099 25034.099 25034.099 25034.1 25034.1 25034.099 25034.099 25034.1 25034.099
n= 7.989103213 7.9891032 7.9891032 7.9891032 7.9891032 7.9891032 7.989103 7.989103 7.9891032 7.9891032 7.989103 7.9891032
K2^2= 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030
K1^2= 0.000115978 0.000232 0.0003479 0.000116 0.000232 0.0003479 0.000116 0.000232 0.0003479 0.000116 0.000232 0.0003479
K1*L/2= 2.692329456 3.8075288 4.6632514 2.6923295 3.8075288 4.6632514 2.692329 3.807529 4.6632514 2.6923295 3.807529 4.6632514
K2*L/2= 2.5 3.5355339 4.330127 2.5 3.5355339 4.330127 2.5 3.535534 4.330127 2.5 3.535534 4.330127
Fy*Rho/(1+2nRho)= 3.621370022 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137
Sfu - above = -3.62137002 -3.62137 -3.62137 -1.91987 -1.91987 -1.91987 -0.21837 -0.21837 -0.21837 3.18463 3.18463 3.18463
K2= 0.01 0.0141421 0.0173205 0.01 0.0141421 0.0173205 0.01 0.014142 0.0173205 0.01 0.014142 0.0173205
K1= 0.010769318 0.0152301 0.018653 0.0107693 0.0152301 0.018653 0.010769 0.01523 0.018653 0.0107693 0.01523 0.018653
Cosh(K1L/2)= 7.416877858 22.530603 52.994771 7.4168779 22.530603 52.994771 7.416878 22.5306 52.994771 7.4168779 22.5306 52.994771
Cosh(K2L/2)= 6.13228948 17.171237 37.98355 6.1322895 17.171237 37.98355 6.132289 17.17124 37.98355 6.1322895 17.17124 37.98355
Sinh(K1L/2)= 7.34915486 22.5084 52.985336 7.3491549 22.5084 52.985336 7.349155 22.5084 52.985336 7.3491549 22.5084 52.985336
Sinh(K2L/2)= 6.050204481 17.142093 37.970384 6.0502045 17.142093 37.970384 6.050204 17.14209 37.970384 6.0502045 17.14209 37.970384
- - - - - - - - - - - - -
X Tau stress Tau stress Tau stress Tau stress Tau stress Tau stress Tau stress Tau stress Tau stress Tau stress Tau stress Tau stress
-250 10.15167604 14.324685 17.538596 9.9804964 14.091373 17.25462 9.809317 13.85806 16.970644 9.4669576 13.39144 16.402691
-200 6.058717725 6.8872504 7.1496826 6.0038161 6.8554463 7.1372815 5.948914 6.823642 7.1248804 5.8391111 6.760034 7.1000781
-150 3.628961482 3.3195417 2.9205115 3.6214913 3.3410165 2.9565485 3.614021 3.362491 2.9925856 3.5990808 3.405441 3.0646596
-100 2.19181798 1.6117052 1.2001241 2.1987152 1.6366736 1.2297106 2.205612 1.661642 1.2592971 2.219407 1.711579 1.3184701
-50 1.349760541 0.8021895 0.5061435 1.3560495 0.81626 0.5206444 1.362339 0.830331 0.5351453 1.3749166 0.858472 0.5641471
0 0.867739267 0.4331222 0.2394831 0.8677393 0.4331222 0.2394831 0.867739 0.433122 0.2394831 0.8677393 0.433122 0.2394831
50 0.607210115 0.2897912 0.1639454 0.6009211 0.2757206 0.1494445 0.594632 0.26165 0.1349436 0.5820541 0.233509 0.1059417
100 0.486165337 0.2751341 0.1958602 0.4792681 0.2501657 0.1662737 0.472371 0.225197 0.1366872 0.4585763 0.17526 0.0775142
150 0.453594908 0.3455462 0.3154513 0.4610651 0.3240713 0.2794143 0.468535 0.302596 0.2433772 0.4834756 0.259647 0.1713032
200 0.470492136 0.4662699 0.5087599 0.5253938 0.498074 0.521161 0.580295 0.529878 0.5335621 0.6900987 0.593486 0.5583643
250 0.490780712 0.5498024 0.6542437 0.6619603 0.7831149 0.9382198 0.83314 1.016427 1.2221959 1.1754991 1.483052 1.7901481
Table 3.
sample data sheet for calculating steel stress (σs).
K(N/mm^3) = 50 100 150 50 100 150 50 100 150 50 100 150
L (mm) = 500 500 500 500 500 500 500 500 500 500 500 500
Fy(N/mm^2)= 420 420 420 420 420 420 420 420 420 420 420 420
Rho % = 1 1 1 1 1 1 1 1 1 1 1 1
Qf (fiber) %= 0 0 0 0.5 0.5 0.5 1 1 1 2 2 2
L/D Fiber = 100 100 100 100 100 100 100 100 100 100 100 100
