2. Literature survey
2.4 Plastic theory
2.4.3 Plastic stress distribution
The last condition states that stresses may be distributed in the most favorable manner within the cross-section, as long as the deformations and plastic stress distribution relate to each other. Figure 2.3 illustrates plasticity in the stress-strain curve by means of a horizontal line, through the yield plateau.
Figure 2.3 Plastic part stress-strain curve [3]
Three cases where single forces are present in the cross-section are displayed in figure 2.4. Every fiber of the cross-section was used, where every fiber reaches the yield strength fy or shear strength τy.
Figure 2.4 Plastic stress distributions by three individual cases [3]
When multiple forces are present instead of a single force, the possible distributions can be shown as figure 2.5. In the case of multiple internal forces the stresses must still remain within the limits of the yield criterion [3].
Figure 2.5 Possible plastic stress distribution in the case of multiple internal forces [3]
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
19 2.5 Classification of cross-sections
NEN-EN 1993-1-1, article 5.5 [1] defines the cross-sectional classes based on the capacity to deform plastically and the ability to redistribute the moments. Figure 2.6 shows the classification in a moment-rotation diagram. Cross-sections will be classified by the c/t ratio of the internal (web) and external (flanges) parts.
t thickness of the flange or web mm
c height of the cross-section minus the flanges and the radii (in case of internal parts) mm
Figure 2.6 Moment rotation behavior for I-shaped beams subjected to bending moment [3]
The four classes of cross-sections are:
- Class 1: plastic sections can form plastic hinges with sufficient rotation capacity to redistribute the moments (strain hardening) before local buckling occurs;
- Class 2: compact sections are able to develop the plastic moment resistance, but the rotation capacity is limited because of local buckling;
- Class 3: semi-compact sections are able to reach the elastic moment resistance, while local buckling prevents the development of the plastic moment resistance;
- Class 4: slender sections are not able to attain the yield stress, because local buckling occurs in one or several parts of the cross-section.
Figure 2.7 shows the different calculation models and corresponding classification of cross-sections.
Figure 2.7 Classification of cross-sections [3]
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
20 2.6 Code requirements
2.6.1 Eurocode 3: EN 1993-1-1
The Eurocode provides rules to determine the resistance of cross-sections for bending moment, shear force, axial forces and all possible combinations of these. The design rules for steel sections are described in NEN-EN 1993-1-1, article 6.2.10 [1].
“Where shear and axial force are present, allowance should be made for the effect of both shear force and axial force on the moment resistance. If the design value of the shear force VEd does not exceed 50% of the design plastic shear resistance Vpl,Rd, no reduction of the resistances defined for bending and axial force need to be made, except where shear buckling reduces the section resistance.
In the case that VEd exceeds 50% of Vpl.Rd, the design resistance of the cross-section to combinations of moment and axial force should be calculated using a reduced yield strength for the shear surface.”
𝑓𝑦,𝑟 = (1 − 𝜌)𝑓𝑦 (2.20) cross-section may be reduced [1].
2.6.2 Dutch National Annex
The Dutch National Annex (NA) of NEN-EN 1993-1-1 article 6.2.10 provides in rules to determine the interaction between bending, shear and normal force. For I-shaped sections of class 1 and 2 the interaction about the strong axis may be checked as follows:
𝑀𝑦,𝐸𝑑
My,Ed design value of the bending moment about the y-axis kNm
My,V,Rd reduced bending resistance due to shear about the y-axis kNm
NEd design value of the axial force kN
NVz,Rd reduced normal force resistance due to shear about the z-axis kN
Npl,Rd design plastic resistance to normal forces of the gross cross-section kN
ᵨ
reduction factor due to shear -Bending moment, shear and normal force interaction of I-shaped steel cross-sections
21
Av shear area mm²
fy yield strength N/mm2
𝛾𝑀0 partial factor for resistance of cross-sections -
2.7 Results from earlier research
2.7.1 Research by Sherbourne & Oostrom, 1972
The investigation of Sherbourne & Oostrom [4] attempted a plastic analysis of castellated beams by developing a computer model which duplicated behavior under incremental, proportional loading.
