2 - Unsymmetric Bending CE14 Lecture Slides

1. Introduction to Geometry of Areas

In civil engineering, geometry of areas is a fundamental topic that bridges mathematics and engineering applications. Whether we are analyzing the strength of a beam, determining the center of gravity of a structural section, computing hydraulic properties of a river channel, or estimating the cross-sectional area of an embankment, engineers rely on accurate determination of areas and their geometric properties.

The study of geometry of areas focuses not only on measuring areas of plane figures but also on calculating related geometric characteristics such as centroids and moments of inertia. These properties serve as the foundation of structural analysis, material mechanics, hydraulics, geotechnical engineering, and surveying.

For instance:

In structural engineering, the bending stress in a beam depends on the second moment of area of its cross-section.

In hydraulics, the flow capacity of a channel depends on its cross-sectional area and wetted perimeter.

In surveying and geodesy, land areas are calculated using geometric methods.

Thus, Geometry of Areas is both a theoretical and practical tool, connecting mathematics to the real-world problems encountered by civil engineers.

1. LECTURE 1: UNSYMMETRIC BENDING Prepared By: Inst. Liezl Tan For CE 14 Lecture @ ICE 303 Jan. 20, 2016 1

2. Discussion Outline Review of Flexural Stress for SYMMETRIC BENDING Significance / Practical Application of the concept of UNSYMMETRIC BENDING Derivation of flexural stress for UNSYMMETRIC BENDING How to determine the LOCATION and the corresponding MAGNITUDE of the MAXIMUM STRESS ? Summary 2

3. SYMMETRIC BENDING 3

4. RECALL: Symmetric Bending FLEXURE FORMULA ASSUMPTIONS : Beam is initially straight and prismatic. Plane section before bending remains plane after bending . Material is Linear-Elastic and homogeneous. Beam does not twist or buckle. Negligible transverse shear stress. Resisting moment is applied in the plane of symmetry 4

5. Why study UNSYMMETRIC BENDING? Inclined members Lateral and vertical loads Sections with no symmetry 5

6. DEFINITION: Unsymmetric Bending Beam sections with a plane of symmetry but with the couple (resisting moment) applied NOT IN or PARALLEL to the section’s plane of symmetry. Beam sections with NO plane of symmetry . 6

7. Flexural Stress: Unsymmetric Bending Consider an arbitrary cross-section of a PRISMATIC and INITIALLY STRAIGHT beam. (ASSUMPTION #1) 7

8. Flexural Stress: Unsymmetric Bending A couple M is applied at the left end of the beam. M 8

9. Flexural Stress: Unsymmetric Bending For the beam segment to be in equilibrium, there must be an equal but opposite couple at the right end (section exposed by traversing a plane) of the beam denoted by Mr (resisting couple) M Mr 9

10. Flexural Stress: Unsymmetric Bending The resisting couple Mr , is the “RESULTANT” moment produced by the normal stress acting on the cross section. M M z y z y 10 Mr σ

11. Flexural Stress: Unsymmetric Bending For the beam segment to be in equilibrium, y z Mr S α Mrz Mry σ y z 11

12. Flexural Stress: Unsymmetric Bending In order to evaluate the equilibrium equations, we have to determine HOW THE NORMAL STRESS IS DISTRIBUTED OVER THE AREA dA ? ASSUMPTION #2: Plane section remains plane. [implies ε is linearly varying] ASSUMPTION #3: Linear-elastic Material. [implies that σ is linearly varying with ε ] BUT!!! since the beam is UNSYMMETRIC, the normal strain, ε , although linearly varying, is NO LONGER dependent on one variable / coordinate. LINEAR ; MULTI-VARIATE 12

13. Flexural Stress: Unsymmetric Bending General form of a LINEAR, MULTI-VARIATE Equation: Applying to normal stress ( σ ): 13

14. Flexural Stress: Unsymmetric Bending Plugging in the above equation into the equilibrium equations: 14

15. Flexural Stress: Unsymmetric Bending To further simplify the evaluation of the integrals, we can set the origin of the coordinate axes (y, z) at the CENTROID of the cross section (ASSUMPTION #4) 15 zero zero zero zero

16. Flexural Stress: Unsymmetric Bending Let: ; ;   16 zero zero zero zero Iyz Iy Iz Iyz

17. Flexural Stress: Unsymmetric Bending The problem now reduces to solving 3 equations with 3 unknowns( k1, k2, k3) provided that Mry & Mrz are known. 17

18. Flexural Stress: Unsymmetric Bending Solving for the coefficients (k1, k2): Substituting the coefficients into the GENERAL EQUATION of stress (Linear, Multi-variate Equation): 18

19. Flexural Stress: Unsymmetric Bending FORM 1: FORM 2: 19

20. EXAMPLE ATTENDANCE CHECK! 20

21. Example 1.1. Determine the stress at point A: y z 1” 2 ” 2 ” 3” 2 ” 300 kip -in A B 21 2 ” 2 ”    

22. MAXIMUM STRESSES [Location & Magnitude] 22

23. Maximum Stresses: Unsymmetric Bending Portion of the cross-section experiences TENSILE stress and a portion experiences COMPRESSIVE stress. Hence, at some point on the cross section we will encounter a ZERO STRESS .  NEUTRAL AXIS As previously discussed, stress is linearly varying. Meaning, the further the element is from the neutral axis, the higher the magnitude. Hence it is just logical to select the point farthest from the neutral axis as the point of maximum stress. In determining the maximum stress, one has to determine the ORIENTATION OF THE NEUTRAL AXIS first. For unsymmetric bending, the NEUTRAL AXIS rotates about the centroid by an angle equal to β . 23

24. Maximum Stresses: Unsymmetric Bending Using FORM 1 of the flexural stress formula for unsymmetric bending, we let σ = 0. 24  

25. Maximum Stresses: Unsymmetric Bending ORIENTATION OF THE NEUTRAL AXIS: Once the orientation of the neutral axis is determined, the point farthest from the neutral axis (longest perpendicular distance) will experience the maximum stress. So, DETERMINE THE COORDINATES ( y,z ) OF THE POINT FARTHEST FROM THE NEUTRAL AXIS. 25  

26. Maximum Stresses: Unsymmetric Bending Then use either FORM 1 or FORM 2 in solving for the MAGNITUDE of the maximum stress. FORM 1: FORM 2: 26

27. EXAMPLE 27

28. Example 1.1. Determine the orientation of the NEUTRAL AXIS: y z 1” 2 ” 2 ” 3” 2 ” 300 kip -in A B 28 2 ” 2 ”    

29. Example 1.1. Determine the MAGNITUDE OF THE MAXIMUM STRESS: y z 1” 2 ” 2 ” 3” 2 ” 300 kip -in A B 29 2 ” 2 ”       C D

30. SUMMARY 30

31. SUMMARY Flexure Formula for UNSYMMETRIC BENDING How to determine the ORIENTATION OF THE NEUTRAL AXIS ? 31  

32. SUMMARY What are the ASSUMPTIONS set forth in the derivation of the stress formula for the unsymmetric bending? How to determine the LOCATION and the corresponding MAGNITUDE of the MAXIMUM STRESS 32

33. END of LECTURE 1 33

34. SEATWORK BY PAIR 34