df fiber = 1 1 1 1 1 1 1 1 1 1 1 1
fc'(N/mm^2)= 28 28 28 28 28 28 28 28 28 28 28 28
Es(N/mm^2)= 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000 200000
bar diameter= 10 10 10 10 10 10 10 10 10 10 10 10
F= 0 0 0 0.5 0.5 0.5 1 1 1 2 2 2
Sigma fu= 0 0 0 1.7015 1.7015 1.7015 3.403 3.403 3.403 6.806 6.806 6.806
u= 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159 31.4159
As= 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975 78.53975
Ec= 25034.099 25034.099 25034.099 25034.099 25034.099 25034.099 25034.1 25034.1 25034.099 25034.099 25034.1 25034.099
n= 7.9891032 7.9891032 7.9891032 7.9891032 7.9891032 7.9891032 7.989103 7.989103 7.9891032 7.9891032 7.989103 7.9891032
K2^2= 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030 0.00010 0.00020 0.00030
K1^2= 0.000116 0.000232 0.0003479 0.000116 0.000232 0.0003479 0.000116 0.000232 0.0003479 0.000116 0.000232 0.0003479
K1*L/2= 2.6923295 3.8075288 4.6632514 2.6923295 3.8075288 4.6632514 2.692329 3.807529 4.6632514 2.6923295 3.807529 4.6632514
K2*L/2= 2.5 3.5355339 4.330127 2.5 3.5355339 4.330127 2.5 3.535534 4.330127 2.5 3.535534 4.330127
Fy*Rho/(1+2nRho)= 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137 3.62137
Sfu - above = -3.62137 -3.62137 -3.62137 -1.91987 -1.91987 -1.91987 -0.21837 -0.21837 -0.21837 3.18463 3.18463 3.18463
K2= 0.01 0.0141421 0.0173205 0.01 0.0141421 0.0173205 0.01 0.014142 0.0173205 0.01 0.014142 0.0173205
K1= 0.0107693 0.0152301 0.018653 0.0107693 0.0152301 0.018653 0.010769 0.01523 0.018653 0.0107693 0.01523 0.018653
Cosh(K1L/2)= 7.4168779 22.530603 52.994771 7.4168779 22.530603 52.994771 7.416878 22.5306 52.994771 7.4168779 22.5306 52.994771
Cosh(K2L/2)= 6.1322895 17.171237 37.98355 6.1322895 17.171237 37.98355 6.132289 17.17124 37.98355 6.1322895 17.17124 37.98355
Sinh(K1L/2)= 7.3491549 22.5084 52.985336 7.3491549 22.5084 52.985336 7.349155 22.5084 52.985336 7.3491549 22.5084 52.985336
Sinh(K2L/2)= 6.0502045 17.142093 37.970384 6.0502045 17.142093 37.970384 6.050204 17.14209 37.970384 6.0502045 17.14209 37.970384
- - - - - - - - - - - - -
- - - - - - - - - - - - -
- - - - - - - - - - - - -
X σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel) σ1(Steel)
-250 420 420 420 420 420 420 420 420 420 420 420 420
-200 261.43153 216.90355 188.47724 263.53332 219.16514 190.79626 265.6351 221.4267 193.11528 269.8387 225.9499 197.75332
-150 166.66209 119.16761 94.025607 169.30994 121.3961 95.936336 171.9578 123.6246 97.847064 177.25351 128.0816 101.66852
-100 109.71441 71.941216 55.369382 112.33169 73.663016 56.58955 114.949 75.38482 57.809717 120.18352 78.82842 60.250053
-50 75.0571 48.815601 39.356332 77.527042 50.136469 40.13422 79.99698 51.45734 40.912109 84.93687 54.09907 42.467885
0 53.344534 36.968057 32.348223 55.747359 38.146588 32.982661 58.15018 39.32512 33.617099 62.955835 41.68218 34.885976
50 38.883111 30.010446 28.533878 41.353054 31.331314 29.311766 43.823 32.65218 30.089654 48.762882 35.29392 31.64543
100 28.132954 24.529967 25.087475 30.750232 26.251767 26.307643 33.36751 27.97357 27.527811 38.602065 31.41717 29.968146
150 18.849334 18.440293 20.117388 21.49719 20.66879 22.028117 24.14505 22.89729 23.938846 29.440758 27.35428 27.760303
200 9.6579908 10.36274 11.959318 11.759785 12.62433 14.278338 13.86158 14.88592 16.597359 18.065167 19.4091 21.235401
250 -2.84E-14 -2.842E-14 0 0 0 0 0 0 0 5.684E-14 0 0

