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Eres Opbul, Dmitrii Dmitriev, Phan Van Phuc — Pages 29-41 PRACTICAL CALCULATION OF FLEXIBLE MEMBERSWITH THE USE OF NON-LINEAR DEFORMATION
MODEL AS EXEMPLIFIED BY TYPICAL GIRDER RGD 4.56-90 DOI: 10.23968/2500-0055-2018-3-3-29-41
PRACTICAL CALCULATION OF FLEXIBLE MEMBERS WITH THE USE OF NON-LINEAR DEFORMATION MODEL AS EXEMPLIFIED BY TYPICAL GIRDER RGD 4.56-90
Eres Opbul1, Dmitrii Dmitriev2, Phan Van Phuc13
1 Saint Petersburg State University of Architecture and Civil Engineering 2-ya Krasnoarmeiskaya st., 4, St. Petersburg, Russia
2 "TEKTON-SPB" LLC
Oktyabrskaya nab., 104, St. Petersburg, Russia
3 Vinh University
182 Le Duan, Ben Thuy, Vinh City, Vietnam 1 [email protected] Abstract
The article is devoted to practical calculation of strength of flexible members made on the basis of non-linear deformation model taking into account experimental diagrams of materials deformation. The relative deformations are determined in the proposed calculation depending on maximum flexion of a member. The calculation of maximum flexion of a member will be performed using two iteration methods. The basic calculation formula, original block diagrams of methods and comparison of calculation parameters proceeding from results of practical calculations made as exemplified by typical girder are provided.
Keywords
Deformation model, non-linear calculation, deformation diagram, iteration, stiffness matrix, stress diagram, neutral line, reinforcement.
Introduction
Obviously, the calculation of construction structures with the use of non-linear deformation model (NLDM) qualifies as check-and-control or testing method. In this case the calculation according to NLDM will be performed with definite force (external or internal) values, reinforcement (of concrete) and member geometrical dimensions (Boujelben, Ibrahimbegovic, 2017; Jagtap et al., 2018).
This article presents two methods of iteration calculations. The purpose of iteration process consists in determining maximum flexion of a member and location of relative deformations in target points of sections
depending on the target flexure (Patni, 2018; Mao, Zhang, 2018).
The target flexure is determined in the first method with finding a new position of neutral line, in the second method it depends on the matrix of stiffening behavior of every small area.
The first iteration method of calculation with respect to steel fiber reinforced-concrete structures has been described in the works (Morozov, Opbul, 2016; Opbul et al., 2017). Described below is the second method of nonlinear calculation with due account of paragraphs 8.1.208.1.29, 3.72-3.75 given in standards (TsRIIPromzdaniy OJSC, 2005; Minstroy RF, 2013) (In order to contract the
volume of article, the first three and the last (eleventh) iterations of calculations only are given).
Diagrams of materials deformation
A typical girder RGD 4.56-90: concrete B40, reinforcement A400 are being considered. Ref. Fig. 3 for girder geometrical dimensions and reinforcement.
The calculation formulae have been acquired according to (TsRIIPromzdaniy OJSC, 2005; Minstroy RF, 2013) and shown in Fig. 1, 2 to determine stresses and stress-strain moduli.
Judgments and assumptions: plane sections hypothesis is true; for determining center of gravity of the given girder cross-section and for the first method of iteration calculation the girder material works in the elastic stage (there are no cracks, the moduli of small areas equal the initial modulus of elasticity).
