By Prof L S Srinath

This e-book is designed to supply an exceptional beginning in Mechanics of Deformable Solids after an introductory path on power of Materials. This variation has been revised and enlarged to make it a accomplished resource at the topic. Exhaustive remedy of crucial subject matters like theories of failure, power equipment, thermal stresses, rigidity focus, touch stresses, fracture mechanics make this an entire providing at the topic.

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**Example text**

The average 2 stresses on these faces (which are assumed to be acting at the mid point of eace face) are On face aa¢d¢d normal stress sr tangential stresses trz and trq On face bb¢c¢c z ( normal stress sr + ∂σ r ∆r ∂r ) Analysis of Stress 47 z r a d Dq d¢ Dr b c y a¢ b¢ c¢ y σ θ′ r τr′ θ r x σ r′ /2 (a) (b) Fig. 32 (a) Geometry of cylindrical element (b) Variation of stresses across faces tangential stresses τr z + ∂τ rz ∆r ∂r and τ rθ + ∂τθr ∆r ∂r The changes are because the face bb¢c¢c is Dr distance away from the face aa¢d¢d.

24. sy sy txy sx txy sx sx sx txy txy sy (a) sy (b) Fig. 24 (a) Plane state of stress (b) Conventional representation Consider a plane with the normal lying in the x y plane. If nx, ny and nz are the direction cosines of the normal, we have nx = cos q, ny = sin q and nz = 0 (Fig. 25). 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.

5 With respect to the frame of reference Oxyz, the following state of stress exists. Determine the principal stresses and their associated directions. Also, check on the invariances of l1, l2, l3. , (s + 1) (s 2 – s + 1) – 3s (s + 1) = 0 or (s + 1) (s 2 – 4s + 1) = 0 Hence, one solution is s = –1. The other two solutions are obtained from the solution of the quadratic equation, which are s = 2 ± 3 . 21) (1 + 1)nx + 2ny + nz = 0 2nx + (1 + 1)ny + nz = 0 nx + ny + (1 + 1)nz = 0 together with 3 ) = –3 nx2 + n2y + nz2 = 1 From the second and third equations above, nz = 0.