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![PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES
Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt
L E C T U R E 1 : B A S I C S O F P L A S T I C D E S I G N 38
Lastly, for other cross-section shapes we have the
moment-curvature relations shown in the following figure.
5.1. Rectangular Sections
5. Moment-Rotation Relationship
𝒇𝒓𝒐𝒎 [𝑪𝒐𝒍𝒊𝒏 𝑪𝒂𝒑𝒓𝒂𝒏𝒊]](https://image.slidesharecdn.com/001-basics-of-plastic-analysis-191107085920/85/Part-1-basics-of-plastic-analysis-38-320.jpg)
![PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES
Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt
L E C T U R E 1 : B A S I C S O F P L A S T I C D E S I G N 50
Example (3):
𝟗𝟐𝟐𝟎
𝟏𝟎𝟕𝟓𝟗
𝑴 𝒑𝒄
𝑴 𝒑
+
𝟑𝟒𝟐
𝟑𝟕𝟏
𝑷
𝑷 𝒄
+
𝟐𝟗
𝟑𝟕𝟏
𝑷
𝑷 𝒄
𝟐
= 𝟏
𝑴 𝒑𝒄
𝑴 𝒑
+
𝟖𝟒𝟏
𝟒𝟔𝟏
𝑷
𝑷 𝒄
𝟐
= 𝟏
𝑷
𝑷 𝒄
> 𝟎. 𝟏𝟓:
𝑴 𝒑𝒄
𝑴 𝒑
= 𝟏. 𝟏𝟖 𝟏 −
𝑷
𝑷 𝒄
𝑷
𝑷 𝒄
≤ 𝟎. 𝟏𝟓:
𝑴 𝒑𝒄
𝑴 𝒑
= 𝟏. 𝟎
𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒆:
6.2. Doubly Symmetric Sections
6. Effect of Axial Force on the Plastic Moment Capacity
𝒇𝒓𝒐𝒎 [𝑪𝒐𝒍𝒊𝒏 𝑪𝒂𝒑𝒓𝒂𝒏𝒊]
200 𝑚𝑚
200 𝑚𝑚
360𝑚𝑚20𝑚𝑚20𝑚𝑚
10 𝑚𝑚](https://image.slidesharecdn.com/001-basics-of-plastic-analysis-191107085920/85/Part-1-basics-of-plastic-analysis-50-320.jpg)
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Part (1) basics of plastic analysis
- 1. PLASTIC ANALYSIS ANDDESIGN OF STEEL STRUCTURES CES 608 Port Said University Faculty of Engineering Civil Engineering Department BASICS OF PLASTIC DESIGN 2019
- 2. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 2 1. Introduction • In the elastic analysis of structures superposition is often used and it is valid. • However, an elastic analysis does not give information about the loads that will collapse a structure. • An indeterminate structure may sustain loads greater than the load that first causes a yield to occur at any point in the structure.
- 3. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 3 • In fact, a structure will stand if it is able to find redundancies to yield. • It is only when a structure has exhausted all its redundancies will extra load causes it to fail. • Plastic analysis is the method through which the actual failure load of a structure is calculated, and as will be seen, this failure load can be significantly greater than the elastic load capacity. 1. Introduction
- 4. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 4 𝝈 𝜺 𝟏 𝑬 Upper yield Lower yield Necking Failure Elastic Zone Plastic Zone Strain Hardening The material can sustain strains far in excess of the strain at which yield occurs before failure. This property of the material is called its ductility. 2. Material Behavior
- 5. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 5 For plastic analysis, we change the typical diagram by its idealized diagram. Different idealized diagrams are considered in engineering practice. Some of idealized models are presented: Ref.: Igor A. Karnovsky and Olga Lebed, “Advanced Methods of Structural Analysis” 2. Material Behavior
- 6. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 6 𝝈 𝜺 𝟏 𝑬 Yield point Necking Failure Elastic Zone Plastic Zone Strain Hardening 𝝈 𝒚 𝝈 𝒖 𝜺 𝒚 𝜺 𝒔𝒉 𝜺 𝒖 2. Material Behavior
- 7. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 7 𝝈 𝜺 𝟏 𝑬 Yield point Elastic Zone Plastic Zone complex models exist to represent the real behavior of the material. The most common, and simplest, model is the idealized stress-strain curve (for an ideal elastic-plastic material - which doesn’t exist). 𝝈 𝒚 𝜺 𝒚 𝜺 𝒖 Failure Perfectly Plastic Material Curve Ideal Elastic-Plastic Material Curve 2. Material Behavior