Tables (1-3) show samples of calculations. 45 datasheets are prepared for each (σc,σs & τ) to cover the effect of all variables.

Figs. (6 and 7) show the variation of concrete stress (σc) along the length of the bar splice (- to ) for concrete without and with steel fiber (Qf=1%). The maximum stress is obtained at center x=0 and minimum at (x = ∓ ).

Figs. (8 and 9) show the variation of bond stress (τ1) between the steel bar and surrounding concrete for concrete without steel fiber and with steel fiber content (Qf=1%). The maximum value is obtained at (x = ∓ ), and minimum value at (x = ).

Figs. (10 and 11) show the variation of steel stress (σs1) in the splice bar for both plain (Qf=0) and fibrous concrete (Qf=1%). The maximum value is (σs1=fy) at (x = - ) and minimum value is (σs1=0) at (x = ).

The value of τ2 is the same as (τ1) but in opposite sides, i.e., at (τ2=τ1). Also (σs2=fy) at (x= ) and equal to zero at (x = - ).

Figs. (12-15) show the effect of fiber content on the maximum concrete stress (σcmax) for steel bar content (ρ=1,5,10% and ). The value of (σcmax) increased linearly with the increase of (ρ), the slope of the lines decreased with the increase in the value of displacement modulus from (k) to , the effect is negligible at (k=150 N/mm3), the same behavior is noticed for (ρ=5&10%), but when (ρ= ), the effect of steel fiber content (Qf) and steel bar reinforcement index (ρ%) are neglected, and concrete stress becomes constant.

Fig. (16) shows the relation between reinforcement ratio (ρ) and maximum concrete stress (σcmax), for steel fiber content (Qf=0%), the concrete stress increased nonlinearly (parabolic) with increasing of (ρ%). Also, increasing value of (k) has a small effect on (σcmax). The same behavior is obtained for other steel fiber content, as shown in Figs. (18-19).

Figs. (20-23) show that the bar diameter has a negligible effect on maximum concrete stress for all steel fiber content (Qf) and steel reinforcement content (ρ).

Fig. (24) shows that the maximum concrete stress (σcmax) increased linearly with (Fy); this effect is increased when the value of (ρ) is increased. The same behavior is noticed in fibrous concrete, as shown in Figs. (25-27) .

Figs. (28-31) show that the maximum concrete stress (σcmax) increased linearly with concrete compressive strength (fc'); the effect is reduced when (ρ) increased from (1& to 5 & 10%) and became negligible at (ρ=).

Fig. (32) shows the relation of maximum bond stress (τmax) and steel fiber content (Qf) for (ρ=1%); value of (τmax) decreased linearly with increasing of (Qf) for all values of (k) and (ρ), as shown in Figs. (33-35).

Figs. (36-39) show that bond stress (τmax) decreased with increasing of (ρ%) for all values of (Qf & k). Also, using of larger bar diameter (db) causes a linear increase in bond stress (τmax) as shown in Figs. (40-43).

Figs. (44-47) show that value of (τmax) increased with increasing of (fy) for (ρ=1,5,10% and ) for plain concrete (Qf=0) and fibrous concrete (Qf=0.5,1&2%).

Figs. (48-51) show the effect of concrete compressive strength (fc1) on the maximum bond stress (τmax) for steel reinforcement content (ρ=1,5,10% and ). The effect is negligible at (ρ=).

