Question

If the beam is subjected to a positive bending moment of M = 100 kN-m, determine the maximum and minimum bending stress. Also determine the shear stress at point, A which is 50 mm above from the bottom. The cross-section of the beam is I-shaped and shown in the figure. 300 mm 30 mm 300 mm . 50 mm 30 mm

Answer 1

300 mm Q Given! 30mm 300 mm A somm 30 mm M = 100 KN.m = 100X10 N.mm To find: (al Maximum and Minimum (b) shear stress at point A bending stress soln: Divide the T-section into two composite figure ♡ Rectangle (b, = 30mm, d, = 300mm) 2 Rectangle (bg=300mm, da = 30mm) Now, Sl. No. E AY Area (mm2) Distance of centroidal) axis from bottom fibre (mm) ( mm ) y, =150 135 0000 0 Rectangle 30x300 E9o0o 30x300 - yooo Y 2 = 300+30=315 2835000 2 Rectangle

y = {AY 1350000 + 2835000 gooot 9000 EA & 185000 1800u = 232.5 = Distance of N.A from bottom fibre = 232.5 mm Now, using Parallel axis theorem b, d, + A, (5-4,)? Ix= 3 12 308 (300)+ 9000 (23215-15012 -(67,500,000 +61256,250 = 128756250 mnt 2 It - b₂ d₂ 13 + A2 (ý -Y) 12 = 300x (301? + gouo ( 232.5– 315 )? =1675000 +61256257 ) 12 = 6193 1250 mmt Ix = Ix, t Ix2 = (128 756250 + 6 1931250) FX = 190687500 mm

using bending equation y E r (a) Maximum bending stress (tensile I will bottom fibre occur at М. x max I C 100x106 190687500 232.5 121.9272 Ni mm² (obending) 121.9272 N/mm² tensile ("bending) compressible -X (330-232-5) I = 100X10b 190687500 X 97.5 51.13 Nimm? at N.A Minimum bending stress will oceur. and - 0 min (82) min comb -51.13 N/mm2 minto N.A (oblmax = 121.92 N/mm2

(b) shear stress at A The resultant sensile force below neutral axis (NA) F = 1 2 t $121. 1.9272 x 232-5X 30? 아 = 425221:11 N Now, shear stress at a brint A - t = FA Ib N.A A Where | 232.5 25mm mm 25mm NA SUMM Ау 30mm Moment of area at point A about N A 50x 30 x (232-5-251 2 311250 mm3 b = 30mm - 23.135 Ni mm2 T = 425221.11 X 311250 190687500x30 ... shear stress at point A = 23.135 Nimme