Moment, shear and axial force interaction surfaces were developed for critical sections of castellated beams and any other I-section beam with a hole in the web. The sections considered the rectangular and T-section. The interaction curves for these shapes were compared with that for the continuous I-section in order to obtain a clearer understanding of the change deriving from a change in shape.
Furthermore, interaction curves were plotted for a complete range of castellated beams in order to assess the effect of changes in size and profile of the web elements and their attendant discontinuities.
Figure 2.8 Castellated beam [4]
To make a 3-dimensional diagram, the shape of the m-v, the m-n and n-v curves as end points were used. The 3-dimensional interaction diagram is shown in Figure 2.9.
Figure 2.9 Geometrical interaction surface [4]
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
22
This research about the castellated beams [4] is not relevant for the current investigation, because old codes where used in the research and the castellated beam differed largely from the I-shaped cross-section. The way of making the 3D-diagram can be used for this research.
2.7.2 Research by Goczek & Supel, 2014
The analytical research of Goczek and Supel [5] includes the resistance of steel cross-sections subjected to bending, shear and axial forces. The aim of their investigation was to analyze the Eurocode 3 rules for the combined bending, shear and axial force interaction. The investigation has been limited to Class 1, 2 and 3 cross-sections. The results showed that the most uncertain design situation arose when bending and axial force were combined with a significant shear force that cannot be neglected. The comparative analysis of interaction diagrams is presented, see Figure 2.10 and 2.11.
Figure 2.10 M-N-V interaction graph – Class 1 and 2 cross-sections[5] Figure 2.11 M-N-V interaction graph – Class 3 cross-sections [5].
The analytical research conducted by Goczek & Supel [5] on the resistance of steel cross-sections subjected to bending, shear and axial forces showed good agreement for Class 1 and 2 cross-sections and poor for Class 3 cross-sections. The pure elastic approach assumed in the Huber-Mises yield criterion, and applied to Class 3 cross-sections excluded partial plastic stress distribution, which is permitted in elastic design by Eurocode 3.
2.7.3 Research by Neal, 1961
Neal, B.G. [6] investigated analytically the effect of shear and normal forces on the fully plastic moment of an I-beam. Therefore a cantilever of I-section subjected to both normal and shear forces at its free end is considered. The results of his investigation are presented in the form of interaction relations between the shear and normal forces and bending moment at the clamped end of the cantilever at failure. The applied forces are shown in Figure 2.12.
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
23
Figure 2.12 Cantilever I-beam with applied forces [6]
In this research only two cases were investigated, namely neutral axis in the web and neutral axis in the flange. The roots were not taken into acount. The relation between normal force, shear force and bending moment is shown in Figure 2.13.
Figure 2.13 Interaction relations between M-N-V [6]
The two curves which are indicated with (i) and (ii) are from earlier research where bending moment is limited to be zero. This research about normal force, shear force and bending is not relevant for current investigation, because the roots are not taken into account.
2.8 Conclusion regarding literature survey
The Eurocode provides rules to determine the resistance of cross-sections for bending moment, shear force, axial forces and all possible combinations of these. The design rules of M-N-V for steel sections are described in NEN-EN 1993-1-1, clause 6.2.10 [1]. There are two cases:
VEd≤ 0,5 ∙ Vpl,Rd the shear force (V) can be neglected
VEd> 0,5 ∙ Vpl,Rd the design resistance of the cross-section to combination of moment and axial force should be calculated using a reduced yield strength for the shear area
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
24
3. Analytical solution
In this chapter an analytical solution is presented for a rolled I-shaped cross-section subjected to combined bending, shear force and normal force. The cross-section was divided in several parts to cope with the shear force, normal force and bending moment, see Figure 3.1. The center of the cross-section is reserved for the shear force. The remaining parts of the cross-section are reserved for the normal force and bending moment. Dependent on the amount of applied shear force the size of these areas changes. Figure 3.1 (right picture) shows the geometrical parameters of an I-shaped cross-section.