- 8. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 8 𝝈 𝜺 𝟏 𝑬 𝝈 𝒚 𝜺 𝒚 Stress-Strain Curve for Elastic- Perfectly Plastic Material Tension Compression − 𝝈 𝒚 −𝜺 𝒚 Horizontal Plateau 2. Material Behavior
- 9. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 9 • Since so much post-yield strain is modelled, the actual material (or cross section) must also be capable of allowing such strains. • That is, it must be sufficiently ductile for the idealized stress-strain curve to be valid. • So, we consider the behavior of a cross section of an ideal elastic-plastic material subject to bending. • In doing so, we seek the relationship between applied moment and the rotation (or more accurately, the curvature) of a cross section. 𝝈 𝜺 𝟏 𝑬 𝝈 𝒚 𝜺 𝒚 • Once the yield has been reached it is taken that an indefinite amount of strain can occur. 3. Behavior of Cross-Section
- 10. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 10 • Consider an arbitrary cross-section with a vertical plane of symmetry, which is also the plane of loading. • Consider the cross section subject to an increasing bending moment and assess the stresses at each stage. 𝐸. 𝑁. 𝐴 𝐸𝑙𝑎𝑠𝑡𝑖𝑐 𝑁𝑒𝑢𝑡𝑟𝑎𝑙 𝐴𝑥𝑖𝑠 𝑧 𝑦 𝐶. 𝐺 3. Behavior of Cross-Section
- 11. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 11 𝐸. 𝑁. 𝐴 𝑃. 𝑁. 𝐴 𝐸. 𝑁. 𝐴 𝑃. 𝑁. 𝐴 𝜀 < 𝜀 𝑦 𝜀 < 𝜀 𝑦 Strain 𝜎 < 𝜎 𝑦 𝜎 < 𝜎 𝑦 𝜀 = 𝜀 𝑦 𝜀 < 𝜀 𝑦 𝜎 = 𝜎 𝑦 𝜎 < 𝜎 𝑦 𝜀 > 𝜀 𝑦 𝜀 = 𝜀 𝑦 𝜎 = 𝜎 𝑦 𝜎 = 𝜎 𝑦 𝜀 > 𝜀 𝑦 𝜀 > 𝜀 𝑦 𝜎 = 𝜎 𝑦 𝜎 = 𝜎 𝑦 𝜀 > 𝜀 𝑦 𝜀 > 𝜀 𝑦 𝜎 = 𝜎 𝑦 𝜎 = 𝜎 𝑦 Stress Elastic Elasto-Plastic Fully Plastic 𝛼=1 𝛼<1 𝛼=0 𝛼<1 α = the ratio of the depth of the elastic to plastic regions. 𝜎 = 𝜎 𝑦 𝛼=0 Strain hardening 3. Behavior of Cross-Section
- 12. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 12 𝑹𝒐𝒕𝒂𝒕𝒊𝒐𝒏 Failure Elastic Elasto- Plastic Strain Hardening 𝑴 𝒚 𝑴 𝒑 𝑴 Plastic 𝜎 = 𝜎 𝑦 𝜎 = 𝜎 𝑦 𝛼<1 𝜎 = 𝜎 𝑦 𝜎 < 𝜎 𝑦 𝛼=1 𝜎 = 𝜎 𝑦 𝜎 = 𝜎 𝑦 𝛼=0 𝜎 = 𝜎 𝜎 = 𝜎 𝑦 𝛼=0 𝜀 > 𝜀 𝑦 𝜀 = 𝜀 𝑦 𝜀 = 𝜀 𝜀 𝑦 < 𝜀 < 𝜀 3. Behavior of Cross-Section
- 13. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 13 𝑹𝒐𝒕𝒂𝒕𝒊𝒐𝒏 Failure Elastic Elasto- Plastic 𝑴 𝒚 𝑴 𝒑 𝑴 Plastic 𝜎 = 𝜎 𝑦 𝜎 = 𝜎 𝑦 𝛼<1 𝜎 = 𝜎 𝑦 𝜎 < 𝜎 𝑦 𝛼=1 𝜎 = 𝜎 𝑦 𝜎 = 𝜎 𝑦 𝛼=0 𝜀 > 𝜀 𝑦 𝜀 = 𝜀 𝑦 𝜀 = 𝜀 𝜀 𝑦 < 𝜀 < 𝜀 Yield point Elastic Zone Plastic Zone 𝝈 𝒚 𝜺 𝒚 𝜺 𝒖 Failure 3. Behavior of Cross-Section
- 14. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 14 Plastic Hinge • With idealized moment-rotation curve, the cross section linearly sustains moment up to the plastic moment capacity of the section and then yields in rotation an indeterminate amount. • Again, to use this idealization, the actual section must be capable of sustaining large rotations – that is it must be ductile. • Once the plastic moment capacity is reached, the section can rotate freely (that is, it behaves like a hinge, except with moment of Mp at the hinge). • This is termed a plastic hinge and it is the basis for plastic analysis. • At the plastic hinge stresses remain constant, but strains and hence rotations can increase. 3. Behavior of Cross-Section
- 15. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 15 Assumptions: • The material obeys Hooke's law until the stress reaches the yield value and thereafter remains constant. • The yield stresses and the modulus of elasticity have the same value in compression as in tension. • The material is homogeneous and isotropic in both the elastic and plastic states. • The plane transverse sections (the sections perpendicular to the longitudinal axis of the beam) remain plane and normal to the longitudinal axis after bending, the effect of shear being neglected. • There is no resultant axial force on the beam. • The cross section of the beam is symmetrical about an axis through its centroid parallel to plane of bending. • Every layer of the material is free to expand and contract longitudinally and laterally under the stress as if separated from the other layers. 4. Analysis of Cross-Section