a) Calculation formulae of stresses and moduli for concrete of class B40
In case of concrete compression:
0 <sbi < 150• 10-
a* = Rb
150•lO-
MPa ;
Figure 2. Diagram of Reinforcement Condition
0<e^< 1.775-10-3 : asi = R, — = 355•
1.775 -10-
-MPa
where Ss1 = ~r =
Rs- = -355- = 1.775 •10-3 E = E E 2-105 s s
• 1.775-10-3 <s. < 25-10-3 : a. = R = 355 MPa,
si si s
E ' = R = — ,MPa
Figure 1. Diagram of Concrete Condition
E,. = const = -
R
= 14666.7 MPa
15040-
• 150 • 10-5 < sbi < 350 -10-5 : a = Rb = 22 MPa ;
Eb. =* = —MPa
Sbi Sbi
In case of concrete expansion:
0 <sbti < 8 -10-5 : abti = Rb, •
8 40-
E - Rbt
' 'I
1.4
b 8 40-5 8-10-5
= 17500 MPA
8 • 10-5 < sbti < 15 • 10-5 : oUi = Rbt = 1.4 MPa ;
, 1.4
Eb,i= — MPa Sbti
Basic calculation formulae and block diagram of the first method
A) Center of gravity of a given cross section:
S.
red
red
B) Reduced static moment:
Sred = X Ajbi A,ysl
(1)
(2)
where Ab,, Asi - area of section of i-n small area (layer) of concrete and i-n reinforcement, accordingly; yb,, ys-distances from extreme stretched fiber, accordingly, to the center of gravity of i-n small area of concrete and i-n reinforcement, a - modular ratio of reinforcement into concrete.
C) Reduced cross-section area:
Ad =Tbbl A, +X a • A,
(3)
where bbi Aw - width and height (thickness) of /-n concrete small area, accordingly. D) New position of neutral line (NL):
^0 j
X^iA^i + -A -y* ZKiAi+XKAi
(4)
b) Calculation formula of stresses and moduli for reinforcement in case of expansion and contraction:
where, E, ,E . - moduli of concrete and reinforcement
' bn si
deformation.
E) Member flexure:
1 _ M __M_
r. ~ EI ■ ' ■ ^ „ . , . ,2 (5)
X E'bAi (yi ) +X EA (ysi )
5 .
5
5
Eres Opbul, Dmitrii Dmitriev, Phan Van Phuc — Pages 29-41 PRACTICAL CALCULATION OF FLEXIBLE MEMBERSWITH THE USE OF NON-LINEAR DEFORMATION
MODEL AS EXEMPLIFIED BY TYPICAL GIRDER RGD 4.56-90 DOI: 10.23968/2500-0055-2018-3-3-29-41
Figure 3. Block Diagram of the First Method
Figure 4. Block Diagram of the Second Iteration Method
Eres Opbul, Dmitrii Dmitriev, Phan Van Phuc — Pages 29-41 PRACTICAL CALCULATION OF FLEXIBLE MEMBERSWITH THE USE OF NON-LINEAR DEFORMATION
MODEL AS EXEMPLIFIED BY TYPICAL GIRDER RGD 4.56-90 DOI: 10.23968/2500-0055-2018-3-3-29-41
where, ybi = yaj - > ysi = y0j - ysi - arm of force couple for small areas of concrete and reinforcement, accordingly.
F) Condition when the maximum design flexure occurs
1
j,calc
8 =
1 1
rj rj-i
100% < 1%
(6)
'j-1
G) Deformations
1 , 1 ,
s = — y-; s. = — y
bi s bi? si s si
r r.
j j
(7)
H) Condition of strength, when compliance is necessary:
• with respect to deformations:
Sbi,calc <\Sb2 ]; Ssi,calc < [ss2 ]
(8)
where ^i, = ——y„i = _ ——ysi - design
rj,cafc ^j ,caic
deformations; [^b2], [es2] - permissible deformations
according to [1].
• with respect to forces according to formula 3.144:
Mcalc _ Xab,Ab,yb,+Xas,As,ys, > Mult
(9)
Figure 3 shows an original block diagram of the first calculation method acquired in the way of iteration.
