APPLYING DIFFERENT DECENTRALIZED LOADINGS ON RC CONTINUOUS DEEP BEAMS USING STM

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International Journal of Civil Engineering and Technology (IJCIET)
Volume 9, Issue 11, November 201
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ISSN Print: 0976-6308 and ISSN Online: 0976
©IAEME Publication
APPLYING DIFFERENT D
LOADINGS ON RC CONTI
Student, University of Diyala
Assistant Prof. University of Technology
Dr.
Prof. University of Diyala
University of Diyala, Civil Engineering,
ABSTRACT
This research presents strut and tie modeling for the reinforced concrete
continuous deep beams that are subj
The different types of loadings area single concentrated force, two concentrated
forces, and a uniformly distributed load.
internal shear span to the effective depth r
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
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
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 str
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
procedures.
IJCIET/index.asp 2752 ed
International Journal of Civil Engineering and Technology (IJCIET)
2018, pp. 2752–2769, Article ID: IJCIET_09_11_276
http://www.iaeme.com/ijciet/issues.asp?JType=IJCIET&VType=
6308 and ISSN Online: 0976-6316
Scopus Indexed
APPLYING DIFFERENT DECENTRALIZED
LOADINGS ON RC CONTINUOUS DEEP BEAMS
USING STM
Ali Mustafa Jalil
University of Diyala, Civil Engineering, Diyala, Iraq
Dr. Mohammed j. Hamood
Prof. University of Technology, Civil Engineering, Baghdad
Dr. Khattab Saleem Abdul-Razzaq
Prof. University of Diyala, Civil Engineering,Diyala Iraq
Dr. Abbas H. Mohammed
University of Diyala, Civil Engineering, Diyala, Iraq
presents strut and tie modeling for the reinforced concrete
continuous deep beams that are subjected to different types of decentralized loadings.
forces, and a uniformly distributed load. Load movement indicates
internal shear span to the effective depth ratio (a/d). Detailed procedures to solve the
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%,
increased the ultimate capacity by about 12%, 17% and 21%, respectively.
dimensions of its section increase significantly which leads to more str
indecisive.
Continuous deep beams, STM, Decent ralized
editor@iaeme.com
IJCIET_09_11_276
asp?JType=IJCIET&VType=9&IType=11
ECENTRALIZED
NUOUS DEEP BEAMS
Iraq
Baghdad, Iraq
Iraq
presents strut and tie modeling for the reinforced concrete
ected to different types of decentralized loadings.
Load movement indicates reducing the
atio (a/d). Detailed procedures to solve the
in case of single concentrated force, reducing a/d from 1.36to 1.09,
capacity by about 13%, 23% and 35%,
%, respectively. The
dimensions of its section increase significantly which leads to more strength. It is true
ralized load, Design

Ali Mustafa Jalil, Mohammed j. Hamood, Khattab Saleem Abdul-Razzaq and Abbas H. Mohammed
http://www.iaeme.com/IJCIET/index.asp 2753 editor@iaeme.com
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://www.iaeme.com/IJCIET/issues.asp?JType=IJCIET&VType=9&IType=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:
= 2 + − 1
where and are the load capacities of interior and exterior concrete struts,
respectively. , 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:
= ℎ −
2
−
2
− − − − − 2
whereh is the overall section depth; and 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 and interior can be calculated by:
=
2
+
2
− − − − − 3
=
2
+
2
− − − − − 4
Where and are the lengths of inclined faces of nodal zone at exterior and interior
support, respectively. and 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:
= 0.85% . &'(.). − − − − − 5
= 0.85% . &'(.). − − − − − 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:
+
,
)-
./
≥ 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 ith
reinforcement layer crossing that strut.
2α
1α
1sA
2sA
2S
1S
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:
F34 = A34 ∗ 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:
F84 = 0.85βn. f<(.A84 − − − − − 9
whereA84 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 > ≤ 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 >? =2000mm d=723.5mm @ =1250mm @ =750mm )- =200mm
f'c=30
MPa
Bearing
Plates=150*200
mmfor external
support
Bearing
Plates=200*200
mmfor internal
support and applied
load
fA=500M
Pa
Vertical web
reinf.= B8mm
@250mm c/c
fAC =450MP
a
Horizontal web
reinf.= B8mm@
250mm c/c
fAD =450
MPa
Bottom and upper
covers=40mm
Main longitudinal top and bottom reinforcement
=6B16mm
Output data
h = jd +
wtJ
2
+
wtK
2
jd = a tan ϑ
ϑ = tanMN
jd
a
wtK = OCc + ∅st. +∅main +
spacing
2
X ∗ 2
= 40 + 8 + 16 + 12.5 ∗ 2 = 153 mm
or
wtK = h − d ∗ 2 = 800 − 723.5 ∗ 2 = 153 mm
wtK = wtJ = 153 mm
becouse number of lyers for top and bottom is equal
Draw STM of continuous deep beams CD.1F, see Figure 5.
