applying different decentralized loadings on

http://iaeme.com/Home/journal/IJCIET 2752 [email protected]

International Journal of Civil Engineering and Technology (IJCIET)

Volume 9, Issue 11, November 2018, pp. 2752–2769, Article ID: IJCIET_09_11_276

Available online at http://iaeme.com/Home/issue/IJCIET?Volume=9&Issue=11

ISSN Print: 0976-6308 and ISSN Online: 0976-6316

©IAEME Publication Scopus Indexed

APPLYING DIFFERENT DECENTRALIZED

LOADINGS ON RC CONTINUOUS DEEP BEAMS

USING STM

Ali Mustafa Jalil

Student, University of Diyala, Civil Engineering, Diyala, Iraq

Dr. Mohammed j. Hamood

Assistant Prof. University of Technology, Civil Engineering, Baghdad, Iraq

Dr. Khattab Saleem Abdul-Razzaq

Prof. University of Diyala, Civil Engineering,Diyala Iraq

Dr. Abbas H. Mohammed

University of Diyala, Civil Engineering, Diyala, Iraq

ABSTRACT

This research presents strut and tie modeling for the reinforced concrete

continuous deep beams that are subjected to different types of decentralized loadings.

The different types of loadings area single concentrated force, two concentrated

forces, and a uniformly distributed load. Load movement indicates reducing the

internal shear span to the effective depth ratio (a/d). Detailed procedures to solve the

current questions are presented besides the detailed numerical examples. It is

concluded that, in case of single concentrated force, reducing a/d from 1.36to 1.09,

0.81, and then to 0.54, increased the ultimate capacity by about 13%, 23% and 35%,

respectively. It is also concluded that, in cases of two concentrated forces and

uniformly distributed load, reducing a/d from 1.09 to 0.81, 0.54 and then to 0.27,

increased the ultimate capacity by about 12%, 17% and 21%, respectively. The

increment in the ultimate capacity occurred because upon moving the load from span

center toward the inner support, the length of the inner strut shortens and the

dimensions of its section increase significantly which leads to more strength. It is true

that, in contrast, the length of the external strut increases and its dimensions decrease,

but this decrease in its dimensions is slight, making the weakness cause is due to that,

ineffective and therefore, indecisive.

Keywords: RC, Continuous deep beams, STM, Decent ralized load, Design

procedures.

https://www.researchgate.net/profile/Abbas_Mohammed7

Ali Mustafa Jalil, Mohammed j. Hamood, Khattab Saleem Abdul-Razzaq and Abbas H. Mohammed


Cite this Article: Ali Mustafa Jalil, Mohammed j. Hamood, Khattab Saleem Abdul-

Razzaq and Abbas H. Mohammed, Applying Different Decentralized Loadings on RC

Continuous Deep Beams Using STM, International Journal of Civil Engineering and

Technology (IJCIET) 9(11), 2018, pp. 2752–2769.

http://iaeme.com/Home/issue/IJCIET?Volume=9&Issue=11

1. INTRODUCTION

Continuous deep beams are divided into two types based on loading conditions; bottom and

top loading. Bottom loaded or indirectly loaded deep beams are widely used as cross-girders

in concrete water tanks or concrete bridges, etc., while top loaded continuous deep beams are

commonly used in reinforced concrete buildings. Also continuous deep beams are divided

into two types based onloading location; centralized or decentralized. Due to the fact that the

stress distribution in the deep beam section is nonlinear, the general beam analysis of linear

elastic theory cannot be used. That is why, ACI code 318 recommends that deep beams

should be designed using non-linear analysis or by using Strut and Tie models (STM). STM is

formulated by straight lines that state resultant forces of compression and tension stresses in

members. According to ACI 318M-14 [1], deep beams are members that are supported on one

face and loaded on the opposite face such that strut-like compression elements can be formed

between the supporting and loading points and that satisfy (a) or (b) [1] below:(a) Clear span

lndoes not exceed four times the overall member depth h. (b) Concentrated loads exist within

a distance 2h from the face of the support. Many investigators have suggested empirical and

semi-empirical expressions to determine the ultimate load capacity of convention allyrein

forced concrete deep beams [2, 3]. Some researchers studied the parameters that affect deep

beam behaviour and capacity [4-11].Since 2002, the ACI-318 Code procedure is based on

empirical equations for the design of deep beams .

According to ACI 318M-14 [1], STM is defined as "a truss model of a structural member

or of a D- region in such a member, made up of struts and ties connected at nodes, capable of

transferring factored loads to the supports or to adjacent B-regions". Provisions for STM have

been taken into considerations for the design purpose. STM complies with the plasticity lower

bound theory, which needs that only yield conditions in addition to equilibrium to be satisfied.

Plasticity lower bound theory states that if the load has such a value that it is possible to find a

distribution of stress corresponding to stresses that keep internal and external equilibrium

within the yield surface, then this load will not cause failure of the body. In other words, the

capacity of a structure as estimated by a lower bound theory will be less than or equal to the

real failure load of the body in question [12].

Strut and tie model is a very useful tool for analyzing and designing reinforced concrete

members in which D-regions exist. The decent realized top loading cases are very common in

structural engineering, while lack of such studies on continuous RC deep beams using STM is

obvious. That is why this study investigates modeling in detail the struts and ties in the

reinforced concrete two-span continuous deep beams under various decentralized loading

cases.

