Learning Objectives - Describe the behavior of beams under lateral loads - Derive the bending equation and discuss the significance of assumptions made - Draw the bending stress distribution diagram and calculate the maximum normal stresses - Explain the behavior of composite beams and determine the stresses in such beams - Derive the equation for shear stress distribution and calculate the maximum shear stress - Design beam sections for bending moment (BM) and shear force (SF) - Explain the concept of shear flow and shear center, and calculate shear flow at any point and locate the shear center for thin-walled sections - Explain the concept of unsymmetrical bending and determine the stresses in such cases
Introduction In this chapter, we will study how internal stress resultants are developed and how the stresses due to BM and SF are distributed over the section.
Behavior of Beams Under the action of external planar loads, a beam bends. At any section of a beam, three internal stress resultants must maintain the equilibrium: bending moment, shear force, and axial force. (Figure 5.1 placeholder)
Behavior of Beams The beam produces an internal resisting couple to balance the BM due to the applied loads. This couple is generated by the normal compressive and tensile forces in the section. (Figure 5.1 placeholder)
Bending Stresses The beams bend due to the effect of external loads and take a curved shape. In the case of a positive or sagging BM, the beam bends concave upwards. (Figure 5.2 placeholder)
Bending Stresses The compressive stresses acting over the top part (area) of the beam cross-section form a resultant compressive force. Similarly, tensile stresses form a tensile force. These forces create an internal resisting couple. (Figure 5.3 placeholder)
Pure Bending Theory of pure bending: In a beam subject to pure bending, the bending moment is constant along the length of the beam and the shear force is zero. (Figure 5.4 placeholder)
Pure Bending Bernoulli's equation: Sections that are plane before bending remain plane after bending. This assumption leads to a linear variation of strain and stress across the depth of the beam. (Figure 5.5 placeholder)
Bending Equation The bending equation is derived from the equilibrium of forces and moments in the beam section: M/I = σ/y = E/R Where: M = Bending Moment I = Moment of Inertia σ = Stress y = Distance from Neutral Axis E = Modulus of Elasticity R = Radius of Curvature
Stress Distribution The stress distribution in a beam varies linearly from the neutral axis. The maximum compressive and tensile stresses occur at the extreme fibers of the beam section. (Figure 5.6 placeholder)
Stress Distribution The total compressive force is equal to the total tensile force, and they form a couple that is equal to the internal resisting moment of the beam section. (Figure 5.7 placeholder)
Design of Beams for Strength Section modulus and modulus of rupture are used to design beam sections for strength. The section modulus is a geometric property that reflects the distribution of material in the cross-section. (Example problems 5.1 to 5.5 placeholders)
Example Problems Example Problems: 1. Bending of a steel bar 2. Moment capacity of a hollow circular section 3. Load capacity of a rectangular section on an SS span 4. Square section with side horizontal and diagonal horizontal 5. Moment capacity of a trapezoidal section
Composite Beams Composite beams consist of two materials rigidly joined together so that they act as one section. The equivalent section approach is used to analyze composite beams. (Example 5.19 placeholder)
Example Problem Example Problem: Load carrying capacity of a flitched beam. (Figure 5.33 placeholder)
Shear Stresses in Beams Shear stresses arise in beams due to shear forces. The distribution of shear stress in a beam section is more complex than bending stress. (Figure 5.35 placeholder)
Shear Stresses in Beams The horizontal shear stress is calculated indirectly by considering the equilibrium of an isolated part of the beam section. (Figure 5.36 placeholder)
Shear Stress Distribution The shear stress distribution in a beam section can be determined using the equation: τ = VQ/It Where: V = Shear Force Q = Statical Moment of Area I = Moment of Inertia t = Width of the section at the point where shear stress is being calculated (Figure 5.37 placeholder)
Shear Stress Distribution The distribution of shear stress varies across the section, with maximum shear stress occurring at the neutral axis. (Figure 5.38 placeholder)
Example Problems Additional Example Problems: 1. Bending stress variation along the length and in the beam section. 2. Effect of shape of beam section on stress induced. 3. Load carrying capacity of a given section.
Summary Summary: - Beams develop internal stresses to resist bending moments and shear forces. - The bending stress distribution is linear, while the shear stress distribution is more complex. - Composite beams can be analyzed using the equivalent section approach. - Design of beam sections involves calculating section modulus and considering both bending and shear stresses.
Additional Examples and Exercises Practice problems from the chapter to reinforce learning.
Additional Examples and Exercises Practice problems from the chapter to reinforce learning.
Additional Examples and Exercises Practice problems from the chapter to reinforce learning.
Additional Examples and Exercises Practice problems from the chapter to reinforce learning.
Additional Examples and Exercises Practice problems from the chapter to reinforce learning.
Additional Examples and Exercises Practice problems from the chapter to reinforce learning.
Additional Examples and Exercises Practice problems from the chapter to reinforce learning.
Additional Examples and Exercises Practice problems from the chapter to reinforce learning.
References References: - Textbook: Title of the Book - Additional resources and reading materials.
Questions and Discussion Questions and Discussion: - Open floor for any questions and discussion points from the lecture.