jd = h −
wtJ
2
−
wtK
2
= 800 − 153 = 647mm
ϑ = tanMN
647
1250
= 27.37°
ϑ = tanMN
647
750
= 40.78°
w3_ = wtK cos ϑ + lse sinϑ
w3_ = 153 cos 27.37 + 150 sin 27.37 = 204.83 mm
w34 = wtK cosϑ + lsi sin ϑ
w34 = 153 cos 40.78 + 200 sin40.78 = 246.49mm
w3`_ = wtJ cosϑ + β ∗ lp sinϑ
w3`_ = 153 cos 27.37 + 0.375 ∗ 200 sin27.37 = 170.35 mm
w3` = wtJ cosϑ + 1 − β ∗ lp sinϑ
w3` = 153 cos40.78 + 1 − 0.375 ∗ 200 sin 40.78 = 197.5 mm
w_3 =
w3_
2
+
w3`_
2
=
204.83
2
+
170.35
2
= 187.59 mm
w43 =
w34
2
+
w3`4
2
=
246.49
2
+
197.5
2
= 222 aa
F8_3 = 0.85βs. f<(.b. w_3, Figure 6.
F843 = 0.85βs.f<(.b. w43, Figure6.
βs = 0.75 when Q ≥ 0.003
βs = 0.6λ when Q < 0.003
f =
,g
) ∗ h
h i1 +
,j
) ∗ h
h i2 =
2 ∗
k
l
∗ 8m
200 ∗ 250
27.37 +
2 ∗
k
l
∗ 8m
200 ∗ 250
90 − 27.37 = 0.0027 < 0.003
f =
,g
) ∗ h
h i1 +
,j
) ∗ h
h i2 =
2 ∗
k
l
∗ 8m
200 ∗ 250
40.78 +
2 ∗
k
l
∗ 8m
200 ∗ 250
90 − 40.78 = 0.0028 < 0.003
f , f < 0.003 → % = 0.6 →
F8_3 = 0.85 ∗ 0.6 ∗ 30 ∗ 200 ∗ 187.59 = 574.03 kN
F8_3 cos ϑ = 509.77 kN
F843 = 0.85 ∗ 0.6 ∗ 30 ∗ 200 ∗ 222 = 679.32 kN
F843 cos ϑ = 514.4kN
[ACI 318M-14, Table 23.4.3] [1]
qrstu qvwvtxyz
{|}s ~ qq• , Figure7 − a.
x€t•x€} ‚vts
F8ƒ = 0.85βn. f<(.A8ƒ
A8ƒ = w3_ ∗ bw
F8ƒ = 0.85 ∗ 0.8 ∗ 30 ∗ 40966 = 835.71 kN
F8ƒ > F8_3 o.k
…|†x‡|yv• ‚vts
σCƒ =
F8_3 sinϑ
150 ∗ 200
=
263.9 ∗ 1000
150 ∗ 200
= 8.8MPa
F<‹ƒ = 0.85βn. f<(
F<‹ƒ = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa
F<‹ƒ > σCƒ The dimension of plate it is o. k
βn = 0.8 for CCT [ACI 318M-14, Table 23.9.2] [1]
A8ƒ = 204.83 ∗ 200 =40966 mmm
{|}s • Žqq••, Figure7 − b.
F8• = 0.85βn. f<(.A8•
A8• = w34 ∗ bw
A8• = 246.49 ∗ 200 = 49298mmm
F8• = 0.85 ∗ 0.8 ∗ 30 ∗ 49298 = 1005.67 kN
F8• > F843 o. k
σC• =
2F843 sin ϑ
200 ∗ 200
=
887.4 ∗ 1000
200 ∗ 200
= 22.19 MPa
F<‹• = 0.85βn. f<(
F<‹• = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa
σC• > F<‹•
‘
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
{|}s q ”•••
••–— ˜ , Figure7 − C.
