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 Vdiagram at any point is equal the negative of the value
of the transverse load intensity at the same point. − Slope of Mdiagram 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 qdiagram between a, b)
xa xb M b  M a = ∫ V dx = area of Vdiagram 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 shearforce 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 superpositionprinciple 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” subdiagrams 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 “DE” ∑M
⇒ @D ∴R e = 1 KN ∑F vl ⇒ (DE) (DE) = 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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This note was uploaded on 02/27/2011 for the course ENGINEERIN 342 taught by Professor Oldham during the Spring '11 term at The University of Akron.
 Spring '11
 oldham
 Shear

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