Calculation Formulae and Block Diagram of the Second Method
A) We use calculation equation 8.26^8.52 (Morozov, Opbul, 2016), where three stiffness characteristics only get defined in case of pure flexure:
Inverse matrix A-1 =-
det A
-x A
(12)
where A =
A11 A12
A A
21 22
matrix of cofactors equal to:
(13)
A11 = (-1)1+1 x A3 = D33 A\2 = (-1)'+2 x = -A,
A21 = (-1)2+' x Dn =-D.3 A22 = (-1)2+2 x Dn = Dn
C) Product of matrices gets calculated in the form of:
A-' x
M X1
x =
0 X2
where x1 =—; x2 =s0 Pj
(14)
(15)
D) Relative deformations in concrete and reinforcement get determined according to formulae 8.29, 8.30 [1] taking into account formulae (16):
sbi=s0 +-
1_ Pj
- ybi
(16)
Zs,_Zo +--ysi
Pj
E) Strength condition:
• with respect to deformations - ref. formula (8).
• with respect to forces according to formula 3.144
[2]:
Mcalc = XaH AiiyH- +Xasi Asiysi * Mul
(17)
Figure 4 shows an original block diagram of the second iteration method.
Determining girder load-carrying ability
Basic data:
Class of concrete B40:
Dn = Z KA ( y'br )2 +Z EA ( yl )2
i i
D13 = D31 = Z EbiAbiy'bi +Z EsiAsiysl i i
D33 = ZÉ,.Ah. +ZE A
33 b b s s
(10)
B) Further, matrix determinant (det A) of
A =
Du A3
D31 D33
type will be defined, wherefrom:
det A = Djj • D33 - D31 • D13
(11)
Figure 5. Girder Transversal Section
1
Rb = 22MPa; Rbt = 1.4MPa;
Eb = 36000MPa
Longitudinal reinforcement of class A- III:
Rs = 355MPa; Rsc = 355MPa; Es = 2 • 105 MPa
Transversal section of 570x450 mm, girder length of 5,400 mm (Fig. 5).
Height of compressed zone:
= 187 mm
R, (A3l + 42 -43 ) = 355 •(3685 + 314 -402) " Rb " 22 • 310
Position of the center of gravity of compressed zone:
y, =
b • y + «• As 3 • as
187
310--+ 5.26 • 402 • 38
2_
= 92 mm
b • x + a As 310 187 + 5.26 • 402
Design load-carrying ability:
Mult = Rs [A,1 (h - y) + As2 (h0 - y -141)] =
= 355 {3685 •( 406 - 92) + 314 •(406 - 92-141)]= 430-106 Nm
Breakdown of transversal section to small areas and finding the center of gravity of every area
In order to calculate the transversal section of a member, we break it down randomly into small areas
(Fig. 6). We have thirteen small areas for our example. The numbering of areas begins from the lower stretched cross-section zone.
Table 1 gives: i - order number of small area; Abl., ybi,bbi, Abi - height, distance from extreme fiber of the lower stretched zone to the center of gravity, width and area of i-n small area of concrete, accordingly; Elbi = 0.85Eb - modulus of concrete deformation; A,Esi = Es - accordingly, distance from extreme fiber
y
Figure 6. Breakdown of Transversal Section
Example of calculation according to the first iteration method