Figure 3.1 Areas reserved for M-N-V (left) and geometrical parameters of a rolled I-shaped cross-section (right)
Where:
Af area of the flanges (2 ∙ 𝑏 ∙ 𝑡𝑓) mm2
Ar area of the roots (4𝑟2(1 − 0.25𝜋)) mm2
Aw area of the web (𝑡𝑤∙ ℎ𝑤) mm2
As an example an HEA240 cross-section was used to demonstrate the analytical solution. To enable the calculation of M-N-V interaction, three regions and several cases were introduced. Table 3.1 shows an overview of those regions & cases.
Table 3.1 All possible cases for the interaction of M-N-V
When distributing shear force, the regions presented themselves namely:
I) The shear force is only in the web II) The shear force is in the web & roots
III) The shear force is in the web, roots and in the flange
Figure 3.2 shows the three regions where the shear force can be situated.
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
25
Figure 3.2 Possible positions of the shear force region I, II and III
When distributing the shear force, normal force and bending moment, three general cases presented themselves, namely:
1) The neutral axis of the cross-section is situated in the web 2) The neutral axis of the cross-section is situated in the roots 3) The neutral axis of the cross-section is situated in the flange Three possible cases are shown in Figure 3.3.
Figure 3.3 Possible positions of the neutral axis [7]
Besides these three general cases two subcases emerged, namely 3A and 3B, see Figure 3.4. When the shear force has reached the flange, which is indicated in black in Figure 3.4, then case 3A and 3B would be applied. Flange1 and flange2 in case 3A and 3B are the distances where the normal force can be varied. The difference between the flange and flange1/flange2 is that in case of the flange the shear force is located in web or in the roots and in case of flange1/flange2 the shear force is located in the flange itself. Table 3.1 shows the possible cases.
Figure 3.4 Possible positions of the neutral axis in the flange, Case 3A (left) and Case 3B (right)
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
26 3.1 M-N-V interaction in region I (case 1, 2 and 3)
Case 1: neutral axis in web
In this case, the neutral line is located in the web. As mentioned earlier the shear force is taken by the center part of the web over the height (ℎ𝑦). The normal force is taken by another part of the web which height is indicated with α. The remaining part of the cross-section beared the bending moment, see Figure 3.5.
Figure 3.5 Yield stress distributions in case 1
To determine which part of the cross-section (hy) is reserved for the shear force, equation (3.1) and (3.2) were used.
𝑛𝑉∙𝑉𝑝𝑙
𝑓𝑦⁄√3 = 𝐴 with 𝐴 = 𝑡𝑤∙ ℎ𝑦 (3.1)
ℎ𝑦 =
𝑛𝑉∙𝑉𝑝𝑙 𝑓𝑦 √3⁄
𝑡𝑤 (3.2)
For region I, ℎ𝑦 ≤ 𝑑
Figure 3.6 shows the distances form the center of the cross-section to the center of the gravity of each surface in Case 1.
Figure 3.6 Distance from the center of the cross-section to the center of the gravity of each surface, in Case 1
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
27
In Equations (3.3) to (3.7) the design bending moment with the corresponding plastic section modulus regarding the shear force, normal force and bending moment is described.
𝑊𝑉=1
The design value of the axial force is given by equation (3.10).
𝑁𝐸𝑑 = 𝑓𝑦∙ 𝐴𝑁 = 𝑓𝑦∙ (2 ∙ 𝑡𝑤∙ 𝛼) (3.5)
With:
hv unknown distance in y-direction where shear force is applied mm Wv plastic section modulus regarding the shear force mm³ WN plastic section modulus regarding the normal force mm³ WM plastic section modulus regarding the bending moment mm³
My,Ed design bending moment kNm
In this case the neutral axis is located in the roots. A function is defined to describe the roots, see equation (3.8). This equation can be used to define the area of the roots or to calculate the neutral point of the root, illustrated by Figure 3.7.
