03 Moment Diagram 1 PDF

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bending moment...

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Moment Diagrams 1 Definition of Bending Moment

1) Bending Moment is the internal result of an externally applied, usually normal (or perpendicular) types of loading (what it is...)

2) Bending Moment is the sum of the moments (force x distance) to the left of an imaginary section cut through the beam at a certain location note: Clockwise = + & Counterclockwise = (convention)

3) A Bending Moment Diagram is a plot of the variation of Moment along the span of the beam from left to right (convention)

4) Bending Moment is given the symbol "M" and is measured in pounds-feet (lbs-ft) or kip-ft since it a sum of moments (forces x distance) (symbol and unit)

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1) Failure due to bending: • Long beams (we will cover this later) (a high proportion of length "L" to depth of member) • Large loads at center concentrated or uniform P

P = applied load

∆ = deflection

= L RL

RR

RL

exaggerated bending of beam

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RR

Properties of Moment Diagrams

1) The variation in Moment is a sloping straight line in the absence of any change in loading 2) The variation in Moment is NOT pronounced where a concentrated load is applied to beam. 3) The variation in Moment is a parabolic curve for a uniform load (curve is concave down). 4) The variation in Moment is always maximum at the location of zero shear. 5) The Moment Diagram should always end up at zero - it always closes. statics 6) The variation of Moment is minimum at the supports. 7) The Moment diagram, particularly the maximum Moment, will lead to the depth (Moment of Inertia) of the member.

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Shear Diagram > Bending Moment P = 5 KIPS

FREE BODY DIAGRAM (FBD) L = 12 ft RL = 2. 5 K

RR= 2. 5 K

VL = + 2. 5 KIPS SHEAR DIAGRAM V= 0 VR = - 2.5 KIPS

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Shear Diagram > Bending Moment P = 5 KIPS

Free Body Diagram (FBD) or Load Diagram STEPS: 1) Find Reactions: since the beam is symmetrical and ∑ loads = 5 k, each reaction is 2.5 k

L = 12 ft

2) V diagram (Left to right) V 1 = + 2.5 k V 2 = + 2.5 k

RL = 2. 5 K

V3 = 0 k

V 4 = - 2.5 k

Shear Diagram

V 5 = - 2.5 k

V6 = 0 k

RR= 2. 5 K

VR = + 2. 5 K

A1 3) Bending Moment or "M" Diagram at section 1-1 V = + 2.5 k and M = + 2.5 k x 2' = 5 k-ft at section 2-2 V = + 2.5 k and M = + 2.5 k x 4' = 10 k-ft

at section 3-3 note a fraction to the left of Z V = + 2.5 k and M = + 2.5 k x 6' = 15 k-ft

Z V= 0 V1

V2

V3

V5

V4

V6 A2

2'-0 increments

VL = - 2. 5 K

Bending Moment Diagram M max = +15 K - ft M2 = +10 K - ft M1 = +5 K - ft

at section 4-4 V = - 2.5 k and M = (+ 2.5 k x 8') + (- 5 k x 2ft) = 10 k-ft from central load P from reaction at section 5-5 V = - 2.5 k and M = (+ 2.5 k x 10') + (- 5 k x 4ft) = 5 k-ft

M3 = +10 K - ft M4 = +5 K - ft M=0

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Relationships between Load, Shear and Bending Moment Diagrams 1) The maximum value of Shear occurs at the supports, while the minimum value of Bending Moment also occurs at the supports. 2) The maximum value of Bending Moment occurs at the location that Shear goes from + to - (or opposite) and thus passes through zero. 3) The sum of the areas under the Shear Diagram to the left of a certain location is equal to the Bending Moment at that location. 4) There is always a sudden drop under a concentrated load in the Shear Diagram There is hardly any change under a concentrated load in the Bending Moment Diagram. 5) If the loading is symmetrical then both the Shear and Bending Moment Diagrams are symmetrical. 6) If the shear diagram over a portion of the beam is a horizontal straight line (no load applied), then this is reflected as sloping straight line in the moment diagram, with the slope equal to the value of the constant shear. 7) The area to the left of the V = 0 (under the shear diagram) is equal to the Mmax. page 6 of 10

