MOS Set_4 - 4300:202 Mechanics of Solids Set # 4 Dr. A. F....

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Unformatted text preview: 4300:202 Mechanics of Solids Set # 4 Dr. A. F. Saleeb Shear and Bending Diagrams in Beams General Relations - Consider the case shown, and separate “FBD” for differential element dx q = q(x) x dx x “FBD” q V M M + dM dx V + dV - With transverse loads only, we have forces and moments on “FBD”, for which the two equilibrium equations give dV = −q dx ……….(1) dM =V dx ………..(2) where q = positive if acting downward V, M = positive as shown (on left/right sides) 1 4300:202 Mechanics of Solids Set # 4 Dr. A. F. Saleeb - The above Eqs. (1) and (2) are valid only away from concentrated forces and moments, respectively, where abrupt changes in the V- and Mdiagrams will occur under such concentrated forces/moments. - Geometrically, we may give the following interpretations for Eqns. (1) and (2): − Slope of the V-diagram at any point is equal the negative of the value of the transverse load intensity at the same point. − Slope of M-diagram at any point is equal to the shear force, V, at the same point. − Also, considering two sections at points “a” and “b”, with xb > xa , we have xb Vb -Va =- ∫ q dx = - (area of q-diagram between a, b) xa xb M b - M a = ∫ V dx = area of V-diagram between a, b xa − At points where V=0, M will attain a stationary value (maximum/minimum). However, M may also be max. or min. locally at locations of abrupt changes in V, M (i.e., at concentrated forces or moments) − The max./min. values of V often occur under concentrated forces. 2 4300:202 Mechanics of Solids Set # 4 Examples Example - 1: 3 Dr. A. F. Saleeb 4300:202 Mechanics of Solids Set # 4 Dr. A. F. Saleeb Important Remarks - As an alternative to the above “formal” method for plotting V , M from their expressions in terms of x, we can more easily make use of “hints” from Eqs. (1), (2); i.e., • For q = constant = 20 in part AC, we know from Eq.(1) that V- diagram in this portion is linear, thus defined uniquely by two points Va = 70 and Vc = -10. • For q = 0 in both CDE and EB, V=const., and we need only one constant shear-force value in each; i.e., Vc=-10(in CDE) and Vb=-90 (in EB) • Also, for q = 0 in CDEB, we know that M has linear segments; each defined by two points; e.g., Mc = 120, and Md (left) = 100 for CD Md (right) = 200, and Me = 180 for DE Me = 180, and Mb = 0 for EB • Since V and M are linear functions of Loads (e.g., q, P, Mo) we can always use superposition-principle to easily obtain loaded by distributed loads q; e.g., consider part AC 4 M in parts 4300:202 Mechanics of Solids Set # 4 Dr. A. F. Saleeb 40 Ma = 0 (here) q =20 A MC = 120KN.m 120KNm M MC C 100 L/2 = 2m L =4m (i) Mc=120 Ma = 0 M(i) (+) 120KNm (End Moments) (“standard”) Symm., 2nd degree parabola q =20 M(ii) (ii) x 1 qL 2 L/2 = 2 (Segment loads) m 1 x2 M(ii) = q L x – q 2 2 nd = symmetric, 2 degree parabola 12 qL = 40KN.m 8 Note : “Standard” sub-diagrams M(i) and M(ii) are far more convenient, to obtain their areas or centroidal points, compared to expressions giving M the final above, and they will be extensively utilized later in chapters 9 and 10. - To find the maximum moment, Mm, occurring within part AC, we proceed as follows: For max.M, dM ≡ V = 0 ; i.e., dx ∴ Vm = V (at x = xm) = 70 – 20 xm = 0 xm = 3.5m OK since 0 < xm < 4 and ⎛ xm 2 ⎞ Mm = 70 xm – 20 ⎜ ⎟ = 122.5 KN.m ⎝2⎠ 5 4300:202 Mechanics of Solids Set # 4 Example – 2: Reactions: -Separate FBD “D-E” ∑M ⇒ @D ∴R e = 1 KN ∑F vl ⇒ (D-E) (D-E) = 0 ∴ R d = 1 KN -Now consider the “usual” beam with overhang, ACD, with known reaction at “D” (1KN downward) and the applied “equivalent” force and moment at “B”: ∑M ⇒ @"A" ∴ R c = 2.5 kN ∑F vl ⇒ (ACD) = 0 (ACD) = 0 ∴ R a = 2.5 kN 6 Dr. A. F. Saleeb 4300:202 Mechanics of Solids Set # 4 7 Dr. A. F. Saleeb 4300:202 Mechanics of Solids Set # 4 8 Dr. A. F. Saleeb ...
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