- 16. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 16 𝐸. 𝑁. 𝐴 𝑃. 𝑁. 𝐴 𝜎 = 𝜎 𝑦 ത𝑦𝑐 𝐶 𝜎 = 𝜎 𝑦𝜎 ≤ 𝜎 𝑦 𝜎 = 𝜎 𝑦 𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑇 𝐶 𝑇 𝑦𝑡 𝑦 𝑏 ത𝑦𝑡 𝐴 𝑐 𝐴 𝑡 𝐶 = 𝑇 𝑀 𝑦 = 𝜎 𝑦 × 𝐼 𝑦𝑡 𝑀 𝑦 = 𝜎 𝑦 × 𝑍 𝑒 𝑀 𝑦 = yield moment 𝑍 𝑒 = elastic section modulus 𝐶 = 𝑇 𝜎 𝑦 𝐴 𝑐 = 𝜎 𝑦 𝐴 𝑡 𝐴 𝑐 = 𝐴 𝑡 = 𝐴/2 𝑀 𝑝 = 𝜎 𝑦 (𝐴 𝑐 ത𝑦𝑐 + 𝐴 𝑡 ത𝑦𝑡) = 𝜎 𝑦 𝐴 2 (ത𝑦𝑐 + ത𝑦𝑡) 𝑀 𝑝 = 𝜎 𝑦 × 𝑍 𝑝 → 𝑀 𝑝 = plastic moment 𝑍 𝑝 = plastic section modulus 4. Analysis of Cross-Sections
- 17. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 17 𝐸. 𝑁. 𝐴 𝜎 𝑦 𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑇 𝑑 𝑏 2 3 𝑑 𝛼𝑑 2 3 𝛼𝑑 ൗ𝑑 2 𝐶 𝐶1 𝐶2 𝑇1 𝑇2 𝑇 𝐶 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦𝑑 1 − 𝛼 4 𝑑 1 + 𝛼 2 𝑃. 𝑁. 𝐴 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 ൗ𝑑 2 ൗ𝑑 2 𝑍 𝑒 = 𝑏 𝑑2 /6 𝑀 𝑦 = 𝜎 𝑦 𝑍 𝑒 𝑀 𝑦 = 𝜎 𝑦 𝑏 𝑑2 6 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝐶 = 𝑇 = 1 2 𝜎 𝑦 𝑑 2 𝑏 = 𝜎 𝑦 𝑏 𝑑 4 𝑀 𝑦 = 𝐶 × 2 3 𝑑 = 𝜎 𝑦 𝑏 𝑑 4 × 2 3 𝑑 𝑀 𝑦 = 𝜎 𝑦 𝑏 𝑑2 6 Alternative method using stress block: M 4.1. Rectangular Cross-Section 4. Analysis of Cross-Sections
- 18. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 18 𝐸. 𝑁. 𝐴 𝜎 𝑦 𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑇 𝑑 𝑏 2 3 𝑑 𝛼𝑑 2 3 𝛼𝑑 ൗ𝑑 2 𝐶 𝐶1 𝐶2 𝑇1 𝑇2 𝑇 𝐶 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦𝑑 1 − 𝛼 4 𝑑 1 + 𝛼 2 𝑃. 𝑁. 𝐴 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 ൗ𝑑 2 ൗ𝑑 2 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝐶1 = 𝑇1 = 𝜎 𝑦 𝑑 (1 − 𝛼) 2 𝑏 , 𝐴𝑟𝑚1 = 𝑑 (1 + 𝛼) 2 𝑀𝑒𝑝 = 𝐶1 × 𝐴𝑟𝑚1 + 𝐶2 × 𝐴𝑟𝑚2 𝑀𝑒𝑝 = 𝜎 𝑦 𝑑 (1 − 𝛼) 2 𝑏 × 𝑑 (1 + 𝛼) 2 + 1 2 𝜎 𝑦 𝛼 𝑑 2 𝑏 × 2 3 𝛼 𝑑 using stress block: 𝐶2 = 𝑇2 = 1 2 𝜎 𝑦 𝛼 𝑑 2 𝑏 , 𝐴𝑟𝑚2 = 2 3 𝛼 𝑑 4.1. Rectangular Cross-Section 4. Analysis of Cross-Sections
- 19. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 19 𝐸. 𝑁. 𝐴 𝜎 𝑦 𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑇 𝑑 𝑏 2 3 𝑑 𝛼𝑑 2 3 𝛼𝑑 ൗ𝑑 2 𝐶 𝐶1 𝐶2 𝑇1 𝑇2 𝑇 𝐶 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦𝑑 1 − 𝛼 4 𝑑 1 + 𝛼 2 𝑃. 𝑁. 𝐴 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 ൗ𝑑 2 ൗ𝑑 2 𝑀𝑒𝑝 = 𝜎 𝑦 𝑑 (1 − 𝛼) 2 𝑏 × 𝑑 (1 + 𝛼) 2 + 1 2 𝜎 𝑦 𝛼 𝑑 2 𝑏 × 2 3 𝛼 𝑑 𝑀𝑒𝑝 = 𝜎 𝑦 𝑏𝑑2 4 1 − 𝛼2 + 2 3 𝛼2 𝑀𝑒𝑝 = 𝜎 𝑦 𝑏𝑑2 4 1 − 𝛼2 3 = 𝜎 𝑦 𝑏𝑑2 6 3 − 𝛼2 2 using stress block:𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 4.1. Rectangular Cross-Section 4. Analysis of Cross-Sections
- 20. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 20 𝐸. 𝑁. 𝐴 𝜎 𝑦 𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑇 𝑑 𝑏 2 3 𝑑 𝛼𝑑 2 3 𝛼𝑑 ൗ𝑑 2 𝐶 𝐶1 𝐶2 𝑇1 𝑇2 𝑇 𝐶 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦𝑑 1 − 𝛼 4 𝑑 1 + 𝛼 2 𝑃. 𝑁. 𝐴 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 ൗ𝑑 2 ൗ𝑑 2 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝐶 = 𝑇 = 𝜎 𝑦 𝑑 2 𝑏 , 𝐴𝑟𝑚 = 𝑑 2 𝑀 𝑝 = 𝜎 𝑦 𝑑 2 𝑏 × 𝑑 2 = 𝜎 𝑦 𝑏 𝑑2 4 = 𝜎 𝑦 𝑍 𝑝 using stress block: 𝑊ℎ𝑒𝑟𝑒: 𝑍 𝑝 = 𝑏 𝑑2 4 4.1. Rectangular Cross-Section 4. Analysis of Cross-Sections
- 21. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 21 𝑓 = 𝑀 𝑝 𝑀 𝑦 = 𝜎 𝑦 𝑍 𝑝 𝜎 𝑦 𝑍 𝑒 = 𝑍 𝑝 𝑍 𝑒 Shape Factor (𝒇): • It is the ratio of elastic to plastic moment capacity: For rectangular section: 𝑓 = 𝑏 𝑑2 4 𝑏 𝑑2 6 = 6 4 = 1.5 𝑓 = 1.5 𝑓 = 1.698 𝑓 = 2.0 𝑓 ≈ 1.12 𝑡𝑜 1.15 𝑓 ≈ 1. 8 𝑓 ≈ 1. 0 Shapefactor(f)increasesasthematerial inthemiddleofsectionincreases 𝑓 ≈ 1.27 4.1. Rectangular Cross-Section 4. Analysis of Cross-Sections
