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# File Lecture4 - Lecture 4 1 Lecture 4 1 Outline-Shear and...

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Unformatted text preview: Lecture 4 1 Lecture 4 1 Outline: -Shear and bending moment diagrams of beams -Bending stress-Bending properties of sections 1. Shear and bending moment diagrams in beams Members that are slender and subjected to loadings that are applied perpendicular to their longitudinal axis are called beams. Often, beams are classified according to how they are supported. For example cantilever is fixed at one end and free at the other hand, simply supported beam is pinned at one end and roller supported at the other end and overhanging beam has one or both ends freely extended over the supports as shown in the Figure below. Note that beams shown in the Figure are all statically determinate beams and thus the equilibrium equations written on the free body diagrams are sufficient to determine the internal forces or moments. Fig.1 Type of statically determinate beams Transverse vertical loading produces internal forces equivalent to a shear force and a moment (force couple). Shear force is obtained by summing forces perpendicular to the beam’s axis up to the end segment. Moment is obtained by summing moments about the end of the segment. Fig.2 Positive sign convention for the applied load, internal shear force and bending moment 1 Mechanics of Solids Dr Emre Erkmen Lecture 4 2 Example Show the internal forces and draw the shear and bending moment diagrams for the cantilever below. P x M V Note that directions of the internal shear and the bending moment are according to the positive sign convention From Equilibrium y F = ∑ V P = − @ B x M = = ∑ M P x = − × Example Show the internal forces and draw the shear and bending moment diagrams for the cantilever below. 1 x ≤ < Y F = ∑ 100 V = − 100 M x = − × @ X M = ∑ 1 2 x ≤ < Y F = ∑ 50 V = − ( ) 100 50 1 M x x =− × + × − @ X M = ∑ 2 3 x ≤ < Y F = ∑ 90 V = − @ X M = ∑ ( ) ( ) 100 50 1 40 2 M x x x =− × + × − − × − Shear force V diagram Moment M diagram x M V Lecture 4 3 In the above two examples, support reactions were not required because the free-bodies for the equilibrium equations were written according to the free-end side of the cantilever. In general, in order to determine the internal forces, support reactions should be determined. Number of unknowns for roller pinned and fixed supports are shown below. Fig. 3 Type of boundary conditions for beams Example Show the internal forces and draw the shear and bending moment diagrams for the simply-supported beam below. Find the support reactions x F + → = ∑ AX R = y F + ↑ = ∑ 100 0 Ay By R R + − = 66.67 Ay R kN = @ A M + ↵ = ∑ 100 1 3 0 By R × − × = 33.33 By R kN = Lecture 4 4 1 x ≤ < Y F = ∑ 66.67 V kN = 66.67 M x = × @ X M = ∑ 1 3 x ≤ < Y F = ∑ 33.33 V kN = − ( ) 66.67 100 1 M x x = × − × − @ X M = ∑ Shear force V diagram Moment M diagram Example Show the internal forces and draw the shear and bending moment diagrams for the simply-supported beam below....
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File Lecture4 - Lecture 4 1 Lecture 4 1 Outline-Shear and...

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