2. RESEARCH METHODOLOGY

According to STM of ACI 318-14 [1], the transfer systems of the main load are tie action of

main longitudinal reinforcement and compressive struts which represent concrete

compression stress fields. Deep beam concrete compressive struts generally considered as

bottle-shaped struts that are commonly idealized as uniformly tapered or prismatic members

in shear spans [13-16]. A tensile tie characterizes one or several steel reinforcement layers.

According to ACI 318-14[1], the main longitudinal reinforcement should be distributed

Applying Different Decentralized Loadings on RC Continuous Deep Beams Using STM


uniformly over thetie width. Strutsand ties meet in nodes, which are considered the joints

where axial forces intersect. There are two main systems of load transfer, one of which is the

strut-and-tie action because of the longitudinal topre in forcement and the other is the strut-

and-tie action developed with the longitudinal bottom reinforcement representing the tie. In a

deep beam that has two spans, the applied loads are transferred from loading to supporting

points through concrete struts of interior and exterior shear spans. Therefore, the continuous

deep beam load capacity PFis:

𝑃𝐹 = 2(𝐹𝑛𝑖𝑠 𝑠𝑖𝑛 𝜗𝑖 + 𝐹𝑛𝑒𝑠 𝑠𝑖𝑛 𝜗𝑒) − 1

where𝐹𝑛𝑖𝑠and 𝐹𝑛𝑒𝑠are the load capacities of interior and exterior concrete struts,

respectively. 𝜗𝑖,𝜗𝑒 are the angles between the interior and exterior concrete strut and

longitudinal axis of the deep beam. These angles can be expressed as tan-1(jd/ai), tan-1(jd/ae),

respectively, where a is the exterior and interior shear span. The distance between the center

of bottom and top nodes jd could be considered as the distance between the center of longitudinal

bottom and top reinforcing bars as below:

𝑗𝑑 = ℎ −𝑤𝑡𝑡

2−

𝑤𝑡𝑏

2− − − − − 2

whereh is the overall section depth;𝑤𝑡𝑡and 𝑤𝑡𝑏 are the heights of nodal zone for top, i.e.

applied load, and bottom, i.e. support, respectively. At the applying load points, they could be

classified as a CCT or CCC type, which are hydrostatic nodes connecting both interior and

exterior compressive struts in sagging zone. In a CCC node type that has equal all in-plane

side stresses, the ratio of each hydrostatic node face width has to be the same as the force

meeting ratio at the node to make the stress state constant in the whole region of the node

[13,14]. The effective width of strut depends on the tie width and loading plate, in addition to

the strut slope. Average effective widths of concrete struts uniformly tapered in exterior shear

spans 𝑤𝑒𝑠and interior 𝑤𝑖𝑠can be calculated by:

𝑤𝑖𝑠 =𝑤𝑠𝑖

2+

𝑤𝑠𝑝𝑖

2− − − − − 3

𝑤𝑒𝑠 =𝑤𝑠𝑒

2+

𝑤𝑠𝑝𝑒

2− − − − − 4

Where 𝑤𝑠𝑒 and 𝑤𝑠𝑖 are the lengths of inclined faces of nodal zone at exterior and interior

support, respectively. 𝑤𝑠𝑝𝑒and𝑤𝑠𝑝𝑖 are the lengths of inclined exterior and interior faces of

nodal zone at applied load, respectively. The load transfer of concrete struts relies on the strut

area and compressive strength of concrete. Therefore, load capacities of interior and exterior

concrete struts are:

𝐹𝑛𝑖𝑠 = 0.85𝛽𝑠. 𝑓𝑐′.𝑏. 𝑤𝑖𝑠 − − − − − 5

𝐹𝑛𝑒𝑠 = 0.85𝛽𝑠. 𝑓𝑐′.𝑏. 𝑤𝑒𝑠 − − − − − 6

where βs, in accordance with Table 23.4.3[1] is a factor that accounts for the cracking

effect and possible transverse reinforcement existence. Values of βs are as follows:

βs = 1.0 for strut with uniform cross-section over its length

βs = 0.75 for sufficiently web-reinforced bottle-shaped strut

βs = 0.6λ for insufficiently web reinforced bottle-shaped strut, whereλ is a correction factor for

lightweight concrete

βs = 0.4 for struts in tension members

βs = 0.6λ for all other cases

The minimum web reinforcement ratios for both horizontal and vertical ones should be

0.0025 with the maximum spacing of d/5 and not more than 300mm [1].



Minimum reinforcement to resist the splitting force and to restrain crack widths in the

struts has to be provided to satisfy the following requirement, Figure 1,of ACI 318-14[1], as

follows:

𝛴𝐴𝑠𝑖

𝑏𝑤𝑠𝑖𝑠𝑖𝑛𝛼𝑖

≥ 0.003 − − − − − 7

whereAsiis the surface reinforcement area in the ith layer crossing the strut,siis the

reinforcing barspacing in the ith layer adjacent to the member surface and αiis the angle

between the axis of the strut and the bars in the ithreinforcement layer crossing that strut.

2

Axis of Strut

1

1sA

2sA

2S

1S

Vertical

Reinforcement

Horizontal

Reinforcement

Figure 1. Reinforcement crossing a strut.

The continuous deep beam failure mode is different from that of simply one. For

continuous, it generally occurs at external strut or internal strut or nodal. Therefore, the

analysis presented by the current study includes the examination of all components of the

model. The nominal strength of tie is:

Fsi = Asi ∗ Fy − − − − − 8

whereAsi is the area of conventional steel reinforcement for top or bottom tie andfy is the

yield stress of conventional steel reinforcement for the top or the bottom tie.

