M m z p2b 18kn 60mm 21 1080nm v 18 kn p2 18kn xy m

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Unformatted text preview: caused by bending moment (M I = σ y) • Ch 6: Shearing stress due in beams (τ = VQ It) • Ch 7 • Ch 8 : Transformation of stress, principal planes, principal normal stresses (σ max,σ min ) and max shear stress (τ max ) : Stress in structural member or machine element due to combination of loads & corresponding σ max,σ min & τ max 20 10 Example 8.01 P = 15kN 1 Member BD of rad 20mm. At pt K, find a) σ & τ b) θ p , σ max , σ min c) τ max • Internal forces at sect k F = P1 = 15kN M y = 750 N .m V = P2 = 18kN T = P2(a) = 18kN( 50mm) = 900N.m M y = P1(a) = 15kN( 50mm) F = 15kN = 750N.m M z = P2(b) = 18kN( 60mm) 21 = 1080N.m V = 18 kN P2 = 18kN τxy M y = 750N.m • Normal stress No F = 15 kN τ xy V = 18 kN (a) σ σ from M z F M y(z = 0.02 ) + A IY = -11 .9 + 119. 3 = 107.4 MPa σx = − σ & τ at K A = π( 0.02 )2 = 1.257 ×10−3 m 2 1 I y = I z = π (0.02) 4 4 = 125.7 ×10−9 m 4 K σ x = 107 .4 MPa 22 11 • Shear stress M y = 750N.m (τ F = 15 kN τ xy (τ xy )V = τ VQ IZt ⎛1 ⎞⎛ 4 c ⎞ Q = A y = ⎜ π c 2 ⎟⎜ ⎟ ⎝2 ⎠ ⎝ 3π ⎠ = 5 . 33 × 10 xy )V = V = 18 kN −6 m 3 (18 × 10 ) Q 3 I Z (t...
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