By Prof L S Srinath
This ebook is designed to supply a very good beginning in Mechanics of Deformable Solids after an introductory direction on energy of Materials. This version has been revised and enlarged to make it a entire resource at the topic. Exhaustive remedy of crucial subject matters like theories of failure, strength tools, thermal stresses, rigidity focus, touch stresses, fracture mechanics make this an entire providing at the topic.
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Extra info for Advanced mechanics of solids
From Eq. 9) Analysis of Stress 39 sy n txy t n sx s sx q q sx txy txy sy sy (a) (b) Fig. 59) n 2 2 2 t 2 = Tx + Ty − σ t= σx −σy sin 2q + tx y cos 2q 2 The principal stresses are given by Eq. 62) The above equation gives two planes at right angles to each other. 63a) 40 Advanced Mechanics of Solids In the xy plane, the maximum shear stress will be τ max = 1 (σ 1 − σ 2 ) 2 and from Eq. 64) DIFFERENTIAL EQUATIONS OF EQUILIBRIUM So far, attention has been focussed on the state of stress at a point.
Recall that we have arranged the principal stresses such that s1 ≥ s2 ≥ s3. There are three cases one can consider. Case (i) s1 > s2 > s3 Case (ii) s1 = s2 > s3 Case (iii) s1 = s2 = s3 We shall consider these cases individually. Case (i) s1 > s2 > s3 For this case, the denominator in Eq. 40) is positive and hence, the numerator must also be positive. In Eq. 41), the denominator being negative, the numerator must also be negative. Similarly, the numerator in Eq. 42) must be positive. Therefore. (s – s2)(s – s3) + t 2 ≥ 0 (s – s3)(s – s1) + t 2 £ 0 (s – s1)(s – s2) + t 2 ≥ 0 The above three inequalities can be rewritten as ⎛ τ 2 + ⎜σ − ⎝ σ 2 + σ3 ⎞ 2 ⎛ σ 2 − σ3 ⎞ ⎟⎠ ≥ ⎜⎝ 2 ⎟⎠ 2 2 28 Advanced Mechanics of Solids σ + σ1 ⎞ ⎛ ⎛ σ − σ1 ⎞ τ + ⎜σ − 3 ≤⎜ 3 2 ⎟⎠ 2 ⎟⎠ ⎝ ⎝ 2 2 2 ⎛ ⎝ τ 2 + ⎜σ − σ1 + σ 2 ⎞ 2 ⎛ σ1 − σ 2 ⎞ ⎟⎠ ≥ ⎜⎝ 2 ⎟⎠ 2 2 According to the first of the above equations, the point (s, t ) must lie on or 1 1 outside a circle of radius (s2 – s3) with its centre at (s2 + s3) along the s axis 2 2 (Fig.
10, it was stated that Eq. 20) has three real roots. The proof is as follows. Dividing Eq. 20) by s 2, l l σ − l1 + 2 − 32 = 0 σ σ For appropriate values of s, the quantity on the left-hand side will be equal to zero. 26) − 3 = f (σ ) σ σ2 Since l1, l2 and l3 are finite, f (s ) can be made positive for large positive values of s. Similarly, f (s ) can be made negative for large negative values of s. Hence, if one σ − l1 + 18 Advanced Mechanics of Solids f (s) plots f (s ) for different values of s as shown in Fig.
Advanced mechanics of solids by Prof L S Srinath