At any section through the nodal zone or at the face of a nodal zone, the nominal strength

Fniis:

Fni = 0.85βn. fc′.Ani − − − − − 9

whereAni is the facearea of the nodal zone and βn is a factor that reflects the sign of forces

acting on the nodal zone. The presence of tensile stresses due to ties decreases the nodal zone

concrete strength. The values of βn shall be in accordance with Table 23.9.2 [1] and as

follows:

3. STRUT AND TIE METHOD (STM) DESIGN PROCEDURES

An emerging design methodology for all D-region types is presented here to predict and

design an internal truss. This truss is consisted of steel tension ties and concrete compressive

struts that meet at nodes, to support the applied loading through the regions of discontinuity.

The STM design procedure includes the well-known general steps summarized below[1]:

i.Define the D-region boundaries and find the applied sectional and local forces. ii.Draw the

internal supporting truss, find equivalent loadings, and calculate the forces of the truss

βn = 1.0 in nodal zones bounded by bearing areas or struts (e.g., C-C-C nodes)

βn = 0.8 in nodal zones anchoring one tie (e.g., C-C-T nodes)

βn = 0.6 in nodal zones anchoring two or more ties (e.g., C-T-T or T-T-T nodes)



member. iii. Choose the reinforcing steel to represent the required capacity of the tie and

guarantee that this tie reinforcement is adequately anchored in the nodal zone, i.e. joints of the

truss.

iv.Calculate the dimensions of the nodes and struts, such that the capacities of these nodes and

struts are adequateto carry the values of the design forces.

v. Select the reinforcement distribution to guarantee the D-region ductile behavior.

It is worth to mention that both non-hydrostatic and hydrostatic nodes are reallyidealized.

The use of either non-hydrostatic or hydrostatic nodes is an assumption; a design tool

proposed to present a direct method for STM proportioning. The classical method of node

dimensioning is by node shape arranging in a way that the applied stresses on all sides of the

node are equal. The stress biaxial state in the node is hydrostatic; so, the in-plane stresses are

homogeneous, isotropic, and equal to those on the node sides. Arranging the node in this

shape can be made by sizing the node boundaries in a way that they become proportional and

perpendicular to the forces that acting on them, i.e. hydrostatic [17]. In the case of

decentralized typeof loading, there is no symmetry in checking nodes, struts and tie, because

the truss formed by loading transferring from the applying to supporting nodes is not

symmetric too. In order to recognize designation of specimens easily, Table 1. demonstrates

the way of this designation.

4. APPLYING DIFFERENT TYPES OF DECENTRALIZED LOADS

4.1. One Decentralized Concentrated Force

Figure 2.shows the principal stress paths and the assumed truss under decentralized 1-

concentrated force for each span in the continuous deep beam specimen CD.1F. The geometry

conforms to the deep beam definition 𝑙𝑛≤ 4ℎ[1]. The capacities are checked of each node face

A, B and C. The capacities of the diagonal struts, which are idealized bottle shape, are also

checked, in addition to the capacities of the top and bottom ties.

To analyze the continuous deep beam with one concentrated force for each span, steps

shown in Figure 3. may be followed. A detailed numerical application example is illustrated

in Table 2.and Figure 4.

Table1. Specimens designation way

Letter or number Meaning

CD ConventionalContinuousDeep Beam

1F Subjected to 1-Concentrated Force

2F Subjected to 2-Concentrated Forces

UL Subjected to Uniformly Distributed Load



Figure 2 The principal stress paths and the assumed truss forthe specimen CD.1F

Figure 3. STM Flow chart for reinforced concrete continuous deep beams subjected to decentralized

1-concentrated force.



Table 2. Numerical Example No. 1, One Decentralized Concentrated Force Input data

h=800mm 𝑙𝑐 =2000mm d=723.5mm 𝑎𝑒 =1250mm 𝑎𝑖=750mm 𝑏𝑤 =200mm f'c=30

MPa

Bearing Plates=150*200

mmfor external

support

Bearing

Plates=200*200

mmfor internal support and applied

load

fy=500M

Pa

Vertical web

reinf.= 𝜙8mm

@250mm c/c

fyv =450MP

a

Horizontal web

reinf.= 𝜙8mm@

250mm c/c

fyh =450

MPa

Bottom and upper

covers=40mm

Main longitudinal top and bottom reinforcement

=6𝜙16mm

Output data

Draw STM of continuous deep beams CD.1F, see Figure 5.

h = jd +wtt

2+

wtb

2

jd = a tan ϑ

ϑ = tan−1(jd

a)

wtb = (Cc + ∅st. +∅main +spacing

2) ∗ 2

= (40 + 8 + 16 + 12.5) ∗ 2 = 153 mm

or

wtb = (h − d) ∗ 2 = (800 − 723.5) ∗ 2 = 153 mm

wtb = wtt = 153 mm

becouse number of lyers for top and bottom is equal

jd = h −wtt

2−

wtb

2= 800 − 153 = 647mm

ϑ𝑒 = tan−1(647

1250) = 27.37°

ϑ𝑖 = tan−1(647

750) = 40.78°

wse = wtb cos ϑ𝑒 + lse sin ϑ𝑒

wse = 153 cos 27.37 + 150 sin 27.37 = 204.83 mm

wsi = wtb cos ϑ𝑖 + lsi sin ϑ𝑖

wsi = 153 cos 40.78 + 200 sin 40.78 = 246.49mm

wspe = wtt cos ϑ𝑒 + β ∗ lp sin ϑ𝑒

wspe = 153 cos 27.37 + 0.375 ∗ 200 sin 27.37 = 170.35 mm

wsp𝑖 = wtt cos ϑ𝑖 + (1 − β) ∗ lp sin ϑ𝑖

wsp𝑖 = 153 cos 40.78 + (1 − 0.375) ∗ 200 sin 40.78 = 197.5 mm

wes =wse

2+

wspe

2=

204.83

2+

170.35

2= 187.59 mm

wis =wsi

2+

wspi

2=

246.49

2+

197.5

2= 222 𝑚𝑚

Fnes = 0.85βs. fc′.b. wes, Figure 6. Fnis = 0.85βs. fc′.b. wis, Figure6.

