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2144: MEGR Introduction to Solid Mechanics
Instructor: Qiuming Wei Grader: Matt Bolen Office: DCH 362 Phone: 704 687 8213 Office Hours: M, W, F: 3:00-4:30pm Other times by appointment
01 of Lecture 09
Review of Lecture 08
Types of beams: Need to know what kinds of reactions can be generated
Simple beam (one pin support + one roller support) Cantilever beam (one fixed end + one free end)
Beam with an over-hang
Types of Loads
Concentrated (vertical or non-vertical) Distributed load q (x) ( intensity in N/m or lb/in.) Couples (moments in N.m or lb.in.) Finding V and M in a beam: basic recipes
1. 2. 3. 4. 5. 6. 7. 8.
Draw the free-body diagram of the beam Identify the reaction forces and moments Find the reactions forces and moments based on static equilibrium (balance of moments and equilibrium of forces) Make a cut thru the beam at a position of interest Draw the free-body diagram of one part of the cut beam Write down static equilibrium equations of forces and balance of moments (about the cut x-section) for that part Solve for V and M. Continue until all V and M are found
MEGR 2144 2
Important topics in Chap 04
What are beams? Types of beams? Types of loads on beams? Shear forces and bending moments in beams? how to calculate them? Relationships between loads, shear forces and bending moments? Shear force and bending moments diagrams?
Chapter 04 is a short one the most difficult part is shear forces and bending moments diagrams!
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Commentary on this lecture
The following slides, including the example problems, are more or less a comprehensive review on the same topics covered in Statics. Please go back to Statics for more details on the topics about relations between loads, V and M, and V & M diagrams (Article 5/7, pages 281-287 of J. L. Meriam and L. G. Kraige, Statics, sixth edition)
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4.4 Relationships between loads, V and M
Consider an elem of a beam cut out between two x-sections dx apart. We may have three types of loads acting on the top surface of the elem distributed load, concentrated load and couple. Sign convention for the loads:
Distributed and concentrated loads: (+)positive when acting downward on the beam; (-)negative when acting upward on the beam; Couple on a beam: (+)positive when ccw; (-)negative when cw.
In general V and M vary along the axis of the beam their values on the right-hand face of the elem may be different from their values on the lefthand face! For distributed load q: increments in V and M are infinitesimal by dV & dM on the right hand face we have V+dV and M+dM. For concentrated load P, increments in V & M may be finite by V1 & M1 on the right-hand face we have V+V1 & M+M1. Based on static equilibrium, we can derive the relationship between the load and V and M.
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(4.4) Relationships between loads, V and M: for distributed loads
The distributed load has intensity q we need to find the relationship between V, M and the load q. FVert 0 V qdx (V dV ) 0
1. Shear force V: Therefore:
That is, the rate of change of shear force at any point on the axis of the beam equals the ve of the intensity of the distributed load at that same point!
Special case I: q=0 dV/dx=0 V=constant! Special case II: q=constant dV/dx=-q V=-qx+a (a is the integration constant) V varies linearly in the part of the beam.
Revisit Example 4-2 of Lecture 08: q=q0x/L (+ve downward) V=-q0x2 /2L dV/dx=-q0x/L=-q dV/dx=-q. What is the relationship between the shear forces at two different x-sections A and B of a beam? q=-dV/dx dV=-qdx integrate w.r.t. x from A to B
This is just the negative of the area of the loading diagram between A & B!
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Again, only q the relationship between M and q?
(4.4) Relationships between loads, V and M: for distributed loads
M=0 (about the axis of the left-hand face of the elem: axis perpendicular to the plane of the figure) -M-qdx(dx/2) (V+dV)dx + M + dM=0 neglect products of differential terms -Vdx+ dM=0 V =dM/dx
dM / dx
That is, the rate of change of the bending moment at any point of the axis of a beam equals the shear force at the same point! this appliesonly to regions of distributed loads! If a concentrated load is applied at a point, a sudden change (or discontinuity) in the shear force occurs and the dM/dx is undefined (singularity!) Again revisit Example 4-2 (page 247, Eqn. 4-3a): M=-q0x3/6L dM/dx=-q0x2 /2L=V! If we integrate dM=Vdx between two points on the beam axis
Or we see that MB-MA = (area of the shear force diagram between A & B).
Note: Eqn (4.7) is valid even when concentrated loads act on the beam between points A & B. But it is not valid if a couple acts between A & B because a couple produces a sudden change in the bending moment (singularity in M).