Fig. (6). Distribution of concrete stress, Rho=1%, Qf=0%.
Fig. (7). Distribution of concrete stress, Rho=1%, Qf=1%.
Fig. (8). Distribution of bond stress, Rho=1%, Qf=0%.
Fig. (9). Distribution of bond stress, Rho=1%, Qf=1%.
Fig. (10). Distribution of steel stress, Rho=1%, Qf=0%.
Fig. (11). Distribution of steel stress, Rho=1%, Qf=1%.
Fig. (12). Relation of Qf versus concrete stress (Rho=1%).
Fig. (13). Relation of Qf versus concrete stress (Rho=5%).
Fig. (14). Relation of Qf versus concrete stress (Rho=10%).
Fig. (15). Relation of Qf versus concrete stress (Rho=Infinity).
Fig. (16). Relation of (Rho %) versus concrete stress (Qf=0%).
Fig. (17). Relation of (Rho %) versus concrete stress (Qf=0.5%).
Fig. (18). Relation of (Rho %) versus concrete stress (Qf=1%).
Fig. (19). Relation of (Rho %) versus concrete stress (Qf=2%).
Fig. (20). Relation of bar diameter versus concrete stress (Qf=0%).
Fig. (21). Relation of bar diameter versus concrete stress (Qf=0.5%).
Fig. (22). Relation of bar diameter versus concrete stress (Qf=1%).
Fig. (23). Relation of bar diameter versus concrete stress (Qf=2%).
Fig. (24). Relation of fy versus concrete stress (Qf=0%).
Fig. (25). Relation of fy versus concrete stress (Qf=0.5%).
Fig. (26). Relation of fy versus concrete stress (Qf=1%).
Fig. (27). Relation of fy versus concrete stress (Qf=2%).
Fig. (28). Relation of fc versus concrete stress (Rho=1%).
Fig. (29). Relation of fc versus concrete stress (Rho=5%).
Fig. (30). Relation of fc versus concrete stress (Rho=10%).
Fig. (31). Relation of fc versus concrete stress (Rho=Infinity).
Fig. (32). Relation of Qf versus concrete stress (Rho=1%).
Fig. (33). Relation of Qf versus concrete stress (Rho=5%).
Fig. (34). Relation of Qf versus concrete stress (Rho=10%).
Fig. (35). Relation of Qf versus concrete stress (Rho=Infinity).
Fig. (36). Relation of (Rho %) versus concrete stress (Qf=0%).
Fig. (37). Relation of (Rho %) versus concrete stress (Qf=0.5%).
Fig. (38). Relation of (Rho %) versus concrete stress (Qf=1%).
Fig. (39). Relation of (Rho %) versus concrete stress (Qf=2%).
Fig. (40). Relation of bar diameter versus concrete stress (Qf=0%).
Fig. (41). Relation of bar diameter versus concrete stress (Qf=0.5%).
Fig. (42). Relation of bar diameter versus concrete stress (Qf=1%).
Fig. (43). Relation of bar diameter versus concrete stress (Qf=2%).
Fig. (44). Relation of fy versus concrete stress (Qf=0%).
Fig. (45). Relation of fy versus concrete stress (Qf=0.5%).
Fig. (46). Relation of fy versus concrete stress (Qf=1%).
Fig. (47). Relation of fy versus concrete stress (Qf=2%).
Fig. (48). Relation of fc versus concrete stress (Rho=1%).
Fig. (49). Relation of fc versus concrete stress (Rho=5%).
Fig. (50). Relation of fc versus concrete stress (Rho=10%).
Fig. (51). Relation of fc versus concrete stress (Rho=Infinity).

CONCLUSION

The results of the analysis show:

  • 1) The variation of concrete stress (σc) along the length of the bar splice (- to ) for concrete without and with steel fiber (Qf=1%). The maximum stress is obtained at center x=0 and minimum at (x = ∓ ).
  • 2) The variation of bond stress (τ1) between the steel bar and surrounding concrete for concrete without steel fiber and with steel fiber content (Qf=1%). The maximum value is obtained at (x = ∓ ), and the minimum value at (x = ).
  • 3) The variation of steel stress (σs1) in the splice bar for both plain (Qf=0) and fibrous concrete (Qf=1%). The maximum value is (σs1=fy) at (x = - ) and the minimum value is (σs1=0) at (x = ). The value of τ2 is the same as (τ1) but in opposite sides, i.e., at (τ2=τ1). Also (σs2=fy) at (x = ) and equal to zero at (x = ).
  • 4) The value of (σcmax) increased linearly with the increase of (ρ), the slope of the lines decreased with an increase of the value of displacement modulus (k) to , the effect is negligible at (k=150 N / mm 3), the same behavior is noticed for (ρ=5&10%), but when (ρ= ), the effect of steel fiber content (Qf) and steel bar reinforcement index (ρ%) are neglected, and concrete stress becomes constant. The concrete stress increased nonlinearly (parabolic) with increasing of (ρ%). Also, increasing value of (k) has a small effect on (σcmax). The same behavior is obtained for other steel fiber content.
  • 5) The bar diameter has a negligible effect on maximum concrete stress for all steel fiber content (Qf) and steel reinforcement content (ρ).
  • 6) The maximum concrete stress (σcmax) increased linearly with (Fy) and concrete compressive strength (fc') and reduced when (ρ) increased from (1& to 5 & 10%) and became negligible at (ρ=). This effect is increased when the value of (ρ) is increased, and the same behavior is noticed in fibrous concrete.
  • 7) The relation of maximum bond stress (τmax) and steel fiber content (Qf) for (ρ=1%), value of (τmax) decreased linearly with increasing of (Qf) for all values of (k) and (ρ) and decreased with increasing of (ρ%) for all values of (Qf & k). Also, using larger bar diameter (dbf) causes a linear increase in bond stress (τmax). The value of (τmax) increased with increasing of (fy) for (ρ=1,5,10% and ) for plain concrete (Qf=0) and fibrous concrete (Qf=0.5,1&2%).

CONSENT FOR PUBLICATION

Not applicable.

AVAILABILITY OF DATA AND MATERIALS

Not applicable.

FUNDING

None.

CONFLICT OF INTEREST

The authors declare no conflict of interest, financial or otherwise.

ACKNOWLEDGEMENTS

Declared none.

REFERENCES

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