βs = 0.75 when Q ≥ 0.003

βs = 0.6λ when Q < 0.003

𝑄𝑒 = (𝐴𝑣

𝑏 ∗ 𝑆)𝑆𝑖𝑛𝛼1 + (

𝐴ℎ

𝑏 ∗ 𝑆)𝑆𝑖𝑛𝛼2 =

2 ∗𝜋

4∗ 82

200 ∗ 250𝑠𝑖𝑛 27.37 +

2 ∗𝜋

4∗ 82

200 ∗ 250𝑠𝑖𝑛 (90 − 27.37) = 0.0027 < 0.003

𝑄𝑖 = (𝐴𝑣


𝐴ℎ


2 ∗𝜋

4∗ 82

200 ∗ 250𝑠𝑖𝑛 40.78 +

2 ∗𝜋

4∗ 82

200 ∗ 250𝑠𝑖𝑛 (90 − 40.78) = 0.0028 < 0.003

𝑄𝑒, 𝑄𝑖 < 0.003 → 𝛽𝑠 = 0.6 →

[ACI 318M-14, Table 23.4.3] [1]

Fnes = 0.85 ∗ 0.6 ∗ 30 ∗ 200 ∗ 187.59 = 574.03 kN

Fnes cos ϑ𝑒 = 509.77 kN

Fnis = 0.85 ∗ 0.6 ∗ 30 ∗ 200 ∗ 222 = 679.32 kN

Fnis cos ϑ𝑖 = 514.4kN

𝐂𝐡𝐞𝐜𝐤 𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐲

𝐍𝐨𝐝𝐞 𝐀 (𝐂𝐂𝐓), Figure7 − a. 𝐢𝐧𝐜𝐥𝐢𝐧𝐝 𝐟𝐚𝐜𝐞

FnA = 0.85βn. fc′.AnA

βn = 0.8 for CCT [ACI 318M-14, Table 23.9.2] [1]

AnA = wse ∗ bw

AnA = 204.83 ∗ 200 =40966 mm2

FnA = 0.85 ∗ 0.8 ∗ 30 ∗ 40966 = 835.71 kN

FnA > Fnes o. k

𝐇𝐨𝐫𝐢𝐳𝐨𝐭𝐚𝐥 𝐟𝐚𝐜𝐞

σvA =Fnes sin ϑ𝑒

150 ∗ 200=

263.9 ∗ 1000

150 ∗ 200= 8.8MPa

FcuA = 0.85βn. fc′


FcuA = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa

FcuA > σvA The dimension of plate it is o. k

𝐍𝐨𝐝𝐞 𝐁 [𝐂𝐂𝐓], Figure7 − b. 𝐢𝐧𝐜𝐥𝐢𝐧𝐝 𝐟𝐚𝐜𝐞

FnB = 0.85βn. fc′.AnB


AnB = wsi ∗ bw

AnB = 246.49 ∗ 200 = 49298mm2

FnB = 0.85 ∗ 0.8 ∗ 30 ∗ 49298 = 1005.67 kN

FnB > Fnis o. k


σvB =2Fnis sin ϑ𝑖

200 ∗ 200=

887.4 ∗ 1000

200 ∗ 200

= 22.19 MPa

FcuB = 0.85βn. fc′


FcuB = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa

σvB > FcuB

(

increasing dimension of plate or using nodal reinforcement

to prevent premature failure)



Figure 4. Details of the specimen CD.1F

Figure 5. Strut – Tie model for 1-concentrated force, each span ofthe specimen CD.1Fis loaded

𝐍𝐨𝐝𝐞 𝐂 [𝑪𝑪𝑪𝑪𝑪𝑻⁄ ] , Figure7 − C.

𝐢𝐧𝐜𝐥𝐢𝐧𝐝 𝐟𝐚𝐜𝐞

FnC = 0.85βn. fc′.AnC

(βn = 1) for CCC [ACI 318M-14, Table 23.9.2] [1]

(βn = 0.8) for CCT [ACI 318M-14, Table 23.9.2] [1]

AnCe = wspe ∗ bw = 34070 mm2

Fnce = 0.85 ∗ 1 ∗ 30 ∗ 34070 = 868.79 kN > Fnes o. k

AnCi = wspi ∗ bw = 39500mm2

Fnci = 0.85 ∗ 0.8 ∗ 30 ∗ 39500 = 805.8 kN > Fnis o. k


σvC =PF/2

area of plate=

707.6 ∗ 1000

200 ∗ 200= 17.7 MPa

FcuC = 0.85βn. fc′


FcuC = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa

σvC < FcuC

The dimension of plate it is o. k

𝐓𝐢𝐞 𝟏

Fsb = Asb ∗ Fy = 6 ∗π

4162 ∗ 500

= 603.19 kN > Fnes cos ϑ𝑒 , Fnis cos ϑ𝑖 O. K

𝐓𝐢𝐞 𝟐

Fst = Ast ∗ Fy = 6 ∗π

4162 ∗ 500 = 603.19 kN

> (Fnis cos ϑ𝑖

− Fnes sin ϑ𝑒) O. K

Load Failure

PF = 2(Fnes sin ϑ𝑒 + Fnis sin ϑ𝑖)