Table 1. Design parameters of small areas
A bi ybi bbi Abi F' bi Vsi Asi E'- si y0j
cm cm cm cm2 MPa cm cm2 MPa cm
1 3 1,5 57 171 3,06E+04
2 2,8 4,4 57 122,8 3,06E+04 4,4 36,8 2,E+05
3 4,5 8,05 57 256,5 3,06E+04
4 4,5 12,55 57 256,5 3,06E+04
5 3,2 16,4 57 182,4 3,06E+04
6 1 18,5 57 53,86 3,06E+04 18,5 3,14 2,E+05
7 3 20,5 57 171 3,06E+04 18,01
8 4,5 24,25 31 139,5 3,06E+04
9 4,5 28,75 31 139,5 3,06E+04
10 4,5 33,25 31 139,5 3,06E+04
11 4,9 37,95 31 151,9 3,06E+04
12 1,6 41,2 31 45,58 3,06E+04 41,2 4,02 2,E+05
13 3 43,5 31 93 3,06E+04
1-st iteration
-99xl0"5 -14.5MPa
o fN œ X X NL -— -180 MPa S o LO
X X 7 ¿ -3.8 MPa
-+=—*-;-*—
X X l.lMPa lOSMPa S
X
64xl0"5
Figure 7. First iteration of the first method
Eres Opbul, Dmitrii Dmitriev, Phan Van Phuc — Pages 29-41 PRACTICAL CALCULATION OF FLEXIBLE MEMBERSWITH THE USE OF NON-LINEAR DEFORMATION
MODEL AS EXEMPLIFIED BY TYPICAL GIRDER RGD 4.56-90 DOI: 10.23968/2500-0055-2018-3-3-29-41
Table 2. First iteration of the first method
A_b E_b E_s y_Oj y_b y_s 1/ , £_b o_b £_S a_s
L 2 cm MPa 2 cm MPa cm cm cm cm MPa MPa
1 171 3.06E+04 16.51 6.41E-04 0.00
2 122.8 3.06E+04 36.8 2.E+05 13.61 13.61 5.29E-04 0.00 5.29E-04 105.7
3 256.5 3.06E+04 9.96 3.87E-04 0.00
4 256.5 3.06E+04 5.46 2.12E-04 0.00
5 182.4 3.06E+04 1.61 6.26E-05 1.10
6 53.86 3.06E+04 3.14 2.E+05 -0.49 -0.49 LO CD -1.89E-05 -0.28 -1.89E-05 -3.7846
7 171 3.06E+04 o CO -2.49 LU 00 00 -9.66E-05 -1.42
8 139.5 3.06E+04 -6.24 m -2.42E-04 -3.55
9 139.5 3.06E+04 -10.74 -4.17E-04 -6.11
10 139.5 3.06E+04 -15.24 -5.92E-04 -8.68
11 151.9 3.06E+04 -19.94 -7.74E-04 -11.35
12 45.58 3.06E+04 4.02 2.E+05 -23.19 -23.19 -9.00E-04 -13.20 -9.00E-04 -180.05
13 93 3.06E+04 -25.49 -9.90E-04 -14.51
Table 3 Second iteration of the first method
A_b E_b E_S y_Oj y_b y_s 1/ , £ b o_b £_S o_s
1 2 cm MPa 2 cm MPa cm cm cm cm MPa MPa
1 171 0.00E+00 19.66 1.88E-03 0.00
2 122.8 0.00E+00 36.8 2.E+05 16.76 16.76 1.61E-03 0.00 1.61E-03 321.2
3 256.5 0.00E+00 13.11 1.26E-03 0.00
4 256.5 0.00E+00 8.61 8.25E-04 0.00
5 182.4 1.75E+04 4.76 4.56E-04 0.00
6 53.86 1.47E+04 3.14 2.E+05 U3 2.66 2.66 LO CD 2.55E-04 0.00 2.55E-04 50.977
7 171 1.47E+04 vH vH CM 0.66 LU 00 un 6.32E-05 1.11
8 139.5 1.47E+04 -3.09 en -2.96E-04 -4.34
9 139.5 1.47E+04 -7.59 -7.27E-04 -10.67
10 139.5 1.47E+04 -12.09 -1.16E-03 -16.99
11 151.9 1.47E+04 -16.79 -1.61E-03 -22.00
12 45.58 1.47E+04 4.02 2.E+05 -20.04 -20.04 -1.92E-03 -22.00 -1.92E-03 -355
13 93 1.47E+04 -22.34 -2.14E-03 -22.00
of the lower stretched zone to the center of gravity, area and modulus of elasticity of i-n reinforcement, accordingly; y0 j = y01 = 180 mm - position of neutral line in the course of the first iteration.