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(4.4) Relationships between loads, V and M: for concentrated loads
This is an beam elem under a concentrated load, P.
From force equilibrium: FVert =0 V-P- (V+V1)=0 V1=-P abrupt change in the shear force at any point where a concentrated load acts! But which side of the elem has higher shear force? What if P is acting upward? From balance of moments (about the left-hand face of the element): -M-P(dx/2) (V+V1)dx + M +M1=0 M1 =P (dx/2) + Vdx +V1dx) = (P/2+V+V1)dx
Since dx is infinitesimally small M1 is also infinitesimally small the bending moment does not change as we pass thru the point of a concentrated load.
However, since dM/dx =V at the right-hand side of the elem. dM/dx =V+ V1 =V-P at the point of a concentrated load P, the rate of change of dM/dx of the bending moment decreases abruptly an by amount of P
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(4.4) Relationships between loads, V and M: loads of couples
The beam elem is only under couples
From force equilibrium: FVert =0 V- (V+V1)=0 V1=0 the shear force does not change at the point of application of a couple. From balance of moments (about the left-hand face of the beam elem) -M + M0 (V+V1)dx +M + M1=0 neglect (V+V1)dx since it is an infinitesimal term we have: M1=-M0!
That is, bending moment decreases by M0 as we move from the left to the right thru the point of couple application the bending moment changes abruptly at the point of application of a couple!
Note: the sign of M0 determines whether the bending moment will increase of decrease!
(4.4) Relationships between loads, V and M: Summary
Concentrated Load: V1 = -P. M does not change.
Couples: V1=0 V does not change; M1 = -M0.
Example 01: Concentrated Load on a simple beam
4.5 Shear-force and bending-moment diagram
Statement of problem: a simple beam under a concentrated load. Questions: (a) Find V and M in the beam; (b) draw the V and M vs. x Solutions: 1st step: Determine the reactions at A & B: RA=Pb/L; RB=Pa/L. 2nd step: (a) Find the V and M within the beam as a function of x. Cut thru the beam at a xsection to the left of P and at a distance x from the pin support A Draw the free-body diagram of the left-hand part of the beam: FVert =0 V-RA=0 V=RA=Pb/L which is a constant! M =0 at A Vx-M =0 M=V x=RA x=Pbx/L (0<x<a) M is linear in x! Then cut thru the beam to the right of the load P (a<x<L) and draw a free-body diagram of the left hand part of the beam: FVert =0 V+P-RA=0 V=RA-P= (Pb/L)-P = P (b/L -1)= - P a/L which is a constant! M =0 at the right end M+ P(x-a) -RA x =0 M=RA x P(x-a) = Pa (L-x)/L (a<x<L) M is linear in x and M=0 when x=L, M=P a b/L when x=a. When can now plot V and M as a function of x! We can find Vmax and Mmax on the diagram
Example 02: Uniform Load on a simple beam
(4.5) Shear-force and bending-moment diagram
Statement of problem: A simple beam AB under a uniform load with intensity q. Question: Find the V and M in AB and draw the V and M diagrams. Solution: Similar to Example 01. In this case, RA=RB=qL/2. V and M at a distance x from the left-hand end are FVert =0 V=RA-qx=qL/2-qx V changes linearly with x and: x=0, V= qL/2; x=L/2, V=0; x=L, V=-qL/2. M =0 at the right end M = RA x - q x (x/2) = q L x/2 q x2 /2 x=0, M=0; x=L, M=0; x=L/2, M is maximum (?) Thus we can plot V and M versus the position of the beam (x).
Check the relations between V, M and Loads!
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(4.5) Shear-force and bending-moment diagram: Example 4-4 (page 287)
Statement of Prob: Part of a simple beam loaded by a uniform load q, with dimensions given in the figure. Question: Draw the V and M diagram. Solution: First, find the reactions RA and RB at pin-support A and roller support B M)A =0 (a+b/2)qb-RB L=0 RB=q b (b+2a)/2L; M)B =0 (c+b/2)qb-RA L=0 RA=q b (b+2c)/2L; Shear forces and bending moments: We have three segments: a, b and c, and we have to consider them individually. Make a cut within a and draw a free-body diagram of the left part (length x, 0<x<a). From static equilibrium: V = RA; M = RA x. Make a cut within b and draw a free-bo...
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