Shear Diagram

Bit of a geometry refresher: VR = + 2. 5 K A1

V1 = 2. 5 k Z

A1

LR = Ltotal/2 = 12/2 ft A1 = L/2 x V1 = 6 ft x 2.5 K = 15 k ft

V1

V= 0 V2

V3

V4

V5

V6 A2

VL = - 2. 5 K

Bending Moment Diagram V1

M max = +15 K - ft L or

M2 = +10 K - ft

V1

M1 = +5 K - ft L Atrianagle =

V1 x L1 2

Vavg =

M4 = +5 K - ft M5

M0

V2

M3 = +10 K - ft

Geometry Notes: (V1 + V2) 2

V1

L Can either take a shape like this and break it into component pieces or (V1 + V2) x L1 A total = 2

• M0 is the algebraic sum of the areas under the V diagram to the left of point 0. (there are not areas thus M = 0) • Mmax is the algebraic sum of the areas under the V diagram to the left of point Z. (area A1 only). • M5 is the algebraic sum of the areas under the V diagram to the left of point 5. (+ A1 - A2 = 0).

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EXAMPLE Free Body Diagram (FBD) or Load Diagram P1 = 2 K

STEP 1: Solve for reactions RL and RR with a free body diagram and Use ∑M = 0 and ∑Fy = 0.

2 ft

P2 = 6 K P3 = 2 K

4 ft

3 ft

RL = 4.2 K

1 ft RR= 5.8 K

Shear Diagram drops = 2 K Step 2: Work through the shear diagram starting from the left - +4.2 K A1 drops = 6 K RL magnitude and direction drops = 2 K +2.2 K A2 give the first value of shear diaZ gram. V= 0 Work across moving the diagram in the direction and magnitude of the applied loads.

A3 -3.8 K

A4 - 5. 8 K

Bending Moment Diagram Step 3: Find areas under shear diagram - A1 etc. these are the corresponding values of M1 etc; M max = +17.2 K - ft watch the sign, in that A3 and A4 are "negative" areas. M1 = +8.4 K - ft slope = 2.2 M3 = +5.8 K - ft slope = 4.2 TO CHECK: slope = - 5.8 in shear diagram M=0 add A1 + A2 + A3 + A4 = 0! NOTE also: slope of M diagram is value of shear diagram page 8 of 10

Recap the steps: 1) Find RL and RR, and check. 2) Determine critical points where loads change. 3) Find the value of shear V at these points. 4) Draw the shear diagram. 5) Determine the number of areas under the shear diagram calculate these areas. 6) Sum the areas to the left at each point. The sum is the moment at that point. 7) Check that M final = 0 8) Draw the Bending Moment Diagram.

P2 = 6 K

P1 = 2 K

2 ft

P3 = 2 K

3 ft

4 ft

1 ft

RL = 4.2 K

RR= 5.8 K

drops = 2 K +4.2 K

A1

drops = 6 K drops = 2 K

+2.2 K A2

Z

V= 0

A3 -3.8 K

A4 - 5. 8 K

M max = +17.2 K - ft M1 = +8.4 K - ft

s l o p e = 2 .2

M3 = +5.8 K - ft

slope = 4.2 M=0

s l o p e = - 5 .8

Other notes: • The M diagram is maximum where the V diagram changes from + values to - values; thus passing the point where V = 0. • The M diagram is a minimum (o in this problem) where the V diagram is maximum (at the support). • Concentrated loads cause a sudden drop in the V diagram and no sudden change in the M diagram. The slope of the M diagram changes. • At any point along the beam, the value of of V and M can be readily found by drawing a line and reading off the values. page 9 of 10

GENERAL INSTRUCTIONS TO DRAW SHEAR AND MOMENT DIAGRAMS 1) If the question is to find Vmax that implies Vmax (+) and Vmax (-) 2) If the question is to find Mmax that implies Mmax (+) and Mmax (-) 3) Perform all calculations to two decimal places. 4) In the process of drawing V and M (or shear and moment ) diagrams, check all your answers along the way. You have a check for every stage 5) "Box" your answers on home work to make them stand out. 6) Be very neat - these problems can be very confusing if you are not. 7) ALWAYS DRAW A FREE BODY DIAGRAM - THEN LINE UP THE SHEAR AND MOMENT DIAGRAMS BELOW. 8) VG & M Diagrams are only complete when all values of V and M are shown at every point where the load changes. 9) Always solve one problem per page.

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