CC BY
0CHECKING THE STRENGTH OF REINFORCED CONCRETE
STRUCTURES
1Urolov U.A., 2Choriev M., 3Eliboyev N.R., 4Umirov Sh.B.
1PhD student of TChTI 2PhD student of TChTI 3Senior Lecturer of Higher Mathematics Department, TChTI 4Assistant at the Department of Higher Mathematics, TChTI https://doi.org/10.5281/zenodo.14024237
Abstract. This article presents a methodology and algorithm for analyzing the tensile and compressive deformations of truss elements made of elastic rods. Numerical results are obtained and analyzed. When static forces are applied to the truss joints, the maximum stresses in the rods are identified.
Keywords: truss, rod, support, reaction force, moment of inertia, cross-sectional area, radius, compression, tension.
Introduction
Today, various metal structures are widely used in the construction industry, including trusses made of metal [1, 2]. In solving such problems, methods of construction mechanics are extensively applied. For this purpose, we use the equilibrium equations known from the statics section of theoretical mechanics [3, 4]. Analyzing truss elements as rods is considered an important issue. Therefore, the ABAQUS software, based on the finite element method, is utilized [5, 6]. This article discusses the problem of calculating the truss. The stresses during compression and tension in each rod are determined, and the conditions under which the strength exceeds the yield limit are examined. The calculations utilize the properties of St3 grade steel.
Let's assume the calculation scheme depicted in Figure 1 is given. We will solve this truss using the ABAQUS software based on the finite element method. As an example, the dimensions of the truss elements are as follows: length l=4 m, connected at a 600 angle. A force PPP is applied at point D. The truss is fixed at point A with a stationary hinge and at point E with a movable hinge. The diagram of the calculation scheme is presented in Figure 1.
Figure 1. Calculation Scheme We will analyze the equilibrium equations of statics separately for the sections at each joint using methods of construction mechanics. By applying the equilibrium conditions for each node, the following equations are obtained.
DC
60°\ D
F
DE
P kN
X
Figure 2. Calculation Scheme for Section at Point D.
In this case, the equilibrium equation will be as follows:
2^3
J1fy =
J1li =
0; -P + FDCcos300 = 0 0 FDE + FDCCOS600 = 0
FDC Fde
3
■PkN
^3
-TpkN
^co c
60°y
€ \
x
Figure 3. Calculation Scheme for Section at Point C.
In this case, the equilibrium equation will be as follows:
1+1Fy = 0
hIF' = 0
10
FCEcos300 - FnrCOs300 = 0
dc1
0
FCE = -^PkN
FdcCOS600 - FCB + FceCOS600 = 0 FCB
y
2^3
P kN
B
60°y V60°
i \ FEE
x
Figure 4. Calculation Scheme for Section at Point B.
In this case, the equilibrium equation will be as follows:
3
î+^Fy = 0 FBEcos300 - FBAcos300 = 0 FBE = FBA ^+^FX = 0 FCB - FbeCOS60° - FbaCOS60° = 0
FBE =
2^3 2^3
P kN Fba = -^^P kN
3
3
BE
60°(
EA
f!!
CE
ï60°
X
'DE
Figure 5. Calculation Scheme for Section at point E.
In this case, the equilibrium equation will be as follows: r+^Fy = 0 Ey- FBEcos300 - FCEcos300 = 0
+
Ey = 2P kN
y - rbe
^FX = 0 Fea - Fde - Fcecos60° + Fbecos60° = 0 FEA = -^P kN The cross-sectional area of the truss rods is S=30 cm2, and for St3 grade steel with a yield strength of O"oq = 140 MPa, the force required for failure (exceeding the yield limit) is
J3
pkr =y~ aoq. s * 0,87-140-106 • 30 -10-4 * 365 kN.
Now, the buckling limit for the truss rods during compression will be determined:
nbh
Pr =
к2 EJ
J =
4l2 64
(in the problem statement, h=b-diameter ( r « 3,1 sm))
■( h2 + b2 )
J * 1,45 -10-6 m4 - Inertia moment P^ * 4,467 kN
For the case where P=8 kN, S=1 m2, E=200*109, the solutions obtained using both the analytical method and the ABAQUS software are as follows:
Fdc = 9.24 kN Fde = -4.62 kN Fbe = 9.24 kN
FBA = 9.24 kN
Ey = 16 kN FEA = 4.62 kN
FCE = 9.24 kN
FCB = 9. 24 kN
Source 1
Frame: Increment 1: Step Time = 1.000 Loc 1: Integration point values from source 1 Output sorted by column "Element Label". Field Output reported at integration points for part: FERMA-1 Element Label Int S.S11
Pt @Loc 1
1 1 9.23760E+03
2 1 -9.23760E+03
3 1 -4.61880E+03
4 1 9.23760E+03
5 1 9.23760E+03
6 1 -9.23760E+03
7 1 -4.61880E+03
Minimum -9.23760E+03
At Element 6
Int Pt 1
Maximum 9.23760E+03
At Element 5
Int Pt 1
Total 0.
Output sorted by column "Node Label". Field Output reported at nodes for part: FERMA-1 Node Label RF.RF1 RF.RF2 @Loc 1 @Loc 1
1 0. 0.
2 0. 0.
3 0. 16.E+03
4 0. 0.
5 1.81899E-12 -8.E+03 Minimum 0. -8.E+03 At Node 4 5 Maximum 1.81899E-12 16.E+03 At Node 5 3
Total 1.81899E-12 8.E+03 CONCLUSION
Thus, when a force P is applied, the rods under compression begin to buckle when it reaches pkr ~ 4,467 kN, and the rods under tension fail at Pkr ~ 365 kN. The results obtained using
both analytical and numerical methods for the case where an 8 kN force is applied at point D of the truss are presented. The solutions confirm consistency with the results obtained through the analytical method.
S, Мая. In-Plane Principal Sj Min. In-Plane Principal S, Out-of-Plane Principal
+9.238e+03
— - +7.698e+03
----+6,158e+G3
--+4.619e+03
----+3.G79e+03
--+1.54Ge+03
-----7.324e-Ü4
---l,540e + 03
-----3.079e+03
---4.619e + 03
— -6,lS8e+03
K™ -7.698e + 03 -9.238e+Ü3
REFERENCES
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