F8™ = 0.85βn. f<(.A8™
A8™_ = w3`_ ∗ bw = 34070 mmm
F8<_ = 0.85 ∗ 1 ∗ 30 ∗ 34070 = 868.79 kN > F8_3 o. k
A8™4 = w3`4 ∗ bw = 39500mmm
F8<4 = 0.85 ∗ 0.8 ∗ 30 ∗ 39500 = 805.8 kN > F843 o.k
σC™ =
PF/2
area of plate
=
707.6 ∗ 1000
200 ∗ 200
= 17.7 MPa
F<‹™ = 0.85βn. f<(
F<‹™ = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa
σC™ < F<‹™
The dimension of plate it is o. k
βn = 1 for CCC [ACI 318M-14, Table 23.9.2] [1]
•xs ›
F3K = A3K ∗ Fy = 6 ∗
π
4
16m
∗ 500
= 603.19 kN > F8_3 cos ϑ , F843 cos ϑ O. K
•xs Ÿ
F3J = A3J ∗ Fy = 6 ∗
π
4
16m
∗ 500 = 603.19 kN
> F843 cos ϑ
− F8_3 sin ϑ O. K
P = 2 F8_3 sin ϑ + F843 sin ϑ
= 2 574.03sin 27.37 + 679.32 sin40.78
Load Failure
=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 > ≤ 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.

Ali Mustafa Jalil, Mohammed j. Hamood, Khattab Saleem Abdul
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
Mohammed j. Hamood, Khattab Saleem Abdul-Razzaq and Abbas H. Mohammed
IJCIET/index.asp 2761 editor@iaeme.com
The principal stress paths and the assumed truss forthe specimen CD.2F
STM Flow chart for reinforced concrete continuous deep beams subjected to decentralized
2-concentrated forces.
Razzaq and Abbas H. Mohammed
editor@iaeme.com
The principal stress paths and the assumed truss forthe specimen CD.2F
STM Flow chart for reinforced concrete continuous deep beams subjected to decentralized

Table 3. Numerical Example No. 2, two decentralized concentrated forces
Input data
h=800mm >? =2000mm d=723.5mm @ =1000mm @ =500mm )- =200mm
f'c=30
MPa
Bearing
Plates=150* 200
mmfor external
support
Bearing
Plates=200* 200
mmfor internal
support and
appliedload
fA=500
MPa
Vertical web
reinf.= B8mm@
250mm c/c
fAC =450
MPa
Horizontal
web
reinf.= B8mm
@250
mm c/c
fAD =450
MPa
Bottom and
upper
covers=
40mm
Main longitudinal top and bottom reinforcement
=6B16mm
Output data
h = jd +
wtJ
2
+
wtK
2
jd = a tan ϑ
ϑ = tanMN
jd
a
wtK = OCc + ∅st. +∅main +
spacing
2
X ∗ 2
= 40 + 8 + 16 + 12.5 ∗ 2 = 153 mm
or
wtK = h − d ∗ 2 = 800 − 723.5 ∗ 2 = 153 mm
wtK = wtJ = 153 mm
becouse number of lyer for top and bottom is equal
Draw STM of continuous deep beams CD.2F, see Figure11.
jd = h −
wtJ
2
−
wtK
2
= 800 − 153 = 647mm
ϑ = tanMN
647
1000
= 32.9°
ϑ = tanMN
647
500
= 52.3°
w3_ = wtK cos ϑ + lse sinϑ
w3_ = 153 cos 32.9°
+ 150 sin 32.9°
= 209.94 mm
w34 = wtK cosϑ + lsi sin ϑ
w34 = 153 cos 52.3°
+ 200 sin 52.3°
= 251.81mm
w3`_ = wtJ cos ϑ + lpe sinϑ
w3`_ = 153 cos 32.9°
+ 200 sin32.9°
= 237.1 mm
w3` = wtJ cosϑ + lpi sinϑ
w3` = 153 cos 52.3°
+ 200 sin52.3°
= 251.81 mm
w_3 =
w3_
2
+
w3`_
2
=
209.94
2
+
237.1
2
= 223.52 mm
w43 =
w34
2
+
w3`4
2
=
251.81
2
+
251.81
2
= 251.81 mm
F8_3 = 0.85βs. f<(.b. w_3
F843 = 0.85βs. f<(.b. w43
βs = 0.75 when Q ≥ 0.003
βs = 0.6λ when Q < 0.003
f =
,g
) ∗ h
h i1 +
,j
) ∗ h
h i2 =
2 ∗
k
l
∗ 8m
200 ∗ 250
32.9°
+
2 ∗
k
l
∗ 8m
200 ∗ 250
90 − 32.9°
= 0.0027 < 0.003
f =
,g
) ∗ h
h i1 +
,j
) ∗ h
h i2 =
2 ∗
k
l
∗ 8m
200 ∗ 250
52.3°
+
2 ∗
k
l
∗ 8m
200 ∗ 250
90 − 52.3°
= 0.0028 < 0.003
f , f < 0.003 → % = 0.6 →
F8_3 = 0.85 ∗ 0.6 ∗ 30 ∗ 200 ∗ 223.52 = 683.97 kN
F8_3 cos ϑ = 574.27 kN
F843 = 0.85 ∗ 0.6 ∗ 30 ∗ 200 ∗ 251.81 = 770.54 kN
F843 cos ϑ = 471.21 kN
[ACI 318M-14, Table 23.4.3] [1]
qrstu qvwvtxyz
…|†x‡|€yv• ¡y†¢y
F£¤ = 0.85βs.f<(.A£¤
A£¤ = 153 ∗ 200 = 30600 mmm
F£¤ = 0.85 ∗ 1 ∗ 30 ∗ 30600 = 780.3 kN
F£¤ > F8_3 cos ϑ o. k
F£¤ + ¥ ¦2 > F843 cosϑ o. k
βs = 1[ACI 318M-14, Table 23.4.3] [1]

Figure 10. Details of the specimen CD.2F
{|}s ~ qq• , Figure12 − a.
