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File Lecture5.pdf - Lecture 51 Outline-Shear stress and...

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Lecture 5 1 Lecture 5 1 Outline: -Shear stress and shear flow -Built-up beams -Shear stress distribution on the cross-section 1. Shear stress Imagine that boards are stacked on top of each other as shown in Figure 1. In the first case there is no bond between the boards so that during deformation they can slip over each other. In the second case the boards are bonded together (by means of glue or nails) so that they doe not slip over each other and act together as one unit. (a) No bond between the boards (b) Boards are bonded together Fig. 1 Beams stacked on top of each other What is preventing slip in between the boards is the shear stress carried by the bond (glue or nails) as shown in Figure 2. (a) Slip between the unbonded boards (b) Shear stress between the bonded boards Fig. 2 Slip and shear stress between the boards As illustrated for the simply supported beam in Fig 1.a, when the boards slip over each other, at the interface of the two boards, the ends of the board on the top gets a head of the ends of the board on the bottom as shown in Fig. 2(b). When the boards are bonded in order to prevent slip, the shear stress acting at the bottom surface of the top board is pulling the ends of the beam back, on the other hand the stresses 1 Mechanics of Solids Dr Emre Erkmen

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Lecture 5 2 acting at the top surface of the bottom beam is pushing the ends of the beam forward as shown in Fig.2 so that they stay together as in Fig1. b. 2.1. How to calculate the shear stresses Consider a small segment x Δ isolated from the loaded beam as shown in the Fig.3. Fig. 3 Loaded beam Vertical loads will cause internal shear force and bending moment. In Fig.4 notice that the internal forces and the applied load are shown according to the positive sign convention on both sides of the beam segment. If at x the internal shear force is V and bending moment is M, then at x x + Δ , they are V V + Δ and M M + Δ because in general, for different coordinates along the axis of the beam there will be different shear force and bending moments. Because x Δ is small, M Δ and V Δ are also small. Fig. 4 Isolated segment of x Δ We can write a moment equilibrium at x (counter-clockwise positive), i.e. ( ) ( ) 0 2 x M M M V V x q x Δ + Δ + Δ Δ − Δ = Because x Δ is small, the terms 2 x Δ and x V Δ Δ are very small so that they are negligible. As a result from the equation above we can obtain a relation between the increment in the moment and the shear force as M V x Δ = Δ In the limit case 0 x Δ → this relation becomes d d M V x = The above equation states that if there is no internal shear force (due to applied loading) there is no change in the bending moment.
Lecture 5 3 Longitudinal shear stress The above equation states that if there is internal shear force (due to applied loading) there is change in the bending moment. By using the bending formula that we derived in lecture 4, i.e.

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This note was uploaded on 08/22/2011 for the course ENG 48331 taught by Professor Brown during the Three '11 term at University of Technology, Sydney.

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File Lecture5.pdf - Lecture 51 Outline-Shear stress and...

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