(3.8)
Figure 3.7 Area and neutral point of the root [7]
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
28
Equation (3.8), (3.9) and (3.10) originate from the literature survey of Rombouts, I.M.J. [7].
For the calculation of the root area which is given in equation (3.10), the integral of equation (3.8) is taken.
(3.10)
This results in a slightly different yield stress distribution, see Figure 3.8.
Figure 3.8 Yield tress distributions in case 2
To calculate the plastic section modulus, the distance from the center of the cross-section to the center of gravity of each surface is needed. These distances are shown in Figure 3.9.
Figure 3.9 Distance from the center of the cross-section to the center of the gravity of each surface, in Case 2
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
29
Equation (3.11) and (3.12) describes the section modulus regarding normal force. In equation (3.13) the normal force is described.
𝑊𝑁= (2 ∙ 𝑡𝑤(𝛽 +𝑑2−ℎ2𝑦)) ∙ (((𝛽+
𝐴𝑟(𝛽) surface area of the roots which is determined by the distance β mm²
In equation (3.14), the section modulus regarding bending moment is described which is used in equation (3.15) to calculate the bending moment.
𝑊𝑀= 𝑊𝑝𝑙− 𝑊𝑉− 𝑊𝑁 (3.14)
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
30
Case 3: neutral axis in flange
In case three the neutral axis is located in the flange. The yield stress distribution in case three is shown in Figure 3.10 (𝛿 ≤ 𝑡𝑓).
Figure 3.10 Yield stress distributions in case 3
The area where normal force and bending moment can be varied is shown in Figure 3.11.
Figure 3.11 Distance from the center of the cross-section to the center of the gravity of each surface, in Case 3
For the calculation of bending moment and axial force equation (3.16) and (3.17) can be used.
𝑀𝑦,𝐸𝑑= 𝑓𝑦∙ ((𝑡𝑓− 𝛿)𝑏 ∙ 2) ∙ ((𝑡𝑓−𝛿)
2 + 𝛿 +ℎ𝑤
2) (3.16)
𝑁𝐸𝑑 = 𝑓𝑦∙ 𝐴𝑁 = 𝑓𝑦∙ ((2 ∙ 𝑡𝑤∙ (ℎ𝑤
2 −ℎ𝑦
2)) + (2 ∙ 𝑏 ∙ 𝛿) + 𝐴𝑟) (3.17)
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
31
Figure 3.12 shows the M-N-V interaction diagrams of HEA240 in region I
Figure 3.12 M-N-V interaction diagrams of HEA240 in Region I
3.2 M-N-V interaction in region II (case 2 and 3)
To determine the shear area in Region II, equation 3.18 is used.
𝑛𝑉∙𝑉𝑝𝑙
𝑓𝑦⁄√3 = (𝑡𝑤∙ d) + 𝐴𝑉,𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 (3.18)
AV,residual= 𝑛𝑓𝑉∙𝑉𝑝𝑙
𝑦⁄√3− (𝑡𝑤∙ d) = (2 ∙ (𝑡𝑤∙ 𝛽)) + 𝐴𝑟(𝛽) (3.19)
Since equation (3.24) cannot be solved for β, the dimensions of an HEA240 cross-section were substituted in equation (3.15) and (3.24). For example, when 𝑛𝑉 = 0.5, the value of β can be found numerically:
(0.5 ∙ 341.6 ∙ 10³
235 √3⁄ ) − (7.5 ∙ 164) = 𝟐𝟖. 𝟕𝟖 𝐦𝐦𝟐 = (2 ∙ (7.5 ∙ 𝛽)) + 𝐴𝑟(𝛽) => 𝜷 = 𝟎. 𝟑𝟐𝟐𝟔 𝒎𝒎
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
32
Case 2: neutral axis in roots (Region II)
The shear area has reached the root which is indicated with β in Figure 3.13. ∆β is the distance where normal force is applied.
Figure 3.13 important parameters in case 2
Where:
Wrec plastic section modulus of the rectangular in Figure 3.13 mm³ WNrec plastic section modulus of the rectangular regarding normal force mm³
The design value of the axial force is given by equation (3.21).
