AStrutTie Design Example – 1. Design of deep beams subjected to concentrated load (3)

1.3 Design example – strut-tie model representing vertical truss mechanism

1.3.1 Problem statement

In this section, the deep beam is designed by the strut-tie model representing the vertical truss load transfer mechanism shown in Fig. 1-1(b). As illustrated in the previous sections, the dimensions and the information on the bearing plates and loads are determined in the same way.

 

1.3.2 Construction of strut-tie model

The ‘Beginning Mode’ is switched to the ‘Modeling Mode’ to construct a strut-tie model for the deep beam. As the shear span-to-effective depth ratio of the deep beam is 1.27, the indeterminate strut-tie model that represents a combined load transfer mechanism is selected automatically from the template for deep beam. In this example, the strut-tie model representing an vertical truss load transfer mechanism as shown in Fig. 1-7 is constructed for design. The top and bottom horizontal elements are placed 127 mm and 102 mm away from the top and bottom surfaces of the deep beam, respectively.

Fig070

Fig. 1-7 Constructed strut-tie model

 

1.3.3 Analysis of strut-tie model

The cross-sectional forces of struts and ties are determined by carrying out the finite element analysis of the constructed strut-tie model.

Fig080

Fig. 1-8 Strut and tie forces

 

1.3.4 Strength verification and required rebars

1.3.4.1 Strength under bearing plates

The strength conditions under bearing plates are examined by the method illustrated in Section 1.2.4.1.

 

1.3.4.2 Required area of rebars

The required areas of main reinforcing bars are determined by the following equation. The requirement on the reinforcing bars is examined in the ‘Design Review’ as shown below.

As,req = Fuφ fy                                                        (1-7)

where,
φ = strength reduction factor of steel tie (= 0.75)
Fu = cross-sectional force of steel tie
fy = yield strength of steel (= 414 MPa)
Fig081
The spacing of shear reinforcing bars is determined by the following equation.

φFn = (φ Av fy weff,tie ) / sv Fu                                           (1-8)

where,
φ = strength reduction factor of steel tie (= 0.75)
Av = area of vertical shear reinforcement
fy = yield strength of steel (= 414 MPa)
weff,tie = effective width of vertical tie (The half of the shear span is a default. A user can define the width in ‘Assign-Outer Element’)
Fu = cross-sectional force of steel tie

Fig082

 

1.3.4.3 Strength verification of struts

The strength condition of a concrete strut is verified by comparing the required width with provided width of the concrete strut, as shown below.
Fig083
The strength conditions of concrete struts can also be verified visually as shown below.
Fig090

Fig. 1-9 Required/proposed area of concrete strut

 

1.3.4.4 Strength verification of nodal zones

The strength condition of a nodal zone is verified by comparing the required width with the provided width of the nodal zone boundary, as shown below.

Fig091

 

1.3.5 Minimum rebars for crack control

Since the effective strength coefficient 0.75 was taken for the two inclined struts of the strut-tie model, the ACI 318M-14 requirement for minimum reinforcing bars for crack control must be satisfied.

∑ (Asi / bssi) sin γi ≥ 0.003                                                    (1-9)

where Asi is the total area of distributed reinforcement at spacing si in the i-th direction of reinforcement crossing a strut at an angle αi to the axis of a strut, and bs is the width of the strut.

Fig092


Relative Contents

AStrutTie Design Example – 1. Design of deep beams subjected to concentrated load (1)

AStrutTie Design Example – 1. Design of deep beams subjected to concentrated load (2)

AStrutTie Design Example – 1. Design of deep beams subjected to concentrated load (3)

AStrutTie Design Example – 1. Design of deep beams subjected to concentrated load (4)


Example File

STM Example_Deepbeam_Vertical Truss Mechanism_SI Unit

Be the first to comment

Leave a Reply