- 22. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 22 Example (1): • Find the plastic moment and shape factor for the shown section. 200 𝑚𝑚 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚 4.2. Doubly Symmetric Sections 4. Analysis of Cross-Sections
- 23. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 23 Example (1): 1- Elastic 𝐸. 𝑁. 𝐴 200 𝑚𝑚 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚 200𝑚𝑚200𝑚𝑚 𝐼 = 10 × 3603 12 +2 × 200 × 203 12 + 200 × 20 × 1902 𝐼 = 327946666.7 𝑚𝑚4 𝑍 𝑒 = 327946666.7 200 = 1639733.333 𝑚𝑚3 𝑀 𝑦 = 𝜎 𝑦 𝑍 𝑒 = 1639733.333 𝜎 𝑦 𝑁. 𝑚𝑚 4.2. Doubly Symmetric Sections 4. Analysis of Cross-Sections
- 24. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 24 Example (1): 2- Plastic 𝐸. 𝑁. 𝐴 𝐶2 200 𝑚𝑚 𝜎 𝑦 𝑃. 𝑁. 𝐴 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚 𝜎 𝑦 180𝑚𝑚 380𝑚𝑚 200𝑚𝑚200𝑚𝑚 𝐶1 𝑇1 𝑇2 4.2. Doubly Symmetric Sections 4. Analysis of Cross-Sections
- 25. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 25 Example (1): 2- Plastic 𝐸. 𝑁. 𝐴 𝐶2 200 𝑚𝑚 𝜎𝑦 𝑃. 𝑁. 𝐴 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚 𝜎𝑦 180𝑚𝑚 380𝑚𝑚 200𝑚𝑚200𝑚𝑚 𝐶1 𝑇1 𝑇2 𝐶1 = 𝑇1 = 200 × 20 × 𝜎 𝑦 = 4000𝜎 𝑦 N 𝑍 𝑝 = 1844000 𝑚𝑚3 𝑓 = 𝑀 𝑝 𝑀𝑒 = 𝑍 𝑝 𝑍 𝑒 = 1844000 1639733.333 𝑓 = 1.124573101 𝐶2 = 𝑇2 = 10 × 180 × 𝜎 𝑦 = 1800𝜎 𝑦 N 𝑀 𝑝 = 𝐶1 𝐴𝑟𝑚1 + 𝐶2 𝐴𝑟𝑚2 = 4000 𝜎 𝑦 × 380 + 1800 𝜎 𝑦 × 180 𝑀 𝑝 = 1844000 𝜎 𝑦 𝑁. 𝑚 3- Shape Factor (f) 4.2. Doubly Symmetric Sections 4. Analysis of Cross-Sections
- 26. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 26 Example (2): • Find the plastic moment and shape factor for the shown section. 160 𝑚𝑚200𝑚𝑚 10 𝑚𝑚 16𝑚𝑚184𝑚𝑚 4.3. Sections Symmetric about Vertical Axis 4. Analysis of Cross-Sections
- 27. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 27 Example (2): 1- Elastic 𝐴 = 160 × 16 + 10 × 184 = 4400 𝑚𝑚2 𝐴. 𝑦𝑒 = 160 × 16 × 8 + 10 × 184 × 108 𝑦𝑒 = 219200 4400 = 49.818 𝑚𝑚 𝐼𝑧 = (160 − 10) × 163 3 + 10 × 2003 3 − 4400 × 49.8182 = 15951321.21 𝑚𝑚4 𝑀 𝑦 = 𝜎 𝑦 𝑍 𝑒 = 106213.3979 𝜎 𝑦 𝑁. 𝑚𝑚 160 𝑚𝑚 200𝑚𝑚 10 𝑚𝑚 16𝑚𝑚184𝑚𝑚 49.818𝑚𝑚 𝐸. 𝑁. 𝐴 𝑦 𝑧 𝑍 𝑒 = 15951321.21 200 − 49.818 = 106213.3979 𝑚𝑚3 4.3. Sections Symmetric about Vertical Axis 4. Analysis of Cross-Sections
- 28. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 28 160 𝑚𝑚200𝑚𝑚 10 𝑚𝑚 16𝑚𝑚184𝑚𝑚 𝐸. 𝑁. 𝐴 𝜎 𝑦 𝐶 𝑇 𝜎 𝑦 𝑦 𝑧 Example (2): 2- Plastic To find plastic center, assume plastic neutral axis (P.N.A) at bottom of flange: 𝐶 = 160 × 16 × 𝜎 𝑦 = 2560 𝜎 𝑦 , 𝑇 = 184 × 10 × 𝜎 𝑦 = 1840 𝜎 𝑦 𝑁 Since 𝐶 > 𝑇, the plastic neutral axis must be shifted upward inside the flange. 4.3. Sections Symmetric about Vertical Axis 4. Analysis of Cross-Sections
- 29. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 29 160 𝑚𝑚200𝑚𝑚 10 𝑚𝑚 16𝑚𝑚184𝑚𝑚 13.75𝑚𝑚 49.818𝑚𝑚 𝐸. 𝑁. 𝐴 𝜎 𝑦 𝐶1 𝑇2 𝜎 𝑦 𝑦 𝑧 𝑃. 𝑁. 𝐴 𝑇1 Example (2): 2- Plastic The compression areas equal to tension area: Ac = At 160 × 𝑦𝑝 = 160 × 16 − 𝑦𝑝 + 10 × 184 → 𝑦𝑝 = 13.75 𝑚𝑚 4.3. Sections Symmetric about Vertical Axis 4. Analysis of Cross-Sections
- 30. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 30 Example (2): 2- Plastic 160 𝑚𝑚 200𝑚𝑚 10 𝑚𝑚 16𝑚𝑚184𝑚𝑚 13.75𝑚𝑚 49.818𝑚𝑚 𝐸. 𝑁. 𝐴 𝜎𝑦 𝐶1 𝑇2 𝜎𝑦 𝑦 𝑧 𝑃. 𝑁. 𝐴 𝑇1 𝐶1 = 160 × 13.75 × 𝜎 𝑦 = 2200𝜎 𝑦 N 𝑍 𝑝 = 188950 𝑚𝑚3 𝑓 = 𝑀 𝑝 𝑀𝑒 = 𝑍 𝑝 𝑍 𝑒 = 188950 106213.3979 = 1.779 𝑇1 = 160 × 2.25 × 𝜎 𝑦 = 360𝜎 𝑦 N 𝑀 𝑝 = 2200 𝜎 𝑦 × 6.875 + 360 𝜎 𝑦 × 1.125 + 1840 𝜎 𝑦 × 94.25 = 188950 𝜎 𝑦 𝑁. 𝑚 3- Shape Factor (f) 𝑇2 = 10 × 184 × 𝜎 𝑦 = 1840𝜎 𝑦 N 4.3. Sections Symmetric about Vertical Axis 4. Analysis of Cross-Sections