F8ƒ = 0.85βn. f<(.A8ƒ
F8ƒ = 0.85 ∗ 0.8 ∗ 30 ∗ 41988 = 856.56 kN
F8ƒ > F8_3 o. k
σCƒ =
F8_3 sinϑ
150 ∗ 200
=
371.52 ∗ 1000
150 ∗ 200
= 12.38 MPa
F<‹ƒ = 0.85βn. f<(
F<‹ƒ = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa
F<‹ƒ > σCƒ The dimension of plate it is o. k
βn = 0.8 for CCT ŽACI 318M-14, Table 23.9.2• Ž1•
A8ƒ = w3_ ∗ bw
A8ƒ = 209.94 ∗ 200=41988
{|}s • Žqq••, Figure12 − b.
F8• = 0.85βn. f<(.A8•
A8• = 251.81 ∗ 200 = 50362mmm
F8• = 0.85 ∗ 0.8 ∗ 30 ∗ 50362 = 1027.38 kN
F8• > F843 o. k
σC• =
2F843 sinϑ
200 ∗ 200
=
1219.34 ∗ 1000
200 ∗ 200
= 30.5 MPa
F<‹• = 0.85βn. f<(
F<‹• = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPa
σC• > F<‹•
‘
increase dimension of plate or
using nodal reinforcement
to prevent premature failure
“
A8• = w34 ∗ bw
{|}s q Žqq••, Figure12 − c.
F8™ = 0.85βn. f<(.A8™
F8< = 0.85 ∗ 0.8 ∗ 30 ∗ 50362 = 1027.18 kN > F843
σC™ =
PF/4
area of plate
=
490.6 ∗ 1000
200 ∗ 200
= 12.3 MPa
F<‹™ = 0.85βn. f<(
F<‹™ = 0.85 ∗ 0.8 ∗ 30 = 20.4 MPaσC™ < F<‹™
A8™ = w3`4 ∗ bw
A8™ = 251.81 ∗ 200 = 50362 mmm
{|}s © Žqqq•, Figure12 − d.
F8ª = 0.85βn. f<(.A8«
A8ª = w3`_ ∗ bw
F8<4 = 0.85 ∗ 1 ∗ 30 ∗ 47420 = 1209.21 kN > F8_3 o.k
σC« =
PF/4
area of plate
=
490.6 ∗ 1000
200 ∗ 200
= 12.3 MPa
F<‹« = 0.85βn. f<(
F<‹™ = 0.85 ∗ 1 ∗ 30 = 25.5 MPa
σC« < F<‹«
βn = 1 for CCC ŽACI 318M-14, Table 23.9.2• Ž1•
A8« = 237.1 ∗ 200 = 47420 mmm
βn = 1 for CCC [ACI 318M-14, Table 23.9.2] [1]
•xs ›
F3K = A3K ∗ Fy = 6 ∗
π
4
16m
∗ 500
= 603.19 kN > F8_3 cosϑ , F843 cos ϑ O. K
•xs Ÿ
F3J = A3J ∗ Fy = 6 ∗
π
4
16m
∗ 500 = 603.19 kN
> F843 cosϑ O. K 780.3 kN + 603.19 kN
P = 2 F8_3 sin ϑ + F843 sinϑ
= 2 683.97sin32.9 + 770.54sin 52.3
= 1962.4 kN
Load Faliure

Applying Different Decentralized Loadings
Figure 11. Strut – Tie model for
a) Faces of supportnodal zone A, CD.2F
c) Faces of support nodal zone C, CD.2F
Figure 12. Nodes in 2-
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 tha
should equal to the uniformly distributed load in value [18
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 th
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
of support nodal zone C, CD.2F d)Faces of load nodal zone D, CD.2F
-concentrated forces loaded each span ofthe specimen CD.2F
Uniformly Distributed Load
Tie model for uniformly distributed loaded continuous beam specimen CD.UL
distributed load, it could be considered as deep beam under two concentrated forces tha
should equal to the uniformly distributed load in value [18-20]. Figure13. shows the principal
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
n RC Continuous Deep Beams Using STM
editor@iaeme.com
concentrated forces loadedin each span ofthe specimen CD.2F
b) Faces of support nodal zone B, CD.2F
d)Faces of load nodal zone D, CD.2F
concentrated forces loaded each span ofthe specimen CD.2F
Tie model for uniformly distributed loaded continuous beam specimen CD.UL
distributed load, it could be considered as deep beam under two concentrated forces that
20]. Figure13. shows the principal
span deep beam subjected to a decentralized uniformly distributed load. It
e 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:
P = 2 F8_3 sinϑe + F843 sin ϑi = 2 683.97 sin 32.9 + 770.54 sin52.3 = 1962.4 kN
W = P / L® = 1 m for two span , ∴ W =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.

