1. ENGR 220 Section 6.3 – 6.4

2. Assumptions
1.Straight prismatic beams
2. Homogenous material
3. Cross sectional area symmetric w.r.t an Axis
4. Bending moment applied perpendicular to this axis of symmetry
5. Neutral surface does not experience any change in length
6. All cross sections remain plane and perpendicular to the longitudinal axis.
7. Any deformation of the cross section in its own plane is neglected.

3. Bending Deformation

4. Bending Deformation

8. Maximum Strain

9. Strain / Stress Profiles

10. Wood Specimen that failed in bending

11. The Flexure Formula Applying Hooke’s Law
A linear variation in strain results in a linear variation in stress.

12. Locating Neutral Axis. Resultant Normal Force on the Cross section = 0
dF =   dA = 0
Substitute for , in terms of max
-(max /c )  y dA = 0
 y dA = 0
 y dA = y = location of centroid.
y = 0 implies the centroidal axes are lying on the neutral axis.

14. Flexure Formula

15. Moment of Inertia Review
Note: Moment of inertia is actually a misnomer. It has been adopted because of its similarity to integrals of the same form related to mass.

16. Parallel Axis Theorem

17. Composite Shapes

18. Determine the area moment of inertia for the rectangle shown about the axis x’ and about the axis xb.

19. Determine the internal moment M at the section caused by the stress distribution (1) using the flexure formula and (2) find the resultant of the stress distribution using basic principles

20. Determine the Moment of Inertia about the z-axis for the following I-Beam.

21. Determine the maximum bending stress at a-a

22. The strut on the utility pole supports the cable having a weight of 600 lb. Determine the absolute maximum bending stress in the strut if A, B, and C are assumed pinned.

23. The simply supported truss is subjected to the central distributed load. Neglect the effect of the diagonal lacing and determine the absolute maximum bending stress in the truss. Top member: 1” OD and ” wall thickness. Bottom member: 0.5” OD Solid Rod.