Analysis of frames|www.BuildingHow.com

VOLUME A
The art of construction and detailing
Introduction
The structural frame
- Introduction
- Structural frame elements
  - Columns
  - Beams
  - Slabs
  - Staircases
  - Foundation
- STRUCTURAL FRAME LOADING
- BEHAVIOR OF THE STRUCTURAL FRAME
  - The behavior and reinforcement of a slab
  - Behavior and reinforcement of beams and columns
Construction methods
- The materials
- The moulds
- Thermal insulation of structural elements
- Concrete
- The steel
- Reinforcement specifications of antiseismic design
- Industrial stirrups - stirrup cages
- Standardized cross-sections of reinforced concrete elements
Reinforcement
- Columns
- Shear walls
- Composite elements
- Beams
- Slabs
- Staircases
- Foundation
Quantity surveying
- Estimation of the concrete’s quantity
- Estimation of the formworks’ quantity
- Estimation of the spacers’ quantity
- Estimation of the reinforcements' quantity
- Total estimation of the materials’ quantities
- Optimization of the reinforcement schedule
- Estimation of the structural frame’s cost
- Electronic exchange of designs - bids - orders
Detailing drawings
- General
- The drawings’ title block
- Carpenter’s drawings
- Steel fixer’s drawings
Tables
- Table 1
- Table 2
- Table 3
Drawings
- Carpenter’s drawings
- Steel fixer’s drawings
Model (exemplary) construction

« Analysis of slab The effect of torsional stiffness on indirect beam-to-beam supports »

Analysis of frames

For both frames of the above structure, the calculation of the stress resultants subjected to static loads (dead and live) and seismic loads is required.

Each frame has a span of 5.0 m and a height of 3.0 m. The moment of inertia of each column bar is I=0.4x0.4³/12=0.0021333 m⁴=21.33x10^-4m⁴.

Τhe effective flange width of the beams is calculated according to §3.1.2: l_o=0.70·l=0.70x5.00=3.50 m, b₁=l _n1/2=3.70/2=1.85 m, b_lim=b_w+b₁=0.30+1.85=2.15 m,

b_eff=min(b_w+b_eff,1, b_lim)=min(0.30+0.70, 2.15)=1.00 m, since b_eff,1=min(0.20·b ₁+0.10·l_o, 0.20·l_o)= min(0.20x1.85+0.10x3.50, 0.20x3.50)=min(0.72 , 0.70)=0.70 m.

From table 2 b_eff=1.0 m, h_f/h=0.17/0.50=0.34, b_w/b_eff=0.30 → μ=0.0417 → I=0.0417x1.0x0.50³=52.125x10 ^-4 m⁴.

Note

According to [EC8, §4.3.1(6),(7)] the stiffnesses of structural elements (beams, columns, walls) are considered as the 50% of their respective elastic ones due to cracking. Provided that stiffnesses of all structural elements are reduced by the same factor, the stress resultants are equal to those of the elastic analysis (i.e. of full stiffnesses). However the deformations of cracked elements are equal to 50% of the respective elastic deformations.

The loads on the crossbar of each frame are:

Self-weight: g_o=0.30mx(0.50-0.17)mx25.00kN/m³= 2.50 kN/m

Dead loads on slab: g_s=2.00mx5.25kN/m²=10.50 kN/m

Additional dead loads: g_w=20.00 kN/m

Total dead loads: g =33.00 kN/m

Live load on slab: q=2.00mx5.00kN/m²=10.00 kN/m

Τhe self-weight of each column is 0.40x0.40x3.00x25.00=12.00 kN. Τhe 1/3 of this load (4.00 kN) is applied on the column head and the rest 2/3 of the load (8.00 kN), are applied on the column base.

The dynamic mass of each crossbar in all seismic combinations corresponds to design load:

P_d=(g+ψ₂·q) ·l+2columnsx4.00kN=(33.00+0.30x10.00)x5.00+8.00=188.00 kN

while the seismic force applied on each crossbar is W= (a_x /g)·P_d=0.65x188.00=122.00 kN.

In the following analyses implemented manually, the function of rigid bodies will not be considered. In each frame three analyses are performed, one for dead loads G, one for live loads Q and one for seismic loads in +X direction.

