LECTURE+24+COE-3001-A

These stresses on a sketch of a stress

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Unformatted text preview: that ma=er) X MC = 0 1 M Ay (L x) + [q · (L x)] · [ (L x)] = 0 Lx VM 2 ⇥ kN ⇤ M = 14kN · (0.8m) 28 · (0.8m) · 0.4m m C M = 2.24 kN · m Ay AlternaYvely, we may invoke: Z Z dM M M (x) = V (x)dx = ( 28x + 14)dx V ( x) = dx M (x) = 14x2 + 14x + C2 M (0) = 0 M ( x) = 14x2 + 14x x = 0.5m 14kN x = 0.8m C Ay x 8.4kN x = 0.5m x = 0.8m 3.5kN · m 2.24kN · m M (0.8) = 2.24kN · m 3/6/13 V M. Mello/Georgia Tech Aerospace x C Ay 20 Obtain moment of inerYa and 1st moment of area above point C 4.  DETERMINE THE MOMENT OF INERTIA OF THE RECTANGULAR CROSS SECTION bh3 I= 12 (25mm)(100mm3 I= = 2083 ⇥ 103 mm4 12 Ac yc 75mm 5.  Determine 1st moment Qc of the cross secYonal area Ac above C Q c = Ac y c Qc = (25mm · 25mm)(37.5mm) = 23440 mm3 3/6/13 M. Mello/Georgia Tech Aerospace 21 Calculate normal stress and shear stress at point C 6.  Determine normal stress at point C: x = x = x = My I (2.24kN · m)(25mm) 2083 ⇥ 103 mm4 26.9 M P a 7.  And finally the shear stress at point C: ⌧= VQ Ib Recall, V = 8.4 kN I = 2083 ⇥ 103 mm4 Q = 23440 mm3 b = 25 mm 3/6/13 ⌧= VQ Ib ⌧ = 3.78 M P a M. Mello/Georgia Tech Aerospace 22 AlternaYve form for soluYon of shear stress at point C 7.  And finally the shear stress at point C (conYnued): ✓ VQ V h2 Try alternaYve formulas, ⌧ = ⌧= Ib V = 8.4 kN I = 2083 ⇥ 103 mm4 h = 100mm y = 25mm ✓ ◆ V h2 ⌧= y2 2I 4 VQ ⌧= Ib 3/6/13 2I 4 y2 ✓ (8.4kN )(23440mm3 ) (100mm)2 ⌧= 2(2083 ⇥ 103 mm3 ) 4 ✓ 3V ⌧= 1 2A y2 c2 ◆ ◆ 25mm2 ◆ ⌧ = 3.78 M P a Could also apply this formula and obtain same result M. Mello/Georgia Tech Aerospace 23 Analysis of beam under four point bend loading X Fv = 0 Ay + By = 2P X MB = 0 Ay L P (L a) Ay L P [(L 2a) (L a)] = 0 PL = 0 ) By = P Ay = P P P a x V x M x<a Ay X x>L X V V = Ay 3/6/13 (V>0, guessed correct sign of shear stress resultant) M a C Ay C Fv = 0 V = P ( x < a) V Fv = 0 2 P + Ay = 0 (V>0, guessed correct sign of shear stress resultant) V =P (x > L M. Mello/Georgia Tech Aerospace a) 24 4 point bend shear force diagram V P 4 point bend shear force diagram P x a x a<x<L V a Ay X V P Fv = 0 P + Ay = 0 V = 0 (a < x < L 3/6/13 C M a) M. Mello/Georgia Tech Aerospace 25 4 point bend configuraYon: Bending moment diagram For a < x < (L For x < a ! V = P V ( x) = P = M ( x) = P x dM dx V ( x) = 0 = a) ! V = 0 dM dx V ( x) = M ( x) = M ( x) = C1 (0 < x < a) Px + C M (L M ( x) = P a a < x < (L a) P (L P (x > L a) + C C = PL M ( x) = Px + PL (x > L M Pa x a L a L 4 point bend...
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