Tables 2-9 and Figure 5-12 show the results of iteration calculations according to the first and the second methods. Thus, the summary limiting moment according to Table
Mac = + X^Ax, = 424.459 • 106 N mm
Height of compressed zone: x = h - y011 = 45 - 21.09 = 24.91cm
Example of calculation according to the second iteration method
2-nd iteration
-214x10 -22.0MPa
188xl0"5
Figure 8. Second iteration of the first method
Table 4. Third iteration of the first method
A_b E_b E_s V_Oj y_b y_s 1/ -, £ b a_b E_S a_s
1 2 cm MPa 2 cm MPa cm cm cm cm MPa MPa
1 171 0.00E+00 19.71 2.08E-03 0.00
2 122.8 0.00E+00 36.8 2.E+05 16.81 16.81 1.77E-03 0.00 1.77E-03 354.41
3 256.5 0.00E+00 13.16 1.39E-03 0.00
4 256.5 0.00E+00 8.66 9.13E-04 0.00
5 182.4 0.00E+00 4.81 5.07E-04 0.00
6 53.86 0.00E+00 3.14 2.E+05 rH 2.71 2.71 t o 2.86E-04 0.00 2.86E-04 57.219
7 171 1.75E+04 IN rH IN 0.71 UÜ LO o 7.53E-05 1.32
8 139.5 1.47E+04 -3.04 rH -3.20E-04 -4.69
9 139.5 1.47E+04 -7.54 -7.94E-04 -11.65
10 139.5 1.47E+04 -12.04 -1.27E-03 -18.60
11 151.9 1.37E+04 -16.74 -1.76E-03 -22.00
12 45.58 1.15E+04 4.02 2.E+05 -19.99 -19.99 -2.11E-03 -22.00 -2.11E-03 -355
13 93 1.03E+04 -22.29 -2.35E-03 -22.00
Figure 10. Eleventh iteration of the first method
Eres Opbul, Dmitrii Dmitriev, Phan Van Phuc — Pages 29-41 PRACTICAL CALCULATION OF FLEXIBLE MEMBERSWITH THE USE OF NON-LINEAR DEFORMATION
MODEL AS EXEMPLIFIED BY TYPICAL GIRDER RGD 4.56-90 DOI: 10.23968/2500-0055-2018-3-3-29-41
Table 5. Eleventh iteration of the first method
A_b E_b E_s y_Oj y_b y_s 1/ -, £ b a_b £_S a_s
1 2 cm MPa 2 cm MPa cm cm cm cm MPa MPa
1 171 0.00E+00 19.59 2.43E-03 0.00
2 122.8 0.00E+00 36.8 2.E+05 16.69 16.69 2.07E-03 0.00 2.07E-03 355
3 256.5 0.00E+00 13.04 1.61E-03 0.00
4 256.5 0.00E+00 8.54 1.06E-03 0.00
5 182.4 0.00E+00 4.69 5.81E-04 0.00
6 53.86 0.00E+00 3.14 2.E+05 <r¡ 2.59 2.59 o 3.21E-04 0.00 3.21E-04 64.175
7 171 1.75E+04 o iH tN 0.59 LU CN 7.32E-05 1.28
8 139.5 1.47E+04 -3.16 rH -3.91E-04 -5.74
9 139.5 1.47E+04 -7.66 -9.48E-04 -13.91
10 139.5 1.47E+04 -12.16 -1.51E-03 -22.00
11 151.9 1.06E+04 -16.86 -2.09E-03 -22.00
12 45.58 8.92E+03 4.02 1.E+05 -20.11 -20.11 -2.49E-03 -22.00 -2.49E-03 -355
13 93 8.01E+03 -22.41 -2.78E-03 -22.00
1-st iteration
_-99xl0"5_-14.5MPa
O IV fNl O CO T—1 — ^ -- - - w -W -- X X X X i NL -7^-180 MPa o / rv r-g , -3.8 MPa o m
X
X X X / £ l.lMPa 106MPa 180
x
64xlCT5
Figure 11. First iteration of the second method
Table 6. First iteration of the second method
i A y_b b_b A_b E_b y_s A_s E_s £ b a_b £ S a_s