- 31. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 31 𝜀 𝑦 𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑑 𝑏 𝛼𝑑 𝜀 𝑦 𝜀 𝑦 𝜀 𝑦 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 ൗ𝑑 2 𝐸. 𝑁. 𝐴 𝜎 𝑦 𝐶. 𝐺 𝑇 𝑑 𝑏 2 3 𝑑 𝛼𝑑 2 3 𝛼𝑑 ൗ𝑑 2 𝐶 𝐶1 𝐶2 𝑇1 𝑇2 𝑇 𝐶 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦 𝜎 𝑦𝑑 1 − 𝛼 4 𝑑 1 + 𝛼 2 𝑃. 𝑁. 𝐴 ൗ𝑑 2 ൗ𝑑 2 ൗ𝑑 2 𝑦 𝜀 𝜅 𝜀 𝜀 𝑦 𝜀 𝜅𝑦 𝜀 Strain Stress 𝜅 𝑦 5.1. Rectangular Sections 5. Moment-Rotation Relationship
- 32. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 32 𝜀 𝑦 𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑑 𝑏 𝛼𝑑 𝜀 𝑦 𝜀 𝑦 𝜀 𝑦 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 ൗ𝑑 2 ൗ𝑑 2 𝜅 𝑦 𝑦 𝜀 𝜅 𝜀 𝜀 𝑦 𝜀 𝜅𝑦 𝜀 Strain 𝑟𝑜𝑡𝑎𝑡𝑖𝑜𝑛 𝑜𝑟 𝑐𝑢𝑟𝑣𝑎𝑡𝑢𝑟𝑒, 𝜅 = 𝜀 𝑦 𝑬𝑳𝑨𝑺𝑻𝑰𝑪: 𝜅 𝑦 = 𝜀 𝑦 𝑑/2 = 2 𝜀 𝑦 𝑑 When the section is subjected to yield moment 𝑀 𝑦 𝜅 = 1 𝑅 = 𝑀 𝐸𝐼 𝜅 𝑦 = 𝑀 𝑦 𝐸𝐼 𝑀 𝑀 𝑦 = 𝜅 𝜅 𝑦 5.1. Rectangular Sections 5. Moment-Rotation Relationship
- 33. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 33 𝜀 𝑦 𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑬𝑳𝑨𝑺𝑻𝑰𝑪 𝑑 𝑏 𝛼𝑑 𝜀 𝑦 𝜀 𝑦 𝜀 𝑦 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 𝑷𝑳𝑨𝑺𝑻𝑰𝑪 ൗ𝑑 2 ൗ𝑑 2 𝜅 𝑦 𝑦 𝜀 𝜅 𝜀 𝜀 𝑦 𝜀 𝜅𝑦 𝜀 Strain 𝑬𝑳𝑨𝑺𝑻𝑶 − 𝑷𝑳𝑨𝑺𝑻𝑰𝑪: 𝜅 = 𝜀 𝑦 𝛼 𝑑/2 = 2 𝜀 𝑦 𝛼 𝑑 For moments applied beyond the yield moment, the curvature can be found by noting that the yield strain, 𝜀 𝑦 , occurs at a distance from the neutral axis of α 𝑑/2, giving: Thus, the ratio curvature to yield curvature is: 𝜅 𝜅 𝑦 = ൘ 2 𝜀 𝑦 𝛼 𝑑 2 𝜀 𝑦 𝑑 = 1 𝛼 → 𝛼 = 𝜅 𝑦 𝜅 5.1. Rectangular Sections 5. Moment-Rotation Relationship
- 34. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 34 𝑀 𝑀 𝑦 = 𝜎 𝑦 𝑏𝑑2 6 3 − 𝛼2 2 𝜎 𝑦 𝑏𝑑2 6 = 3 − 𝛼2 2 Also, the ratio of elasto-plastic moment to yield moment is: If we now substitute the value 𝛼 = 𝜅 𝑦 𝜅 we find: 𝑀 𝑀 𝑦 = 1 2 3 − 𝜅 𝑦 𝜅 2 = 1.5 − 0.5 𝜅 𝜅 𝑦 −2 5.1. Rectangular Sections 5. Moment-Rotation Relationship
- 35. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 35 0.00 0.25 0.50 0.75 1.00 1.25 1.50 1.75 0 1 2 3 4 5 6 7 8 / / 𝑀 𝑀 𝑦 = 1.5 − 0.5 𝜅 𝜅 𝑦 −2 𝑀 𝑀 𝑦 = 𝜅 𝜅 𝑦 𝑑 𝑏 𝑦 𝑧 𝑀 = 𝑀 𝑦 𝑀 = 𝑀 𝑝 𝑓 = 5.1. Rectangular Sections 5. Moment-Rotation Relationship
- 36. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 36 There are some important observations to be made from this graph: • To reach the plastic moment capacity of the section requires large curvatures. Thus the section must be ductile. • The full cross-section plasticity associated with the plastic moment capacity of a section can only be reached at infinite curvature (or infinite strain). Since this is impossible, we realize that the full plastic moment capacity is unobtainable. 5.1. Rectangular Sections 5. Moment-Rotation Relationship
- 37. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 37 To show that the idea of the plastic moment capacity of section is still useful. Firstly we note that strain hardening in mild steel begins to occur at a strain of about (10 𝜀 𝑦). At this strain, the corresponding moment ratio is: 𝑀 𝑀 𝑦 = 1.5 − 0.5 10 −2 = 1.495 Since this is about 99.7% of the plastic moment capacity, we see that the plastic moment capacity of a section is a good approximation of the section’s capacity. These calculations are based on a ductility ratio of 10. This is about the level of ductility a section requires to be of use in any plastic collapse analysis. 5.1. Rectangular Sections 5. Moment-Rotation Relationship
- 38. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 38 Lastly, for other cross-section shapes we have the moment-curvature relations shown in the following figure. 5.1. Rectangular Sections 5. Moment-Rotation Relationship 𝒇𝒓𝒐𝒎 [𝑪𝒐𝒍𝒊𝒏 𝑪𝒂𝒑𝒓𝒂𝒏𝒊]