𝐴𝑁,𝑟𝑜𝑜𝑡𝑠 = 𝐴𝑁𝑟(∆𝛽)+ 𝐴𝑁𝑟𝑒𝑐 = 𝐴𝑁𝑟(∆𝛽)+ (2 ∙ 𝑡𝑤∙ (∆𝛽)) (3.20)
𝑁𝐸𝑑 = 𝑓𝑦∙ 𝐴𝑁,𝑟𝑜𝑜𝑡𝑠 = 𝑓𝑦∙ (𝐴𝑁𝑟(∆𝛽)+ (2 ∙ 𝑡𝑤∙ (∆𝛽))) (3.21)
Equation (3.22) to (3.25) gives the plastic section modulus regarding M-N-V.
𝑊𝑉=1
4∙ 𝑡𝑤∙ 𝑑2+ 𝑊𝑟(𝛽)+ 𝑊𝑟𝑒𝑐 (3.22)
𝑊𝑟𝑒𝑐 = (2 ∙ 𝑡𝑤∙ (𝛽)) ∙ (𝛽
2+𝑑
2) (3.23)
𝑊𝑀= 𝑊𝑝𝑙− 𝑊𝑉− 𝑊𝑁,𝑟(∆𝛽)− 𝑊𝑁𝑟𝑒𝑐 (3.24)
𝑊𝑁𝑟𝑒𝑐 = (2 ∙ 𝑡𝑤∙ (∆𝛽)) ∙ (∆𝛽
2 + 𝛽 +𝑑
2) (3.25)
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
33
Using equation (3.22) to (3.25), the bending moment can be calculated in equation (3.26)
𝑀𝑦,𝐸𝑑= 𝑓𝑦∙ 𝑊𝑀 (3.26)
𝑀𝑦,𝐸𝑑= 𝑓𝑦∙ (𝑊𝑝𝑙− 𝑊𝑉− 𝑊𝑁,𝑟(∆𝛽)− 𝑊𝑁𝑟𝑒𝑐)
= 𝑓𝑦∙ ((𝑏 ∙ 𝑡𝑓(ℎ − 𝑡𝑓) +1
4∙ 𝑡𝑤(ℎ − 2 ∙ 𝑡𝑓)2+ (4𝑟2(1 − 0.25𝜋)) ∙ (ℎ
2− 𝑡𝑓− 0.223 ∙ 𝑟))
−(1
4∙ 𝑡𝑤∙ 𝑑2+ 𝑊𝑟(𝛽)+ (2 ∙ 𝑡𝑤∙ (𝛽)) ∙ ((𝛽 2⁄ ) + (𝑑
2)))−(𝑊𝑁,𝑟(∆𝛽))
− (2 ∙ 𝑡𝑤∙ (∆𝛽)) ∙ ((∆𝛽
2 ) + 𝛽 + (𝑑 2)))
Case 3: neutral axis in flange (Region II)
Equation (3.27) was used to obtain the surface of the normal force. For the calculation of normal force, equation (3.28) was used.
𝐴𝑁,𝑓𝑙𝑎𝑛𝑔𝑒 = ((2 ∙ 𝑡𝑤∙ ∆𝛽) + (𝐴𝑟− 𝐴𝑟(𝛽)) + (2 ∙ 𝑏 ∙ 𝛿)) (3.27)
𝑁𝐸𝑑 = 𝑓𝑦∙ 𝐴𝑁,𝑓𝑙𝑎𝑛𝑔𝑒 = 𝑓𝑦∙ ((2 ∙ 𝑡𝑤∙ ∆𝛽) + (𝐴𝑟− 𝐴𝑟(𝛽)) + (2 ∙ 𝑏 ∙ 𝛿)) (3.28)
Equatio (3.29) and (3.30) described the plastic section modulus and applied bending moment.