cm cm cm cm2 MPa cm cm2 MPa MPa MPa
1 3 1.5 57 171 3.06E+04 6.41E-04 0.00
2 2.8 4.4 57 122.8 3.06E+04 4.4 36.8 2.E+05 5.29E-04 0.00 5.29E-04 105.70
3 4.5 8.05 57 256.5 3.06E+04 3.87E-04 0.00
4 4.5 12.6 57 256.5 3.06E+04 2.12E-04 0.00
5 3.2 16.4 57 182.4 3.06E+04 6.26E-05 1.10
6 1 18.5 57 53.86 3.06E+04 18.5 3.14 2.E+05 -1.89E-05 -0.28 -1.89E-05 -3.78
7 3 20.5 57 171 3.06E+04 -9.66E-05 -1.42
8 4.5 24.3 31 139.5 3.06E+04 -2.42E-04 -3.55
9 4.5 28.8 31 139.5 3.06E+04 -4.17E-04 -6.11
10 4.5 33.3 31 139.5 3.06E+04 -5.92E-04 -8.68
11 4.9 38 31 151.9 3.06E+04 -7.74E-04 -11.35
12 1.6 41.2 31 45.58 3.06E+04 41.2 4.02 2.E+05 -9.00E-04 -13.20 -9.00E-04 -180.05
13 3 43.5 31 93 3.06E+04 -9.90E-04 -14.51
Table 7. Second iteration of the second method
1 y_b b_b A_b E_b y_s E_s E_b a_b £ S a_s
cm cm 2 cm MPa cm 2 cm MPa MPa MPa
1 1.5 57 171 0.00E+00 1.88E-03 0.00
2 4.4 57 122.8 0.00E+00 4.4 36.8 2.E+05 1.61E-03 0.00 1.61E-03 321.20
3 8.05 57 256.5 0.00E+00 1.26E-03 0.00
4 12.6 57 256.5 0.00E+00 8.25E-04 0.00
5 16.4 57 182.4 1.75E+04 4.56E-04 0.00
6 18.5 57 53.86 1.47E+04 18.5 3.14 2.E+05 2.55E-04 0.00 2.55E-04 50.98
7 20.5 57 171 1.47E+04 6.32E-05 1.11
8 24.3 31 139.5 1.47E+04 -2.96E-04 -4.34
9 28.8 31 139.5 1.47E+04 -7.27E-04 -10.67
10 33.3 31 139.5 1.47E+04 -1.16E-03 -16.99
11 38 31 151.9 1.47E+04 -1.61E-03 -22.00
12 41.2 31 45.58 1.47E+04 41.2 4.02 2.E+05 -1.92E-03 -22.00 -1.92E-03 -355.00
13 43.5 31 93 1.47E+04 -2.14E-03 -22.00
188x10 -
Figure 12. Second iteration of the second method
Table 8. Third iteration of the second method
i y_b b_b A_b E_b y_s A_s E_s £ b ob £ S a_s
cm cm 2 cm MPa cm 2 cm MPa MPa MPa
1 1.5 57 171 0.00E+00 2.08E-03 0.00
2 4.4 57 122.8 0.00E+00 4.4 36.8 2.E+05 1.77E-03 0.00 1.77E-03 354.41
3 8.05 57 256.5 0.00E+00 1.39E-03 0.00
4 12.6 57 256.5 0.00E+00 9.13E-04 0.00
5 16.4 57 182.4 0.00E+00 5.07E-04 0.00
6 18.5 57 53.86 0.00E+00 18.5 3.14 2.E+05 2.86E-04 0.00 2.86E-04 57.22
7 20.5 57 171 1.75E+04 7.53E-05 1.32
8 24.3 31 139.5 1.47E+04 -3.20E-04 -4.69
9 28.8 31 139.5 1.47E+04 -7.94E-04 -11.65
10 33.3 31 139.5 1.47E+04 -1.27E-03 -18.60
11 38 31 151.9 1.37E+04 -1.76E-03 -22.00
12 41.2 31 45.58 1.15E+04 41.2 4.02 2.E+05 -2.11E-03 -22.00 -2.11E-03 -355.00
13 43.5 31 93 1.03E+04 -2.35E-03 -22.00
Eres Opbul, Dmitrii Dmitriev, Phan Van Phuc — Pages 29-41 PRACTICAL CALCULATION OF FLEXIBLE MEMBERSWITH THE USE OF NON-LINEAR DEFORMATION
MODEL AS EXEMPLIFIED BY TYPICAL GIRDER RGD 4.56-90 DOI: 10.23968/2500-0055-2018-3-3-29-41
Figure 13. Third iteration of the second method Table 9. Eleventh iteration of the second method
