Strength of Materials Problem 1: Derivation of Shear stress in rectangular crosssection Problem 2: Computation of Shear stresses Problem 3: Computation of Shear stresses Problem 4: Computation of Shear stresses Indian Institute of Technology Madras Prof. M. S. Sivakumar Prof. M. S. Sivakumar Strength of Materials Problem 1: Derivation of Shear stress in rectangular crosssection Derive an expression for the shear stress distribution in a beam of solid rectangular crosssection transmitting a vertical shear V. The cross sectional area of the beam is shown in the figure. A longitudinal cut through the beam at a distance y1, from the neutral axis, isolates area klmn. (A1). Shear stress, VQ It V y.dA = It ∫ τ= A1 = = V Ib d/2 ∫ by dy y1 V⎡ ( d / 2 )2 − ( y1 )2 ⎤⎥⎦ ---------------------- (1) ⎢ ⎣ 2I The Shear Stress distribution is as shown below Max Shear Stress occurs at the neutral axis and this can be found by putting y = 0 in the equation 1. Indian Institute of Technology Madras Prof. M. S. Sivakumar Strength of Materials Vd 2 8I 3 V = 2 bh 3V = 2A τmax = Top Indian Institute of Technology Madras Prof. M. S. Sivakumar Strength of Materials Problem 2: Computation of Shear stresses A vertical shear force of 1KN acts on the cross section shown below. Find the shear at the interface (per unit length) Solution: Formula used: q = VQ/I We first find the distance of the neutral axis from the top fiber. All dimensions in mm y NA = 20 × 100 × 10 + 20 × 100 × 70 = 40mm 20 × 100 + 20 × 100 Q = ∫ ydA of shaded area about neutral axis. Q = 20 x 100 x 30 = 6 x 104 Indian Institute of Technology Madras Prof. M. S. Sivakumar Strength of Materials V = 1KN 3 20 × 1003 2 100 × 20 I= + 20 × 100 × 30 + + 100 × 20 × 302 12 12 = 5.33 × 106 q= VQ 103 × 6 × 104 × φ(10−3 )3 N = = 1.125 × 104 4 I m 5.33 × 106 × 10−3 ( ) = 11.25 KN m Top Indian Institute of Technology Madras Prof. M. S. Sivakumar Strength of Materials Problem 3: Computation of Shear stresses A 6m long beam with a 50 mm × 50 mm cross section is subjected to uniform loading of 5KN/m. Find the max shear stress in the beam Solution: τmax = 3V 2A We first find the section of maximum shear force. We know this is at the supports and is equal to 5× 6 = 15KN 2 We also know that max.shear stress occurs at the centre (for a rectangular cross section) and is 1.5 times the average stress. So, τmax = 3 × 15 × 103 2 × 50 × 50 × 10−6 = 9 Mpa Top Indian Institute of Technology Madras Prof. M. S. Sivakumar Strength of Materials Problem 4: Computation of Shear stresses The cross section of an I beam is shown below. Find the max.shear stress in the flange if it transmits a vertical shear of 2KN. Solution: τ= Formula used: VQ It V = 2KN ⎞ 10 × 1003 ⎛ 100 × 103 I= +⎜ + 100 × 10 × 552 ⎟ × 2 = 6.9 × 106 mm 4 ⎜ 12 ⎟ 12 ⎝ ⎠ Q is maximum at the midpoint as shown below Q = 50 × 10 × 55 τmax = 2 × 103 × 50 × 10 × 55 × 10−3 ( ) ( 6.9 × 10 )(10 ) −3 6 −3 4 Indian Institute of Technology Madras × 10 × 10 3 = 0.79 MPa Strength of Materials Prof. M. S. Sivakumar Top Indian Institute of Technology Madras