𝑊𝑀,𝑓𝑙𝑎𝑛𝑔𝑒= ((𝑡𝑓− 𝛿) ∙ 𝑏 ∙ 2) ∙ ((𝑡𝑓2−𝛿)+ 𝛿 +ℎ2𝑤) (3.29)
𝑀𝑦,𝐸𝑑= 𝑓𝑦∙ 𝑊𝑀,𝑓𝑙𝑎𝑛𝑔𝑒 (3.30)
𝑀𝑦,𝐸𝑑= 𝑓𝑦∙ (((𝑡𝑓− 𝛿) ∙ 𝑏 ∙ 2) ∙ ((𝑡𝑓− 𝛿)
2 + 𝛿 +ℎ𝑤 2 ))
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
34
Figure 3.14 shows the M-N-V interaction diagrams of HEA240 in region II
Figure 3.14 M-N-V interaction diagrams of HEA240 in Region II
3.3 M-N-V interaction in region III (case 3A and 3B)
In region III the neutral axis varies only in the flange. There are 2 cases, namely 3A and 3B, see Figure 3.4.
𝑛𝑉∙𝑉𝑝𝑙
𝑓𝑦⁄√3 = (𝑡𝑤∙ ℎ𝑤) + 𝐴𝑟+ 𝐴𝑉,𝑟𝑒𝑠𝑖𝑑𝑢𝑎𝑙 (3.31)
AV,residual= 𝑛𝑓𝑉∙𝑉𝑝𝑙
𝑦⁄√3− (𝑡𝑤∙ ℎ𝑤) − 𝐴𝑟 = ((2𝑟 + 𝑡𝑤) ∙ 2) ∙ 𝛿 (3.32)
𝛿 = ( (𝑛𝑓𝑉∙𝑉𝑝𝑙
𝑦⁄√3− (𝑡𝑤∙ ℎ𝑤) − 𝐴𝑟)
((2𝑟 + 𝑡𝑤) ∙ 2)
⁄ ) (3.33)
For example, when 𝑛𝑉= 0.8, the value of δ is:
(0.8 ∙ 341.6 ∙ 103
235 √3⁄ ) − (7.5 ∙ 206) − 378.56 = 𝟗𝟎. 𝟔𝟑 𝐦𝐦𝟐= ((2 ∙ 21 + 7.5) ∙ 2) ∙ δ =>
𝜹 = 𝟎. 𝟗𝟏𝟒𝟎 𝒎𝒎
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
35
Case 3A: neutral axis in flange1 (distance between hw/2 and (hw/2+δ)
For the case that the neutral axis is in the flange where also shear force is present, the normal force should be varied in the flange at distance ∆𝛿1, see Figure 3.15. The remaining part of the flange would carry the bending moment.
Figure 3.15 important distances in case 3A
The design value of the axial force is given by equation (3.35).
𝐴𝑁,𝑓𝑙𝑎𝑛𝑔𝑒,1= (2 ∙ (𝑏 − (2𝑟 + 𝑡𝑤)) ∙ ∆𝛿1) (3.34)
𝑁𝐸𝑑 = 𝑓𝑦∙ 𝐴𝑁,𝑓𝑙𝑎𝑛𝑔𝑒,1= 𝑓𝑦∙ (2 ∙ (𝑏 − (2𝑟 + 𝑡𝑤)) ∙ ∆𝛿1) (3.35)
For the case that the neutral axis is located in the flange, equations (3.36) to (3.38) show the plastic section modulus.
Equation (3.39) describes the bending moment in case 3A.
𝑀𝑦,𝐸𝑑= 𝑓𝑦∙ 𝑊𝑀,𝑓𝑙𝑎𝑛𝑔𝑒,1 (3.39)
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
36
Case 3B: neutral axis in flange2 (distance between (hw/2+δ) and h/2)
For the case that the neutral axis is in the flange where also shear force is present, the normal force should be varied in the flange at distance ∆𝛿2, see Figure 3.16. The remaining part of the flange would carry the bending moment.
Figure 3.16 important distances in case 3B
The design value of the axial force is given by equation (3.41).