Figure 3.3.3-1: The structural model of the frame considering loadings: g and q and the seismic W

The general analysis of the frame subjected to vertical uniform load w, from table 1 is:

The general analysis of the frame subjected to horizontal load W, from table 39 is:

For load g=33.0 kN/m:

M_A,g=-M_B,g =0.60m² x33.0kN/m=19.8 kNm

M_C_A_,g =M_C_D_,g =M_D_C_,g =-M_D_B_,g =-1.20m² x33.0kN/m =-39.6 kNm,

H_A_,g =-H_B_,g =0.60m x33.0kN/m=19.8 kN, V_A,g =V_B,g =2.50mx33.0kN/m=82.5 kN

For load q=10.0 kN/m:

M_A_,_q=-M_B_,_q =0.60m² x10.0kN/m=6.0 kNm, M_C_A,_q =M_C_D,_q =M_D_C,_q =-M_D_B,_q =-1.20m²x10.0kN/m=-12.0 kNm,

H_A,_q =-H_B,_q =0.60m x10.0kN/m=6.0 kN, V_A_,_q =V_B_,_q =2.50mx10.0kN/m=25.0 kN

For load W=122.0 kN:

M_A_,_W=M_B_,_W =-0.826m x122.0kN=-100.8 kNm, M_C_A,_W =M_C_D,_W =-M_D_C,_W =M_D_B,_W =0.674x122.0=82.2 kNm,

H_A_,W =H_B_,W =-0.50 x122.0kN=-61.0 kN, V_A,W =-V_B,W =-0.27x122.0kN=-32.9 kN

The actions examined previously will be used for the calculation (a) of combinations of actions without seismic loads according to §2.3.1 and (b) of the combination of actions with seismic loads according to §2.3.2.

(a) Combination of actions without seismic loads:

The only combination examined is ο γ_g·G+ γ_q·Q

(b) All seismic combinations are examined when the gravity loads are

G+ ψ₂·Q

The two seismic combinations examined in this case, are in +x direction and in -x direction

Therefore, the total three combinations are [*] Note Noteaccording to 6.4 there are more combinations required by ec8, but those three are the important ones. :

1^st combination: γ_g·G+ γ_q·Q

2^nd combination: G+ ψ₂·Q + "Ε_x" [*] Note NoteThe results of earthquake acting in x direction is symbolized as Ex?. In general, there are components resulted in both x and y directions. In this example, the results are only in x direction.

3^rd combination: G+ ψ₂·Q - "Ε_x"

In this example ψ₂=0.30 based on paragraph §2.2.3.4, γ_g=1.35 and γ_q=1.50. Therefore the combinations are expressed as follows.

1^st combination : w=1.35g+1.50q=1.35x33+1.50x10.0=59.55 kN/m

M_A=-M_B=1.35M_A,g+1.50M_A,q=1.35x19.8+1.50x6.0=35.7 kNm

M_C_A=M_C_D=M_D_C=-M_D_B=1.35M_CA,g+1.50M_CA,q=-1.35x39.6-1.50x12.0=-71.5 kNm,

H_A=-H_B=1.35H_A,g+1.50H_A,q=1.35x19.8+1.50x6.0=35.7 kN,

V_A=V_B=1.35H_A,g+1.50H_A,q =1.35x82.5+1.50x25.0=148.9 kN

V_CD=w·l/2+(M_DC-M_CD)/l=59.55x5.0/2+(-71.5+71.5)/5.0=148.9 kN,

V_DC=V_CD-w·l=148.9 -59.55x5.0=-148.9 kN [*] Note NoteThis stress resultant, as well as most of the following, could be calculated by the simple observation of the symmetry both of structure and loading. However this general process is handling asymmetries as well, e.g. as in the 2nd loading.

N_Α=-V_CD-1.35·(self-weight of column)=-148.9-1.35x12.0=-165.1 kN

N_B=V_DC- 1.35·( self-weight of column)=-148.9-1.35x12.0=-165.1 kN

x=V_CD/w=148.9/59.55=2.50 m [*] Note Notex is the point of zero shear force corresponding to position of maximum bending moment. , M_max= M_CD + (V_CD·x)/2=-71.5+(148.9x2.50)/2=114.6 kNm [*] Note NoteFrom structural analysis it is known that the maximum bending moment M_max in a span i,j is located at the point m at distance x from the end i, where shear forces become zero. The moment at that point is given by the expression M_max=Mi,j + A_V where A_V is the area under the shear forces diagram from the point i to the point m. In this example, since there is only uniform load, the area is A_V=(V_i,j · x)/2 ,

w·l²/8=59.55x5.0²/8=186.1 kNm

Figure 3.3.3-2 Figure 3.3.3-2	Figure 3.3.3-3 Figure 3.3.3-3
Figure 3.3.3-4 Figure 3.3.3-4	Figure 3.3.3-5 Figure 3.3.3-5