= 2(574.03 sin 27.37 + 679.32 sin 40.78)

=1415.21 kN



Figure 6. Reinforcement crossing strut of the specimen CD.1F

a) Faces of support nodal zone A, CD.1F b) Faces of load nodal zone B, CD.1F

c) Faces of support nodal zone C, CD.1F

Figure 7. Nodes in 1-concentrated force, each span ofthe specimen CD.1Fis loaded

4.2. Two Decentralized Concentrated Forces

Figure 8. shows the principal stress paths and the assumed truss underthe decentralized2-

concentrated forces in continuous deep beam CD.2F. According to the shear provisions of the

ACI 318M-14 design code, same as in the case of 1-concentrated force, the geometry

conforms to the deep beam definition 𝑙𝑛≤ 4ℎ[1]. The capacities are checked of each node face

A, B and C. The capacities of the diagonal struts, which are idealized bottle shape, are also

checked, in addition to the capacities of the top and bottom ties.

To analyze the continuous deep beam with two concentrated forces for each span, the

steps shown in Figure 9.may be followed. A detailed numerical application example is shown

in Table 3.and Figure 10.



Figure 8. The principal stress paths and the assumed truss forthe specimen CD.2F

Figure 9. STM Flow chart for reinforced concrete continuous deep beams subjected to decentralized

2-concentrated forces.



Table 3. Numerical Example No. 2, two decentralized concentrated forces

Input data

h=800mm 𝑙𝑐 =2000mm d=723.5mm 𝑎𝑒 =1000mm 𝑎𝑖=500mm 𝑏𝑤 =200mm f'c=30

MPa

Bearing

Plates=150* 200

mmfor external support

Bearing

Plates=200* 200

mmfor internal

support and

appliedload

fy=500

MPa

Vertical web

reinf.= 𝜙8mm@

250mm c/c

fyv =450

MPa

Horizontal

web

reinf.= 𝜙8mm

@250

mm c/c

fyh =450

MPa

Bottom and

upper

covers=

40mm

Main longitudinal top and bottom reinforcement

=6𝜙16mm

Output data

Draw STM of continuous deep beams CD.2F, see Figure11.

h = jd +wtt

2+

wtb

2

jd = a tan ϑ

ϑ = tan−1(jd

a)

wtb = (Cc + ∅st. +∅main +spacing

2) ∗ 2

= (40 + 8 + 16 + 12.5) ∗ 2 = 153 mm

or

wtb = (h − d) ∗ 2 = (800 − 723.5) ∗ 2 = 153 mm

wtb = wtt = 153 mm

becouse number of lyer for top and bottom is equal

jd = h −wtt

2−

wtb

2= 800 − 153 = 647mm

ϑ𝑒 = tan−1(647

1000) = 32.9°

ϑ𝑖 = tan−1(647

500) = 52.3°

wse = wtb cos ϑ𝑒 + lse sin ϑ𝑒

wse = 153 cos 32.9° + 150 sin 32.9° = 209.94 mm

wsi = wtb cos ϑ𝑖 + lsi sin ϑ𝑖

wsi = 153 cos 52.3° + 200 sin 52.3° = 251.81mm

wspe = wtt cos ϑ𝑒 + lpe sin ϑ𝑒

wspe = 153 cos 32.9° + 200 sin 32.9° = 237.1 mm

wsp𝑖 = wtt cos ϑ𝑖 + lpi sin ϑ𝑖

wsp𝑖 = 153 cos 52.3° + 200 sin 52.3° = 251.81 mm

wes =wse

2+

wspe

2=

209.94

2+

237.1

2= 223.52 mm

wis =wsi

2+

wspi

2=

251.81

2+

251.81

2= 251.81 mm

Fnes = 0.85βs. fc′.b. wes

Fnis = 0.85βs. fc′.b. wis

βs = 0.75 when Q ≥ 0.003

βs = 0.6λ when Q < 0.003

𝑄𝑒 = (𝐴𝑣


𝐴ℎ


2 ∗𝜋

4∗ 82

200 ∗ 250𝑠𝑖𝑛 32.9° +

2 ∗𝜋

4∗ 82

200 ∗ 250𝑠𝑖𝑛 (90 − 32.9°) = 0.0027 < 0.003

𝑄𝑖 = (𝐴𝑣


𝐴ℎ


2 ∗𝜋

4∗ 82

200 ∗ 250𝑠𝑖𝑛 52.3° +

2 ∗𝜋

4∗ 82

200 ∗ 250𝑠𝑖𝑛 (90 − 52.3°) = 0.0028 < 0.003

𝑄𝑒, 𝑄𝑖 < 0.003 → 𝛽𝑠 = 0.6 →

[ACI 318M-14, Table 23.4.3] [1]

Fnes = 0.85 ∗ 0.6 ∗ 30 ∗ 200 ∗ 223.52 = 683.97 kN

Fnes cos ϑ𝑒 = 574.27 kN

Fnis = 0.85 ∗ 0.6 ∗ 30 ∗ 200 ∗ 251.81 = 770.54 kN

Fnis cos ϑ𝑖 = 471.21 kN

𝐂𝐡𝐞𝐜𝐤 𝐂𝐚𝐩𝐚𝐜𝐢𝐭𝐲

𝐇𝐨𝐫𝐢𝐳𝐨𝐧𝐭𝐚𝐥 𝐒𝐭𝐫𝐮𝐭

FHS = 0.85βs. fc′.AHS

βs = 1[ACI 318M-14, Table 23.4.3] [1]

AHS = 153 ∗ 200 = 30600 mm2

FHS = 0.85 ∗ 1 ∗ 30 ∗ 30600 = 780.3 kN

FHS > Fnes cos ϑ𝑒 o. k

FHS + 𝑇𝑖𝑒2 > Fnis cos ϑ𝑖 o. k



Figure 10. Details of the specimen CD.2F

𝐍𝐨𝐝𝐞 𝐀 (𝐂𝐂𝐓), Figure12 − a.