1 y_b b_b A_b E_b y_s E_s £_b a_b £ S a_s
cm cm 2 cm MPa cm 2 cm MPa MPa MPa
1 1.5 57 171 0.00E+00 2.40E-03 0.00
2 4.4 57 122.8 0.00E+00 4.4 36.8 2.E+05 2.04E-03 0.00 2.04E-03 355.00
3 8.05 57 256.5 0.00E+00 1.59E-03 0.00
4 12.6 57 256.5 0.00E+00 1.04E-03 0.00
5 16.4 57 182.4 0.00E+00 5.70E-04 0.00
6 18.5 57 53.86 0.00E+00 18.5 3.14 2.E+05 3.12E-04 0.00 3.12E-04 62.40
7 20.5 57 171 1.75E+04 6.66E-05 1.17
8 24.3 31 139.5 1.47E+04 -3.94E-04 -5.77
9 28.8 31 139.5 1.47E+04 -9.46E-04 -13.87
10 33.3 31 139.5 1.47E+04 -1.50E-03 -21.97
11 38 31 151.9 1.07E+04 -2.07E-03 -22.00
12 41.2 31 45.58 8.99E+03 41.2 4.02 1.E+05 -2.47E-03 -22.00 -2.47E-03 -355.00
13 43.5 31 93 8.07E+03 -2.76E-03 -22.00
240x10
Figure 14. Eleventh iteration of the second method
The relative deformations get determined depending on stiffness characteristics of every small area. The determination of height of compressed zone and graphical presentation of strain-stress state of cross section in the course of every iteration have been built depending on the calculated values of stresses and deformations with adherence to scale.
Table 10. Comparison of calculation parameters.
Calculation method Deformations Load-carrying ability, kNm Height of compressed zone, cm
With respect to limiting conditions - 430.0 18.7
Deformation calculation 1 = 2.78-10-3<[sH = 3.5-10-3] Ss,clc = 0.00207 < [sl2 = 0.025] 424.5 24.91
Deformation calculation 2 stc„ic = 2.76-10-3 < [st2 = 3.5-10-3] = 0.00204 < [sa = 0.025] 427.8 24.06
The height of compressed zone following a principle of "similarity of triangles" will be determined according to: h
xj =■
r +1
'13
9:
Thus, the summary limiting moment according to Table
Mcalc =ZvtiAtiyt, i = 427.795-106 Nmm
We determine a height of the compressed zone through deformations using "indicators of similarity of triangles":
x = -
h
45
■ = 24.06 cm
1.86956 1.86956 Conclusion
The results presented in Table 10 testify to the fact that the acquired design values of deformations feature insignificant differences in case of calculations according to methods 1 and 2 and become adequately consistent with calculation data with respect to limiting conditions, which also proves adequacy of the developed methods.
Eres Opbul, Dmitrii Dmitriev, Phan Van Phuc — Pages 29-41 PRACTICAL CALCULATION OF FLEXIBLE MEMBERSWITH THE USE OF NON-LINEAR DEFORMATION
MODEL AS EXEMPLIFIED BY TYPICAL GIRDER RGD 4.56-90 DOI: 10.23968/2500-0055-2018-3-3-29-41
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