- 39. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 39 𝐸. 𝑁. 𝐴𝐶. 𝐺 𝑃𝑙𝑎𝑠𝑡𝑖𝑐 𝐶𝑒𝑛𝑡𝑒𝑟 𝑑 𝑏 ൗ𝑑 2 𝑇 𝐶 𝜎 𝑦 𝜎 𝑦 𝑃. 𝑁. 𝐴 ൗ𝑑 2 ൗ𝑑 2 𝐶 = 𝑇 = 𝜎 𝑦 𝑑 2 𝑏 + 2 𝜎 𝑦 𝐶` = 𝜎 𝑦 𝜎 𝑦 𝛽 𝑑 𝐶` = 2 𝜎 𝑦 𝛽 𝑑 𝑏 , 𝐴𝑟𝑚 = 𝛽 𝑑 2 the plastic ‘squash load’ of the section is given by: 𝑃𝑐 = 𝜎 𝑦 𝑏 𝑑 𝑇ℎ𝑢𝑠, 𝑃 = 2 𝛽𝑃𝑐 𝑃 = 𝐶 − 𝑇 + 𝐶` = 𝐶` 𝑀 𝑝𝑐 = 𝑀 𝑝 − 𝑃 𝛽 𝑑 2 The plastic moment is given by: 6.1. Rectangular Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 40. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 40 𝑈𝑠𝑖𝑛𝑔, 𝑀 𝑝 = 𝜎 𝑦 𝑏 𝑑2 4 𝑀 𝑝𝑐 = 𝜎 𝑦 𝑏 𝑑2 4 − 2 𝜎 𝑦 𝛽 𝑑 𝑏 𝛽 𝑑 2 = 𝜎 𝑦 𝑏 𝑑2 4 1 − 4𝛽2 𝑀 𝑝𝑐 = 𝑀 𝑝 1 − 4𝛽2 𝑀 𝑝𝑐 = 𝑀 𝑝 − 𝑃 𝛽 𝑑 2 The plastic moment is given by: 𝑁𝑜𝑡𝑖𝑛𝑔 𝑡ℎ𝑎𝑡, 2 𝛽 = 𝑃/𝑃𝑐 𝑀 𝑝𝑐 𝑀 𝑝 = 1 − 4𝛽2 = 1 − 𝑃 𝑃𝑐 2 The interaction equation is given by: 𝑀 𝑝𝑐 𝑀 𝑝 + 𝑃 𝑃𝑐 2 = 1 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 P/Pc Mpc/Mp 𝑀 𝑝𝑐 𝑀 𝑝 + 𝑃 𝑃𝑐 2 = 1 6.1. Rectangular Sections 6. Effect of Axial Force on the Plastic Moment Capacity 𝑑 𝑏 𝑦 𝑧
- 41. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 41 Example (3): • Deduce and plot the interaction equation between axial and bending of the shown section. 200 𝑚𝑚 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 42. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 42 Example (3): Crushing Load: 𝑃𝑐 = 𝐴. 𝜎 𝑦 = 10 × 360 + 2 × 200 × 20 𝜎 𝑦 = 11600 𝜎 𝑦 𝑁 𝐸. 𝑁. 𝐴 𝐶2 200 𝑚𝑚 𝜎 𝑦 𝑃. 𝑁. 𝐴 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚 𝜎 𝑦 200𝑚𝑚200𝑚𝑚 𝐶1 𝑇1 𝑇2 𝑃𝑐 Pure Moment Pure Axial Load 𝜎 𝑦 𝜎 𝑦 Plastic Moment (deduced before): 𝑀 𝑝 = 1844000 𝜎 𝑦 𝑁. 𝑚 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 43. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 43 Example (3): Stress distribution: 1) 𝛽 𝑑 ≤ 180 𝑚𝑚 𝐸. 𝑁. 𝐴 𝐶2 200 𝑚𝑚 𝜎 𝑦 𝑃. 𝑁. 𝐴 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚 𝜎 𝑦 200𝑚𝑚200𝑚𝑚 𝐶1 𝑇1 𝑇2 𝛽 𝑑 2 𝜎 𝑦 Pure Moment Effect of Axial Force (P) + = 𝜎 𝑦 𝜎 𝑦 𝐶` Combined Effect 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 44. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 44 Example (3): Modified Plastic Moment: 𝑀 𝑝𝑐 𝑀 𝑝 = 1844000 − 1600000 𝛽2 1844000 = 1 − 400 461 𝛽2 𝑀 𝑝𝑐 𝑀 𝑝 = 1 − 400 461 29 20 𝑃 𝑃𝑐 2 = 1 − 841 461 𝑃 𝑃𝑐 2 𝑏𝑢𝑡 𝛽 = 29 20 𝑃 𝑃𝑐 𝑀 𝑝𝑐 𝑀 𝑝 + 841 461 𝑃 𝑃𝑐 2 = 1Interaction Equation: Load: 𝑃 = 2𝜎 𝑦 𝛽 𝑑 𝑡 𝑤 = 2 × 400 × 10 = 8000 𝛽 𝜎 𝑦 𝑁 Load ratio: 𝑃 𝑃𝑐 = 8000 𝛽 𝜎 𝑦 11600 𝜎 𝑦 = 20 29 𝛽 → 𝛽 = 29 20 𝑃 𝑃𝑐 𝑀 𝑝𝑐 = 𝑀 𝑝 − 𝑃 𝛽𝑑 2 𝑀 𝑝𝑐 = 𝑀 𝑝 − 400 2 𝛽 × 8000 𝛽 𝜎 𝑦 = 1844000 − 1600000 𝛽2 𝜎 𝑦 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 45. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 45 Example (3): Stress distribution: 2) 𝛽 𝑑 ≥ 180 𝑚𝑚 𝐸. 𝑁. 𝐴 200 𝑚𝑚 𝜎 𝑦 𝑃. 𝑁. 𝐴 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚 𝜎 𝑦 200𝑚𝑚200𝑚𝑚 𝛽 𝑑 2 𝜎 𝑦 Pure Axial Force (Pc) Effect of Axial Force (P) 𝜎 𝑦 𝜎 𝑦 𝐶` Combined Effect 𝑃𝑐 =+ 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 46. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 46 Example (3): Modified Plastic Moment: Load: Load ratio: 𝑃 𝑃𝑐 = 1 − 80000 1 − 2 𝛽 𝜎 𝑦 11600 𝜎 𝑦 = 1 − 200 29 1 − 2 𝛽 𝑃 𝑃𝑐 = 400 29 𝛽 − 171 29 → 𝑀 𝑝𝑐 = 𝐶` × 𝐴𝑟𝑚 = 2𝜎 𝑦 𝑏 𝑑 2 − 𝛽 𝑑 × 1 2 𝑑 2 + 𝛽 𝑑 𝑃 = 𝑃𝑐 − 2𝜎 𝑦 𝑑 2 − 𝛽 𝑑 𝑏 = 𝑃𝑐 − 𝜎 𝑦 × 400 × 200 1 − 2 𝛽 𝑃 = 𝑃𝑐 − 80000 1 − 2 𝛽 𝜎 𝑦 𝑁 𝛽 = 29 400 𝑃 𝑃𝑐 + 171 400 𝑀 𝑝𝑐 = 𝐶` × 𝐴𝑟𝑚 = 𝜎 𝑦 𝑏 𝑑2 4 1 − 4 𝛽2 = 8 × 106 1 − 4 𝛽2 𝜎 𝑦 𝑀 𝑝𝑐 𝑀 𝑝 = 8 × 106 1 − 4 𝛽2 𝜎 𝑦 1844000𝜎 𝑦 = 2000 461 1 − 4 𝛽2 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 47. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 47 Example (3): This yields: But 𝛽 = 29 400 𝑃 𝑃𝑐 + 171 400 𝑀 𝑝𝑐 𝑀 𝑝 = 2000 461 1 − 4 29 400 𝑃 𝑃𝑐 + 171 400 2 𝑀 𝑝𝑐 𝑀 𝑝 = 10759 9220 − 9918 9220 𝑃 𝑃𝑐 − 841 9220 𝑃 𝑃𝑐 2 9220 10759 𝑀 𝑝𝑐 𝑀 𝑝 + 342 371 𝑃 𝑃𝑐 + 29 371 𝑃 𝑃𝑐 2 = 1 The interaction equation: 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 48. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 48 Example (3): There are 2 interaction equations: 𝑀 𝑝𝑐 𝑀 𝑝 = 1 − 841 461 𝑃 𝑃𝑐 2 = 1 − 841 461 9 29 2 = 380 461 = 0.8243 𝑃 𝑃𝑐 = 20 29 𝛽 = 20 29 × 180 400 = 9 29 = 0.3103 • The first one is valid for 𝛽𝑑 = 0 𝑡𝑜 180 • The other is valid for 𝛽𝑑 = 180 𝑡𝑜 200 Both equations are valid for 𝛽𝑑 = 180 , 𝑖. 𝑒. 𝛽 = Τ180 400 By substituting into By substituting into first interaction equation Same result can be obtained by substituting into second equation: 𝑀 𝑝𝑐 𝑀 𝑝 = 10759 9220 − 9918 9220 × 9 29 − 841 9220 × 9 29 2 = 380 461 = 0.8243 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity
- 49. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 49 0.00 0.10 0.20 0.30 0.40 0.50 0.60 0.70 0.80 0.90 1.00 0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1 P/Pc Mpc/Mp Example (3): 𝟗𝟐𝟐𝟎 𝟏𝟎𝟕𝟓𝟗 𝑴 𝒑𝒄 𝑴 𝒑 + 𝟑𝟒𝟐 𝟑𝟕𝟏 𝑷 𝑷 𝒄 + 𝟐𝟗 𝟑𝟕𝟏 𝑷 𝑷 𝒄 𝟐 = 𝟏 𝑴 𝒑𝒄 𝑴 𝒑 + 𝟖𝟒𝟏 𝟒𝟔𝟏 𝑷 𝑷 𝒄 𝟐 = 𝟏 𝟎. 𝟑𝟏𝟎𝟑 𝟎.𝟖𝟐𝟒𝟑 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity 200 𝑚𝑚 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚
- 50. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 50 Example (3): 𝟗𝟐𝟐𝟎 𝟏𝟎𝟕𝟓𝟗 𝑴 𝒑𝒄 𝑴 𝒑 + 𝟑𝟒𝟐 𝟑𝟕𝟏 𝑷 𝑷 𝒄 + 𝟐𝟗 𝟑𝟕𝟏 𝑷 𝑷 𝒄 𝟐 = 𝟏 𝑴 𝒑𝒄 𝑴 𝒑 + 𝟖𝟒𝟏 𝟒𝟔𝟏 𝑷 𝑷 𝒄 𝟐 = 𝟏 𝑷 𝑷 𝒄 > 𝟎. 𝟏𝟓: 𝑴 𝒑𝒄 𝑴 𝒑 = 𝟏. 𝟏𝟖 𝟏 − 𝑷 𝑷 𝒄 𝑷 𝑷 𝒄 ≤ 𝟎. 𝟏𝟓: 𝑴 𝒑𝒄 𝑴 𝒑 = 𝟏. 𝟎 𝑨𝒑𝒑𝒓𝒐𝒙𝒊𝒎𝒂𝒕𝒆: 6.2. Doubly Symmetric Sections 6. Effect of Axial Force on the Plastic Moment Capacity 𝒇𝒓𝒐𝒎 [𝑪𝒐𝒍𝒊𝒏 𝑪𝒂𝒑𝒓𝒂𝒏𝒊] 200 𝑚𝑚 200 𝑚𝑚 360𝑚𝑚20𝑚𝑚20𝑚𝑚 10 𝑚𝑚
- 51. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 51 𝑧 𝑃 𝑀𝑒𝑝 𝛼 𝑑 𝑀 𝑝 𝑀 𝑦 𝑀 = 0 𝑙 𝑝 𝐿/2 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑒𝑙𝑎𝑠𝑡𝑜 − 𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑑 𝑃 < 𝑃𝑦 𝑃 = 𝑃𝑦 𝑃 = 𝑃𝑝 𝐵. 𝑀. 𝐷 𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑧𝑜𝑛𝑒 𝐿/2 7.1. Simply-Supported Beam Subjected to Midspan Point Load 7. Plastic Hinge Formation
- 52. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 52 The bending moment at center of beam is given by: 𝑀𝑐 = 𝑃𝐿 4 → 𝑃 = 4𝑀𝑐 𝐿 The bending moment at first yield (at center of beam) is given by: 𝑀𝑐 = 𝑀 𝑦 = 𝑃𝑦 𝐿 4 → 𝑃𝑦 = 4𝑀 𝑦 𝐿 The collapse load occurs when the moment at the center reaches the plastic moment capacity: 𝑀𝑐 = 𝑀 𝑝 = 𝑃𝑝 𝐿 4 → 𝑃𝑝 = 4𝑀 𝑝 𝐿 7.1. Simply-Supported Beam Subjected to Midspan Point Load 7. Plastic Hinge Formation
- 53. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 53 The ratio collapse to yield load is: 𝑃𝑃 𝑃𝑦 = 4𝑀 𝑝 𝐿 4𝑀 𝑦 𝐿 = 𝑀 𝑝 𝑀 𝑦 𝑏𝑢𝑡, 𝑀 𝑝 𝑀 𝑦 = 𝑓 𝑡ℎ𝑢𝑠, 𝑃𝑝 𝑃𝑦 = 𝑓 This is a general result: the ratio of collapse load to first yield load is the shape factor of the member, for statically determinate prismatic structures. 7.1. Simply-Supported Beam Subjected to Midspan Point Load 7. Plastic Hinge Formation
- 54. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 54 Shape of the Plastic Hinge: We are also interested in the plastic hinge, and the zone of Elasto- plastic bending. As can be seen from the diagram, the plastic material zones extend from the center out to the point where the moment equals the yield moment. Using similar triangles from the bending moment diagram at collapse, we see that: 𝑀 𝑃 𝐿 = 𝑀 𝑝 − 𝑀 𝑦 𝑙 𝑝 = 𝑀 𝑝 − 𝑀𝑒𝑝 2𝑧 Equating the first two equations gives: 𝑙 𝑝 = 𝐿 𝑀 𝑝 − 𝑀 𝑦 𝑀 𝑃 = 𝐿 1 − 𝑀 𝑦 𝑀 𝑝 = 𝐿 1 − 1 𝑓 For a beam with a rectangular cross section (𝑓 = 1.5) the plastic hinge extends for a length: 𝑙 𝑝 = 𝐿 1 − 1 1.5 = 𝐿 3 7.1. Simply-Supported Beam Subjected to Midspan Point Load 7. Plastic Hinge Formation