2^nd combination : w=g + 0.30 · q + "Ε_x"=33.0+0.30x10.0 + "E_x"=36.0 kN/m + "E_x"

M_A=M_A,g+0.30M_A,q+ M_A,W =19.8+0.30x6.0-100.8 =-79.2 kNm, M_B =-19.8-0.30x6.0-100.8 =-122.4 kNm

M_C_A= M_C_D=-39.6-0.30x12.0+82.2=39.0 kNm,

M_D_C=-39.6-0.30x12.0-82.2=-125.4 kNm, M_D_B=39.6+0.30x12.0+82.2=125.4 kNm,

H_A=H_A,g+0.30H_A,q+ H_A,W =19.8+0.30x6.0-61.0=-39.4 kN, H_B=-19.8-0.30x6.0-61.0=-82.6 kN,

V_A= V_A,g+0.30V_A,q+ V_A,W =82.5+0.30x25.0-32.9=57.1 kN, V_B=82.5+0.30x25.0+32.9=122.9 kN

V_CD=w·l/2+(M_DC-M_CD)/l=36.0x5.0/2+(-125.4-39.4)/5.0=90.0-33.0=57.0 kN,

V_DC=V_CD-w · l=57.0-36.0x5.0=-123.0 kN

N_Α=-V_CD- self-weight of column =-57.0-12.0=-69.0 kN

N_B=V_DC- self-weight of column =-123.0-12.0=-135.0 kN

x=V_CD/w=57.0/36.0=1.58 m, M_max= M_CD + (V_CD·x)/2=39.0+(57.0x 1.58)/2=84.0 kNm,

w·l²/8=36.0x5.0²/8=112.5 kNm

Figure 3.3.3-6 Figure 3.3.3-6	Figure 3.3.3-7 Figure 3.3.3-7
Figure 3.3.3-8 Figure 3.3.3-8	Figure 3.3.3-9 Figure 3.3.3-9

3^rd combination : w=g + 0.30·q - "Ε _x "=33.0+0.30x10.0 - "E_x"=36.0 kN/m - "E_x"

The values of stress resultants are mirror values of those of 2^nd combination.

Figure 3.3.3-10 Figure 3.3.3-10	Figure 3.3.3-11 Figure 3.3.3-11
Figure 3.3.3-12 Figure 3.3.3-12	Figure 3.3.3-13 Figure 3.3.3-13

Combinations and Envelopes of stress resultants

Figure 3.3.3-14 Figure 3.3.3-14	Figure 3.3.3-15 Figure 3.3.3-15
Figure 3.3.3-16 Figure 3.3.3-16

Calculations performed by using the related software (project <B_331>):

Figure 3.3.3-17: The model of the structure, created by the software, consists of slabs, nodes, bars, rigid bodies and diaphragms

The moments and shears of the slab are displayed by pressing the corresponding buttons on the left of the screen. The results are the same with the theoretical results of the analysis of the previous paragraph.

The model of the structure, created by the software, consists of bars, located on the centroidal axes of the columns and on the central axes at the top faces of the beams. The column-to-beam connections are modelled by rigid bodies, placed along the direction of beams.

The diaphragm of the slab connects the two frames.

The Interface allows the user to see the numbering of master and slave nodes. Moreover the user can select a node, or a bar and retrieve their geometrical and loading data. Most of their properties can be modified by selecting 'Properties' from the menu [*] Note NoteThe file, frame.sfd (which can be modified optionally by the user) contains the values of the interface passed to the solver.

Τhe software displays the effective area of the slab on each frame, the total dead linear load g=34.5 kN/m and the total live linear load q=10.8 kN/m. The slightly augmented values of g and q used by the software are due to corrections made for taking into account the slab loads that extend beyond both frame ends.

Τhe results of the three combinations from the software are given in the following diagrams of bending moments and shear forces (the elastic line is also shown).

The small discrepancies between practical and automatic (via the software) calculations are due to the fact that the software uses rigid bodies and takes into account work done by shears and axial forces.

1^st combination : 1.35g+1.50q