By D. W. A. Rees (auth.)

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**Extra resources for Basic Solid Mechanics**

**Example text**

P = 120 mm. X Note that through the use of different safety factors in tension and shear the mode of failure can be controlled. The joint efficiency is defined as: 41 Basic Stress and Strain · ffi . 10) The numerator in Eq. 046 MN X (ii) Resistance of the rivets to shearing per pitch: = No. 046 MN. 043 MN. It appears from (iii) that the rivets will crush first. From Eq. 8 %. In practice, however, as the shear mode factor associated with double shear is nearer 1. 75 than 2, then either the plate tearing strength or the rivet shear strength would dictate joint efficiency for this design.

5 X 500) = 10RB, :. RB = 395 N. RA + RB = 900, :. RA = 505 N. F-Diagram. e. maximum M) is found from: F = 0 = -505 + 400 + 100 (z - 3), :. 05 m. e. w = OF I&). 23 Static Equilibrium M-diagram. The ordinates are found by taking moments at points A, B, C and D working either to the left or right as shown. 2 Contrajlexure In the previous examples the bending moment diagrams were all of the same sign. With other forms of loading, the bending moment diagrams may contain both positive and negative portions.

H. H. end). 1. 18 Find the magnitude and position of Mnl4X and the points of contraflexure for the beam in Fig. 38. H. H. H. 19 Sketch, for the beam in Fig. 39, the F and M diagrams. Determine the position and magnitude of M111n and the point of contraflexure. H. H. 20 Find w in tonne/m for the simply supported beam in Fig. H. end. Draw the F and M diagrams and find the points of contraflexure. (Ans 1113 tim, 0. 21 Calculate, for the beam in Fig. 41, the position and magnitude of the maximum hogging and sagging bending moments.