Residual Stresses in Hot Rolled
Wide-Flange Steel Members
Yaze Chena, Thomas Hookerb, Ming Songc
Civil Engineering Master of Engineering (Structural)a [email protected] b [email protected]
What is “Hot Rolling?”
Direction of rolling
What are residual stresses?
What are residual stresses?
Rolling process
Straightening procedures
Nonuniform cooling Cross-sectional geometry
Cooling conditions
Steel material properties
What are residual stresses?
} Rolling process
Straightening procedures
Nonuniform cooling Cross-sectional geometry
Cooling conditions
Steel material properties
Rolling process
Straightening procedures
Nonuniform cooling Cross-sectional geometry
Cooling conditions
Steel material properties
What are residual stresses?
𝑡𝑤𝑒𝑏
𝑡 𝑓𝑙𝑎𝑛𝑔𝑒
Why are we interested?
Partial yielding of cross-section
Why are we interested?
Partial yielding of cross-section
Why are we interested?
Partial yielding of cross-section
Basic Equation of Transient heat Problems The temperature distribution inside the body is
varies with time. The basic equation of transient thermal problems is
Where [C] is the specific heat matrix [k] is the thermal conductivity matrix
Heat Conduction Equation A basic law of heat conduction
A heat flow is controlled by:
Governing equation of temperature:
Generalized Finite-Element Method
The boundary conditions:
Green formula:
Finite-Element Formulation Temperature at any point:
Temperature gradient at any point:
Heat flux in each point:
The Expression of Matrix The basic equation of transient thermal problems is:
The element matrices and external heat load vector:
Mode Superposition Step 1: Find the eigenvalues λn, and the associated eigenvectors from establish matrix [A], whose columns are the eigenvectors. Step 2: Calculate the elements Cnn in matrix [C], using
Mode Superposition Step 3: Solve differential equation as below, to obtain the vector {a}.
Step 4: Use equation below to obtain the nodal temperature solution {T(t)}.
Time Integration θ-family of approximation A weighted average of the time derivatives at two
consecutive time step is approximated by linear interpolation of the values of the variable at the two steps.
From
∆ 𝑡 ¿
(𝐶+𝜃∆ 𝑡𝐾 )𝑇𝑛+1={∆ 𝑡 [𝜃𝑝𝑛+1+(1−𝜃 )𝑝𝑛 ]+[𝐶− (1−𝜃 )∆ 𝑡𝐾 ]𝑇𝑛}
Where,
Time Integration θ-family of approximation
Rearranging terms as form of
Plugging and in, we arrive at,
Time Integration θ-family of approximation Different time integration schemes: Θ= 0, the forward difference scheme (conditionally stable); , the Crank-Nicolson scheme (unconditionally stable); , the Galerkin method (unconditionally stable); 1, the back ward difference scheme (unconditionally stable)
Thermal Stresses σ = E ε = E α dt
σ = stress due to temperature expansion E = Young’s Modulus ε = strain α = temperature expansion coefficient dt = temperature difference
ANSYS SIMULATION
Dimensions in inches
A36W14x730
Young’s modulus vs. Temperature
202GPa
25GPa
Yield stress & Tangent modulus
Temperature
Yield stress
Tangent modulus
C MPa MPa38 248 9929
149 219 8770343 184 7364454 161 6454593 136 5443900 78 3107
Thermal Properties
Density 7832kg/m Isotropic Thermal Conductivity 60W/(m*°C) Specific Heat 434J/(kg*°C) Film Coefficient 193W/m *°C
3
2
Analysis ProcessTransient Heat Transfer Analysis Initial Temperature: Uniform 900 °C Ambient Convection: 20 °C End Time: 5000 Sec. Substeps: 250Thermal Stress Analysis (Static Structural)σ = E ε = E α dt
Mesh W14X730 element size: 1’’ 0.5’’ 0.25’’ 0.125’’
Temperature
Normal Stress
Verification Empirical (W14x730)Maximum compression stress: Analysis 10.3 ksi AISC Steel Construction Manual 0.3Fy=0.3*36=10.8ksi Error: 4.63% Mesh Convergence (W14x730)
1 2 3 40
2
4
6
8
10
12
7.989.02
10.1 10.3
Trial of Element Size (1'', 0.5'', 0.25'', 0.125'')
Max
Nor
mal
Str
ess
(ksi
)
Max Stress vs Size
2.11 2.51 2.91 3.31 3.71 4.11 4.51 4.91
-6
-4
-2
0
2
4
6
8
10
12
-4.99-4.59-4.15-4.13-3.58-2.64-2.39-2.31
10.39.28
8.437.37
6.444.93
4.073
Max Normal Stress vs Flange Thickness
max C (ksi)
Flange thickness (in)
Norm
al S
tress
(ksi)
(W14X730)
Flange thickness/Web thickness=1.6
Thank You!