- 55. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 55 Shape of the Plastic Hinge: Lastly, the shape of the hinge follows from the first and third equation: 𝑀 𝑃 𝐿 = 𝑀 𝑝 − 𝑀𝑒𝑝 2𝑧 → 𝑧 𝐿 = 1 2 1 − 𝑀𝑒𝑝 𝑀 𝑝 𝑧 𝐿 = 1 2 1 − 𝜎 𝑦 𝑏𝑑2 6 3 − 𝛼2 2 𝜎 𝑦 𝑏𝑑2 4 = 1 2 1 − 1 3 3 − 𝛼2 𝑧 𝐿 = 𝛼2 6 , 𝑎𝑡 𝛼 = 1, 𝑧 = 𝑙 𝑝 2 = 𝐿 6 , 𝑙 𝑝 = 𝐿 3 From expressions for the elasto-plastic and plastic moments, we have: 𝐹𝑖𝑛𝑎𝑙𝑙𝑦, 𝛼 = 6𝑧 𝐿 7.1. Simply-Supported Beam Subjected to Midspan Point Load 7. Plastic Hinge Formation
- 56. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 56 𝑧 𝑤 𝑀𝑒𝑝 𝛼 𝑑 𝑀 𝑝 𝑀 𝑦 𝑀 = 0 𝑙 𝑝 𝑒𝑙𝑎𝑠𝑡𝑖𝑐 𝑒𝑙𝑎𝑠𝑡𝑖𝑐𝑒𝑙𝑎𝑠𝑡𝑜 − 𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑑 𝑤 < 𝑤 𝑦 𝑤 = 𝑤 𝑦 𝑤 = 𝑤 𝑝 𝐵. 𝑀. 𝐷 𝑝𝑙𝑎𝑠𝑡𝑖𝑐 𝑧𝑜𝑛𝑒 𝐿/2 𝐿/2 7.2. Simply-Supported Beam Subjected to Uniform Load 7. Plastic Hinge Formation
- 57. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 57 The bending moment at center of beam is given by: 𝑀𝑐 = 𝑤𝐿2 8 → 𝑤 = 8𝑀𝑐 𝐿2 The bending moment at first yield (at center of beam) is given by: 𝑀𝑐 = 𝑀 𝑦 = 𝑤 𝑦 𝐿2 8 → 𝑤 𝑦 = 8𝑀 𝑦 𝐿2 The collapse load occurs when the moment at the center reaches the plastic moment capacity: 𝑀𝑐 = 𝑀 𝑝 = 𝑤 𝑝 𝐿2 8 → 𝑤 𝑝 = 8𝑀 𝑝 𝐿2 7.2. Simply-Supported Beam Subjected to Uniform Load 7. Plastic Hinge Formation
- 58. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 58 The ratio collapse to yield load is: 𝑤 𝑃 𝑤 𝑦 = 8𝑀 𝑝 𝐿2 8𝑀 𝑦 𝐿2 = 𝑀 𝑝 𝑀 𝑦 𝑏𝑢𝑡, 𝑀 𝑝 𝑀 𝑦 = 𝑓 𝑡ℎ𝑢𝑠, 𝑤 𝑝 𝑤 𝑦 = 𝑓 This is a general result: the ratio of collapse load to first yield load is the shape factor of the member, for statically determinate prismatic structures. 7.2. Simply-Supported Beam Subjected to Uniform Load 7. Plastic Hinge Formation
- 59. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 59 Shape of the Plastic Hinge: We are also interested in the plastic hinge, and the zone of Elasto- plastic bending. As can be seen from the diagram, the plastic material zones extend from the center out to the point where the moment equals the yield moment. Using parabolic proportion from the bending moment diagram at collapse, we see that: 𝑀 𝑃 𝐿 2 2 = 𝑀 𝑝 − 𝑀 𝑦 𝑙 𝑝 2 2 = 𝑀 𝑝 − 𝑀𝑒𝑝 𝑧2 Equating the first two equations gives: 𝑙 𝑝 2 = 𝐿2 𝑀 𝑝 − 𝑀 𝑦 𝑀 𝑃 = 𝐿2 1 − 𝑀 𝑦 𝑀 𝑝 = 𝐿2 1 − 1 𝑓 → For a beam with a rectangular cross section (𝑓 = 1.5) the plastic hinge extends for a length: 𝑙 𝑝 = 𝐿 1 − 1 𝑓 𝑙 𝑝 = 𝐿 1 − 1 1.5 = 𝐿 3 𝑙 𝑥 𝑏 𝑎 𝑎 𝑙2 = 𝑏 𝑥2 7.2. Simply-Supported Beam Subjected to Uniform Load 7. Plastic Hinge Formation
- 60. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 60 Shape of the Plastic Hinge: Lastly, the shape of the hinge follows from the first and third equation: 𝑧 𝐿 = 1 2 1 − 𝜎 𝑦 𝑏𝑑2 6 3 − 𝛼2 2 𝜎 𝑦 𝑏𝑑2 4 = 1 2 1 − 1 3 3 − 𝛼2 𝑧 𝐿 = 𝛼 2 3 , 𝑎𝑡 𝛼 = 1, 𝑧 = 𝑙 𝑝 2 = 𝐿 2 3 , 𝑙 𝑝 = 𝐿 3 From expressions for the elasto-plastic and plastic moments, we have: 𝐹𝑖𝑛𝑎𝑙𝑙𝑦, 𝛼 = 2 3 𝑧 𝐿 𝑀 𝑃 𝐿 2 2 = 𝑀 𝑝 − 𝑀𝑒𝑝 𝑧2 → 𝑧 𝐿 = 1 2 1 − 𝑀𝑒𝑝 𝑀 𝑝 Which is an equation of straight line 7.2. Simply-Supported Beam Subjected to Uniform Load 7. Plastic Hinge Formation
- 61. PLASTICANALYSISANDDESIGNOFSTEELSTRUCTURES Assoc.Prof.AshrafI.M.El-Sabbagh,FacultyofEngineeringPortSaid,Egypt L E CT U R E 1 : B A S I C S O F P L A S T I C D E S I G N 61 • Colin Caprani “Plastic Analysis 3rd Year Structural Engineering”, 2009, Monash University (Australia). • T.H.G. Megson “Structural and Stress Analysis”, 2nd ed., 2005, Elsevier. • “Chapter 35: PLASTIC ANALYSIS”, Version 2, INSTITUTE FOR STEEL DEVELOPMENT & GROWTH, http://www.steel-insdag.org. REFERENCES