𝐴𝑁,𝑓𝑙𝑎𝑛𝑔𝑒,2= ((2 ∙ (𝑏 − (2𝑟 + 𝑡𝑤))) ∙ ∆𝛿1) + (2 ∙ 𝑏 ∙ ∆𝛿2) (3.40)
𝑁𝐸𝑑 = 𝑓𝑦∙ 𝐴𝑁,𝑓𝑙𝑎𝑛𝑔𝑒,2= 𝑓𝑦∙ ((2 ∙ (𝑏 − (2𝑟 + 𝑡𝑤))) ∙ ∆𝛿1) + (2 ∙ 𝑏 ∙ ∆𝛿2) (3.41) For the case that the neutral axis is located in the flange, equations (3.42) to (3.44) show the design bending moment.
𝑊𝑁= ((2 ∙ (𝑏 − (2𝑟 + 𝑡𝑤)) ∙ ∆𝛿1) ∙ (∆𝛿1
2 +ℎ2𝑤)) + ((2 ∙ 𝑏 ∙ ∆𝛿2) ∙ (∆𝛿22 + ∆𝛿1 +ℎ2𝑤)) (3.42)
𝑊𝑀,𝑓𝑙𝑎𝑛𝑔𝑒,2= 𝑊𝑝𝑙− 𝑊𝑉− 𝑊𝑁 (3.43)
𝑀𝑦,𝐸𝑑= 𝑓𝑦∙ 𝑊𝑀,𝑓𝑙𝑎𝑛𝑔𝑒,2 (3.44)
𝑀𝑦,𝐸𝑑= 𝑓𝑦∙ (𝑊𝑝𝑙− 𝑊𝑉− 𝑊𝑁)
= 𝑓𝑦∙ ((𝑏 ∙ 𝑡𝑓(ℎ − 𝑡𝑓) +1
4∙ 𝑡𝑤(ℎ − 2 ∙ 𝑡𝑓)2+ (4𝑟2(1 − 0.25𝜋)) ∙ (ℎ
2− 𝑡𝑓− 0.223 ∙ 𝑟))
−(1
4∙ 𝑡𝑤∙ ℎ𝑤2+ 𝑊𝑟,𝑡𝑜𝑡+ ((2 ∙ (2 ∙ 𝑟 + 𝑡𝑤) ∙ 𝛿) ∙ (𝛿 2+ℎ𝑤
2 )))
− ((2 ∙ (𝑏 − (2𝑟 + 𝑡𝑤)) ∙ ∆𝛿1) ∙ (∆𝛿1 2 +ℎ𝑤
2 ))
+ ((2 ∙ 𝑏 ∙ ∆𝛿2) ∙ (∆𝛿2
2 + ∆𝛿1 +ℎ𝑤 2 )))
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
37
Figure 3.17 shows the M-N-V interaction diagrams of HEA240 in region III
Figure 3.17 M-N-V interaction diagrams of HEA240 in Region III
To combine the bending moment, shear force and normal force, the utilization ratio 𝑛𝑉 is chosen from 0.01 up to 1.00. The design values for bending moment (𝑀𝐸𝑑) and normal force (𝑁𝐸𝑑) is calculated for every location over the height of the beam. After determining the design values, the ratios 𝑛𝑀 and 𝑛𝑁 can be calculated, which give the interaction diagram, see equation (3.45) and (3.46).
𝑛𝑀= 𝑀𝐸𝑑
𝑀𝑝𝑙,𝑦,𝑅𝑑 (3.45)
𝑛𝑁 = 𝑁𝐸𝑑
𝑁𝑝𝑙,𝑦,𝑅𝑑 (3.46)
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
38
To make the interaction diagram the plastic resistance of the cross-section for the moment, normal force and shear force is needed. These plastic resistances are described by equation (3.47) to (3.49).
