Question

The horizontal force of P = 90 kN acts at the end of the plate. The plate has a thickness of 10 mm and Pacts along the centerline of this thickness such that d = 50 mm. The position of P is shown with a red spot on the centerline in a side view. a) Determine normal stress at point A (10 points). b) Determine normal stress at point B (10 points). c) Plot the distribution of normal stress acting along section a-a and FBD using accurate scale (10 points). a A A 200 mm d. B B -300 mm 10 mm

Answer 1

cawy direct normal stres The given force P = 90 kn (sw) and bending normal stres. 6) Area = 10 X 200 - 2000MM on = Р A 90,000 = 45 mpa 2000 (Tensile) Bending ofre Distribution: ez eccentricity c 100-d d=50mm = e = 50mm A moment, m= Pxe Me 90,000x50 me 4500,000 Nmm. 3 Moment bertia 8 Is bh3 O x 200 6.666, 666 67mmt 12 orlog Eln y = - 100mm Gb = my 67.49 mpa S 4500000 x 100 6666 666.67 I ob is tensile at B and compressive at A.

(a) Bending stress is compression at A. so, Net normal stress - ON-OG (oNet) 45-67-49 = -22.499 mpi (orvet), - 22.499 MPa (compressive) (1) Conte 45+ 67.49 = 112-49 mpa C Tensile). Stress Distribution Diagram y Scale lmm in Diagram = 2 mm in Body. e А e > 22.49. 112.49. 33-34 FBD MR P R B