FnA = 0.85βn. fc′.AnA


AnA = wse ∗ bw

AnA = 209.94 ∗ 200=41988

FnA = 0.85 ∗ 0.8 ∗ 30 ∗ 41988 = 856.56 kN

FnA > Fnes o. k


σvA =Fnes sin ϑ𝑒

150 ∗ 200=

371.52 ∗ 1000

150 ∗ 200= 12.38 MPa

FcuA = 0.85βn. fc′


FcuA = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa

FcuA > σvA The dimension of plate it is o. k

𝐍𝐨𝐝𝐞 𝐁 [𝐂𝐂𝐓], Figure12 − b.


FnB = 0.85βn. fc′.AnB


AnB = wsi ∗ bw

AnB = 251.81 ∗ 200 = 50362mm2

FnB = 0.85 ∗ 0.8 ∗ 30 ∗ 50362 = 1027.38 kN

FnB > Fnis o. k


σvB =2Fnis sin ϑ𝑖

200 ∗ 200=

1219.34 ∗ 1000

200 ∗ 200

= 30.5 MPa

FcuB = 0.85βn. fc′


FcuB = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa

σvB > FcuB

(

increase dimension of plate or using nodal reinforcement

to prevent premature failure)

𝐍𝐨𝐝𝐞 𝐂 [𝐂𝐂𝐓], Figure12 − c.


FnC = 0.85βn. fc′.AnC

(βn = 0.8) for CCT [ACI 318M-14, Table 23.9.2] [1]

AnC = wspi ∗ bw

AnC = 251.81 ∗ 200 = 50362 mm2

Fnc = 0.85 ∗ 0.8 ∗ 30 ∗ 50362 = 1027.18 kN > Fnis


σvC =PF/4

area of plate=

490.6 ∗ 1000

200 ∗ 200= 12.3 MPa

FcuC = 0.85βn. fc′


FcuC = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPaσvC < FcuC


𝐍𝐨𝐝𝐞 𝐃 [𝐂𝐂𝐂], Figure12 − d.


Fn𝐷 = 0.85βn. fc′.AnD

(βn = 1) for CCC [ACI 318M-14, Table 23.9.2] [1]

An𝐷 = wspe ∗ bw

AnD = 237.1 ∗ 200 = 47420 mm2

Fnci = 0.85 ∗ 1 ∗ 30 ∗ 47420 = 1209.21 kN > Fnes o. k


σvD =PF/4

area of plate=

490.6 ∗ 1000

200 ∗ 200= 12.3 MPa

FcuD = 0.85βn. fc′

βn = 1 for CCC [ACI 318M-14, Table 23.9.2] [1]

FcuC = 0.85 ∗ 1 ∗ 30 = 25.5 MPa

σvD < FcuD


𝐓𝐢𝐞 𝟏

Fsb = Asb ∗ Fy = 6 ∗π

4162 ∗ 500

= 603.19 kN > Fnes cos ϑ𝑒 , Fnis cos ϑ𝑖 O. K

𝐓𝐢𝐞 𝟐

Fst = Ast ∗ Fy = 6 ∗π

4162 ∗ 500 = 603.19 kN

> Fnis cos ϑ𝑖 O. K 780.3 kN + 603.19 kN

Load Faliure

PF = 2(Fnes sin ϑ𝑒 + Fnis sin ϑ𝑖)

= 2(683.97 sin 32.9 + 770.54 sin 52.3)

= 1962.4 kN



Figure 11. Strut – Tie model for 2-concentrated forces loadedin each span ofthe specimen CD.2F

a) Faces of supportnodal zone A, CD.2F b) Faces of support nodal zone B, CD.2F

c) Faces of support nodal zone C, CD.2F d)Faces of load nodal zone D, CD.2F

Figure 12. Nodes in 2-concentrated forces loaded each span ofthe specimen CD.2F

4.3. Decentralized Uniformly Distributed Load

Figure 13. Strut – Tie model for uniformly distributed loaded continuous beam specimen CD.UL

Many researchers went to the conclusion that when deep beam is subjected to uniformly

distributed load, it could be considered as deep beam under two concentrated forces that

should equal to the uniformly distributed load in value [18-20]. Figure13. shows the principal

stress paths in two-span deep beam subjected to a decentralized uniformly distributed load. It

is worth to mention that this substitution is allowed only if the equality of the maximum



moments, the most fundamental value in the Strut-Tie model application of the both systems

is guaranteed.Figures 14-a. and 14-b. show how the bending moment for the two equivalent

equal concentrated forces are closer to the bending moment of uniformly distributed load than

the bending moment when the two forces are unequal, Figure 14-c. That is why, in this study,

the two equivalent equal two forcesare considered as a substitution forequivalent uniformly

distributed load.Based on that, the prediction of strength capacity for the reinforced concrete

continuous deep beam subjectedto decentralized uniformly distributed loading, CD.UL shown

in Figure 14-b. and Figure 10.can be obtained by the same procedure shown in Figure9.