𝑀𝑝𝑙,𝑦,𝑅𝑑= 𝑓𝑦∙ 𝑊𝑝𝑙,𝑦 (3.47)
With 𝑊𝑝𝑙,𝑦 = (𝑏 ∙ 𝑡𝑓∙ (ℎ − 𝑡𝑓) +14∙ 𝑡𝑤(ℎ − 2 ∙ 𝑡𝑓)2+ (4𝑟2(1 − 0.25𝜋)) ∙ (ℎ
2− 𝑡𝑓− 0.223 ∙ 𝑟))
𝑀𝑝𝑙,𝑦,𝑅𝑑= 𝑓𝑦(𝑏 ∙ 𝑡𝑓∙ (ℎ − 𝑡𝑓) +1
4∙ 𝑡𝑤(ℎ − 2 ∙ 𝑡𝑓)2+ (4𝑟2(1 − 0.25𝜋)) ∙ (ℎ
2− 𝑡𝑓− 0.223 ∙ 𝑟))
𝑁𝑝𝑙,𝑦,𝑅𝑑= 𝑓𝑦∙ 𝐴 (3.48)
with 𝐴 = (((ℎ − 2 ∙ 𝑡𝑓) ∙ 𝑡𝑤) + (4𝑟2(1 − 0.25𝜋)) + 2 ∙ 𝑏 ∙ 𝑡𝑓)
𝑁𝑝𝑙,𝑦,𝑅𝑑= 𝑓𝑦(((ℎ − 2 ∙ 𝑡𝑓) ∙ 𝑡𝑤) + (4𝑟2(1 − 0.25𝜋)) + 2 ∙ 𝑏 ∙ 𝑡𝑓)
𝑉𝑝𝑙,𝑦,𝑅𝑑 = 𝐴𝑉∙𝑓𝑦
√3 (3.49)
With 𝐴𝑉 = (𝐴 − 2 ∙ 𝑏 ∙ 𝑡𝑓+ (𝑡𝑤+ 2 ∙ 𝑟)𝑡𝑓)
𝑉𝑝𝑙,𝑦,𝑅𝑑 = (𝐴 − 2 ∙ 𝑏 ∙ 𝑡𝑓+ (𝑡𝑤+ 2 ∙ 𝑟)𝑡𝑓) 𝑓𝑦
√3 With:
AV shear area mm²
Npl,y,Rd design plastic resistance to normal force of the gross cross-section kN
Vpl,y,Rd plastic resistance of shear force kN
Figure 3.18 shows the M-N-V interaction diagrams of HEA240 (𝑛𝑉 = 0.01 up to 𝑛𝑉 = 1.00) 2D (left) and 3D (right).
Figure 3.18 M-N-V interaction diagrams of HEA240 2D (left) and 3D (right)
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
39 3.4 Conclusion regarding analytical solution
An analytical solution made for a rolled I-shaped cross-section subjected to combined bending, shear force and normal force. This solution was divided into 3 cases which all represent another general location of the neutral axis, namely:
The neutral axis of the cross-section is situated in the web
The neutral axis of the cross-section is situated in the roots
The neutral axis of the cross-section is situated in the flange
First of all the utilization ratio 𝑛𝑉 was chosen from 0.01 up to 1.00 and with that the corresponding 𝑛𝑁 and 𝑛𝑀 was calculated. For each cross-section (𝑛𝑉 = 0.01 up to 𝑛𝑉 = 1.00), 100 2D graphs were made in Excel and by using Matlab the data formed a 3D scatter plot. The 3D scatter plot represents the interaction between bending, normal force and shear force.
Bending moment, shear and normal force interaction of I-shaped steel cross-sections
40 4. Numerical investigation
In this chapter the numerical model of HEA240 beam is described which may also be used for other I-shaped cross-sections. The combined bending moment, shear force and normal force interaction is investigated by means of Finite Element (FE) Analyses, using the software package ABAQUS/CAE version 6.13 [9]. To investigate the bending moment, shear force and normal force interaction
In this chapter the numerical model of HEA240 beam is described which may also be used for other I-shaped cross-sections. The combined bending moment, shear force and normal force interaction is investigated by means of Finite Element (FE) Analyses, using the software package ABAQUS/CAE version 6.13 [9]. To investigate the bending moment, shear force and normal force interaction