Ali Mustafa Jalil, Mohammed j. Hamood, Khattab Saleem Abdul
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
33.99 40.13 48
Load(kN)
Internal Angle (degree)
Load(kN/m)
Table 4. Effect of load decentralization on ultimate capacity
Specimen
Group
Load
Type
Shear Span
(mm)
A1
A
Single
Concentrated
Force
1250 1250
A2 1500 1000
A3 1750 750
A4 2000 500
B1
B
Two
Concentrated
forces
1000 1000
B2 1250 750
B3 1500 500
B4 1750 250
C1
C
Uniformly
Distributed
Load
1000 1000
C2 1250 750
C3 1500 500
C4 1750 250
Mohammed j. Hamood, Khattab Saleem Abdul-Razzaq and Abbas H. Mohammed
Figure 16. Effect of internal angle on the
ultimate capacity of specimens in group
Effect of internal angle on the ultimate
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
Load(kN)
Internal Angle (degree)48.34 59.33
0
500
1000
1500
2000
2500
3000
3500
40.13 48.34 59.33 73.48
Effect of load decentralization on ultimate capacity
Angle
(degrees)
Ultimate capacity
°±/d
²e ²i ³´(kN)
Wu
(kN/m)
33.99 33.99 1968.5 - 1.36
29.34 40.13 2232.5 - 1.63
25.72 48.34 2421.4 - 1.90
22.86 59.33 2655.7 - 2.17
40.13 40.13 2716.3 - 1.09
33.99 48.34 3043 - 1.36
29.33 59.33 3183.6 - 1.63
25.72 73.48 3280.2 - 1.90
40.13 40.13 - 2716.3 1.09
33.99 48.34 - 3043 1.36
29.33 59.33 - 3183.6 1.63
25.72 73.48 - 3280.2 1.90
Razzaq and Abbas H. Mohammed
editor@iaeme.com
Effect of internal angle on the
ultimate capacity of specimens in group B
Effect of internal angle on the ultimate capacity of specimens in group C
59.33 73.48
Effect of load decentralization on ultimate capacity
d
°µ/d
²
%
Increasing
in
1.36 1 Reference
1.09 1.37 13.41
0.81 1.88 23.00
0.54 2.59 34.91
1.09 1 Reference
0.81 1.42 12.03
0.54 2.02 17.20
0.27 2.86 20.75
1.09 1 Reference
0.81 1.42 12.03
0.54 2.02 17.20
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
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-Cons.Force(kN)
1900
2000
2100
2200
2300
2400
2500
2600
2700
30 40 50 60 70
1-Cons.Force(kN)
2600
2700
2800
2900
3000
3100
3200
3300
3400
30 40 50 60 70 80
UniformDist.Load(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–
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[3] Rasheed, Mohammed M. "Modified Softened Strut and Tie Model for Concrete Deep
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[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
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[10] Abdul-Razzaq, Khattab Saleem, and Sarah FarhanJebur. "Suggesting Alternatives for
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[13] Peter Marti. “Basic Tools of Reinforced Concrete Beam Design.” ACI Journal
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[14] Comité Euro-International du Beton CEB-FIP. “CEB-FIP model code 1990 for concrete
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[19] Sam-Young, Noh, Lee Chang-Yong, and Lee Kyeong-Min. "Deep Beam Design Using
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