It was considered that the equivalent two concentrated forces are equal, so the strength

capacity can be calculated by the followings:

PF = 2(Fnes sin ϑe + Fnis sin ϑi) = 2(683.97 sin 32.9 + 770.54 sin 52.3) = 1962.4 kN

WF = PF/(LW = 1 m for two span), ∴ WF =1962.4 kN/m, this is similar to the numerical example

No.2.

a): Uniformly distributed load

b): Equivalent two equal concentrated forces= 2(133.65) Kn



c): Equivalent two unequal concentrated forces

Figure 14. Moment diagrams for the specimen CD.UL

5. EFFECT OF LOADING DECENTRALIZATION

Applying decentralized load on a span leads to formation of unsymmetrical stress paths that

joining loading and supporting points. It is observed that moving load from the centre of a

span towards the internal support leads to increasing the ultimate capacity of the beam as

shown in Table 4. In case of a single decentralized force, moving load from span center to the

internal support, that is to say, decreasing internala/dfrom 1.36 to 1.09, 0.81 and 0.54 led to

increase ultimate capacity by about 13%, 23%, and 35%, respectively. On the other hand, in

cases of 2-concentrated decentralized forces and the uniformly distributed load, moving load

from the span centre to the internal support, let's to say decreasing internala/dfrom 1.09 to

0.81, 0.54 and 0.27 leads to increase ultimate capacity by about 12%, 17%, and 21%,

respectively.

In other words, decreasing the ratio a/d leads to increase the value of the internal strut-tie

angle ϑi. In more detail,in cases of 1-concentrated decentralized force, increasing ϑi from

33.99 to 40.13, 48.34, and 59.33degrees led to increase the ultimate capacity by about 13%,

23%, and 35%, respectively as shown in Figures 15and18.In cases of the 2-concentrated

decentralized forces and the uniformly distributed load, moving load from the span centre to

the internal support, let's to say increasing ϑi from 40.13 to 48.34,59.33and 73.48degrees led

to increase ultimate capacity by about 12%, 17%, and 21%, respectively as shown in Figures

16and17in addition to Figures 19and20.

These differences in ultimate capacity take place due to the facts that a non-concentric

load causes an asymmetry in terms of the quantity of stresses and the form of their

distribution. Thus, the truss drawn between the points of loading and the points of supporting

is not symmetrical. As shown in Table 4.,it is observed that at each load movement toward the

internal support, an increase in ultimate capacity takes place.



Figure 16. Effect of internal angle on the

ultimate capacity of specimens in group B

Figure 15. Effect of internal angle on the ultimate

capacity of specimens in group A

Figure 17. Effect of internal angle on the ultimate capacity of specimens in group C

0

500

1000

1500

2000

2500

3000

3500

40.13 48.34 59.33 73.48

Lo

ad (

kN

)

Internal Angle (degree)

0

500

1000

1500

2000

2500

3000

33.99 40.13 48.34 59.33

Lo

ad (

kN

)


0

500

1000

1500

2000

2500

3000

3500

40.13 48.34 59.33 73.48

Lo

ad (

kN

/m)


Table 4. Effect of load decentralization on ultimate capacity

Sp

ecim

en

Gro

up

Load

Type

Shear Span

(mm)

Angle

(degrees) Ultimate capacity

𝒂𝒆/d

𝒂𝒊/d

𝝑𝒊/𝝑𝒆

%

Increasing

in

𝑷𝒖 𝒂𝒆 𝒂𝒊 𝝑e 𝝑i 𝑷𝑭(kN)

Wu

(kN/m)

A1

A

Single

Concentrated

Force

1250 1250 33.99 33.99 1968.5 - 1.36 1.36 1 Reference

A2 1500 1000 29.34 40.13 2232.5 - 1.63 1.09 1.37 13.41

A3 1750 750 25.72 48.34 2421.4 - 1.90 0.81 1.88 23.00

A4 2000 500 22.86 59.33 2655.7 - 2.17 0.54 2.59 34.91

B1

B

Two

Concentrated

forces

1000 1000 40.13 40.13 2716.3 - 1.09 1.09 1 Reference

B2 1250 750 33.99 48.34 3043 - 1.36 0.81 1.42 12.03

B3 1500 500 29.33 59.33 3183.6 - 1.63 0.54 2.02 17.20

B4 1750 250 25.72 73.48 3280.2 - 1.90 0.27 2.86 20.75

C1

C

Uniformly

Distributed

Load

1000 1000 40.13 40.13 - 2716.3 1.09 1.09 1 Reference

C2 1250 750 33.99 48.34 - 3043 1.36 0.81 1.42 12.03

C3 1500 500 29.33 59.33 - 3183.6 1.63 0.54 2.02 17.20

C4 1750 250 25.72 73.48 - 3280.2 1.90 0.27 2.86 20.75



Figure 19. Effect of internal angle on the

ultimate capacity of specimens in group B

Figure 18. Effect of internal angle on the ultimate

capacity of specimens in group A

Figure 20. Effect of internal angle on the ultimate capacity of specimens in group C

6. CONCLUSIONS

This study presents a detailed theoretical study on the effect of moving different types of

loadings on the continuous reinforced concrete deep beams. The applied load is moved from

the center of span towards the inner support. It is found that the remaining distance between

the applied load and the inner support is inversely proportional to the deep beam ultimate

strength. That takes place because the length of the inner strut shortens and the dimensions of

its section increase significantly when the applied load be closer to the inner support, which

gives more strength. It is true that, in contrast, the length of the external strutincreases, but its

dimensions decrease slightly, making the weakness caused due to that ineffective and

therefore unregulated. This also applies to internal and external nodes. In more detail, moving

the load from the span center toward the internal support leads to a remarkable increase in the

size of the internal node, while the external node is less influenced and therefore does not

control design.

REFERENCES

[1] ACI Committee and American Concrete Institute. "Building code requirements for

structural concrete (ACI 318-14) and commentary." American Concrete Institute, 2014.

2600

2700

2800

2900

3000

3100

3200

3300

3400

30 40 50 60 70 80

2-

Co

ns.

Fo

rce

(kN

)


1900

2000

2100

2200

2300

2400

2500

2600

2700

30 40 50 60 70

1-

Co

ns.

Fo

rce

(kN

)


2600

2700

2800

2900

3000

3100

3200

3300

3400

30 40 50 60 70 80

Un

ifo

rm D

ist.

Lo

ad (

kN

/m)




[2] Zhang, Ning, and Kang-Hai Tan. "Direct Strut-and-Tie Model for Single Span and

Continuous Deep Beams. " Engineering Structures 29, no. 11 (November 2007): 2987–

3001. doi:10.1016/j.engstruct.2007.02.004.

[3] Rasheed, Mohammed M. "Modified Softened Strut and Tie Model for Concrete Deep

Beams." Journal of Engineering and Sustainable Development 16, no. 1 (2012): 348-361.

[4] Ashour, A. F. “Experimental behaviour of reinforced concrete continuous deep beams.”

Department of Civil and Environmental Engineering, University of Bradford,

(1996):1743-3509.

[5] Abdul-Razzaq, Khattab Saleem, "Effect of heating on simply supported reinforced

concrete deep beams." Diyala Journal of Engineering Sciences 8, no. 2 (2015): 116-133.

[6] Abdul-Razzaq, Khattab Saleem, Alaa Hussein Abed, and HayderIhsan Ali. "Parameters

affecting load capacity of reinforced self-compacted concrete deep beams." International

Journal of Engineering 5, no. 05 (2016).

[7] Abdul-Razzaq, Khattab Saleem, and SarahFarhanJebur. "Experimental Verification of

Strut and Tie Method for Reinforced Concrete Deep Beams under Various Types of

Loadings." Journal of Engineering and Sustainable Development 21, no. 6 (2018): 39-55.

[8] Abdul-Razzaq, Khattab Saleem, Hayder I. Ali, and Mais M. Abdul-Kareem. "A New

Strengthening Technique for Deep Beam Openings Using Steel Plates." International

Journal of Applied Engineering Research 12, no. 24 (2017): 15935-15947.

[9] Abdul-Razzaq, Khattab Saleem, and AliMustafa Jalil. "Behavior of Reinforced Concrete

Continuous Deep Beams-Literature Review." Second Conference of Post Graduate

Researches (CPGR'2017), College of Engineering, Al-Nahrain Univ., Baghdad, Iraq-4th

Oct. 2017.

[10] Abdul-Razzaq, Khattab Saleem, and Sarah FarhanJebur. "Suggesting Alternatives for

Reinforced Concrete Deep Beams by Reinforcing Struts and Ties." MATEC Web of

Conferences 120 (2017): 01004. doi:10.1051/matecconf/201712001004.

[11] Abdul-Razzaq, Khattab Saleem, Sarah FarhanJebur, and Abbas H. Mohammed. "Concrete

and Steel Strengths Effect on Deep Beams with Reinforced Struts." International Journal

of Applied Engineering Research 13, no. 1 (2018): 66-73.

[12] Nielsen, M. P., M. W. Braestrup, B. C. Jensen, and Finn Bach. "Concrete plasticity, beam

shear–shear in joints–punching shear." Special Publication (1978): 1-129.

[13] Peter Marti. “Basic Tools of Reinforced Concrete Beam Design.” ACI Journal

Proceedings 82, no. 1 (1985). doi:10.14359/10314.

[14] Comité Euro-International du Beton CEB-FIP. “CEB-FIP model code 1990 for concrete

structures.” CEB-FIP 90, Bulletin d’Information No. 213-214, Lausanne, Switzerland,

(1993).

[15] Canadian Standards Association, ed. A23. 3-94: “Design of Concrete Structures for

Buildings.” Canadian Standards Association, (1994).

[16] American Association of Highway and Transportation Officials (AASHTO), L. R. F. D.

"Bridge Design Specifications." (1998).

[17] Sanders, David H. "Verification and Implementation of Strut-and-Tie Model in LRFD

Bridge Design Specifications." AASHTO, Highway Subcommittee on Bridge and

Structures (2007).

[18] Nagarajan, Praveen, and T. M. Madhavan Pillai. "Analysis and Design of Simply

Supported Deep Beams Using Strut and Tie Method." Advances in Structural Engineering

11, no. 5 (October 2008): 491–499. doi:10.1260/136943308786412050.

[19] Sam-Young, Noh, Lee Chang-Yong, and Lee Kyeong-Min. "Deep Beam Design Using

Strut-Tie Model."

[20] Brown, Michael D. and OguzhanBayrak. "Investigation of Deep Beams with Various

Load Configurations." ACI Structural Journal 104, no. 5 (2007). doi:10.14359/18863.

applying different decentralized loadings on

Documents