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We will analyze bending in a beam due to two types of loads: dead loads and live loads. Dead loads typically represent the weight of the structure itself. Live loads typically represent the weight of the structure's contents (people in an office building, merchandise in a warehouse, trucks on a bridge). Live loads differ from dead loads in that the location of live loads is uncertain. This will be discussed further in Step 3 of the example problem below. Example Problem 1 Given the loads and beam configuration shown below, calculate the maximum moment due to all loads. •

uniform distributed dead load (wD) = 0.50klf applied to entire beam

•

uniform distributed live load (wL) = 1.00klf applied along either Span 1 only, Span 2 only, or Spans 1 and 2 Span 1 17 ft

Span 2 8 ft

The most accurate solution would be to calculate the bending moment for the following three cases: wL = 1.0klf

Case I: DL + LL on Span 1 only

wD = 0.5klf

wL = 1.0klf wD = 0.5klf

Case II: DL + LL on Span 2 only

wL = 1.0klf

Case II: DL + LL on Spans 1 & 2

wD = 0.5klf

Structural engineers use computer programs to do just this. For this hand-calculation example, we will calculate a bending moment which should be close to the actual maximum moment due to dead plus live loads. We will use the following procedure: 1. Draw the beam to scale (horizontally). The resulting shear and moment diagrams will also be drawn to this horizontal scale. These scaled drawings can be used to check our calculations for reasonableness.

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2. Draw the shear and moment diagram due to dead load. Note the magnitude and location of the maximum bending moment, MD. 3. Calculate the moment due to live load, ML. We will assume that the maximum moment due to dead plus live loads (MD+L) occurs at the location of the maximum moment due to dead load (MD). This assumption is often correct, and even when not, yields a maximum total moment (MD+L) that is reasonably close to the actual total moment. 4. Calculate the total moment due to dead + live load, MD+L

Solution. 1. Determine the horizontal scale for your sketch (use maximum): • 7” across engineering paper, 5 little squares per inch, = 35 little squares • 25 foot-long-beam / 35 little squares = 0.71 feet per little square • Round 0.71 up to nearest of: 1, 2, 5, 10, 20, 50, …

Use 1 little square per foot ( or 1” = 5’)

See sketch on next page. 2. Calculate the max moment due to dead load (max MD) 2.1 Calculate reactions 25 ft ) − ( Rright support )(17 ft ) = 0, Rright support = 9.19 k 2 )( 25 ft )( 4.5 ft ) − ( Rrleft support t )(17 ft ) = 0, R left support = 3.31k

Σ M left support = 0, (0.5 klf )( 25 ft )( Σ M right support = 0, (0.5 klf

ΣFv = 0, 9.19 k + 3.31k − (0.5 klf )( 25 ft ) = 0, Checks

12.5

12.5 Resultant = (0.5klf)(25ft) = 12.5k

4.5 ft

2.2 Draw the shear diagram to scale (see sketch below). 2.3. Draw the moment diagram to scale (see below). Note: Change in moment = area under shear diagram.

Shear & Moment Diagrams Examples

CE 331, Fall 2007

17 ft

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8 ft W = 0.5 klf

3.31k

3.31

V, k

9.19k

1 = (6.62 ft )(3.31k ) = 10.96 k − ft 2 10.38ft 6.62ft

=

4.0

16.0k-ft

which it should

26.94k-ft

3.31k 0.5k / ft

V = 0k at beam end,

-5.19

10.96k-ft

M = 0k at beam end,

M, k-ft

which it should

-16.0k-ft

max MD = 16.0k-ft at Support 2 3. Calculate the max. moment due to live load (ML) at the location of the max. moment due to dead load (MD). 3.1 Determine where to place the live load to cause the max ML at the middle of Span 1. As mentioned on Page 1, the location of live loads is variable. Although it's physically possible that live loads are located on only part of a span, structural

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engineers typically apply distributed live loads to the entire span. The choices are therefore to apply live load to: • Span 1 only, • Span 2 only, or • Spans 1 and 2. To determine which of the above span loading patterns will cause the max ML at Support 2, we look at the deflected shape of the beam that will cause the max. bending moment at this location (see below). Bend beam over Support 2 to cause negative bending (since MD = -16.0k-ft)

Load spans to cause the deflected shape above. wL = 1.0klf

3.2 Calculate the moment at Support 2 due to live load (ML) We could calculate the reactions, draw the V and then the M diagram; OR use one of the formulas below: Simply-supported Beam with Uniform Load: L w

w L2 / 8 M

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Cantilever Beam with Uniform Load: w

L

M w L2 / 2

Therefore, ML = wL L2 / 2 = (1.0klf)(8ft)2 / 2 = 32.0k-ft ML = -32.0k-ft

4. Calculate the total moment at Support 2, MD+L. MD+L = MD + ML = -16.0k-ft + -32.0k-ft = -48.0k-ft MD+L = -48.0-ft

Check: Envelope of maximum moments due to MD and ML from computer-aided analysis:

Max. M = -48k-ft at Support 2 (same as hand-calcs.)

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Example Problem 2. (same as Problem 1 except overhang = 7ft)

•

uniform distributed dead load (wD) = 0.50klf applied to entire beam

•

uniform distributed live load (wL) = 1.00klf applied along either Span 1 only, Span 2 only, or Spans 1 and 2 Span 2 7 ft

Span 1 17 ft

MD: 17 ft

7 ft W = 0.5 klf

3.53k

8.47k

ΣM A = 0, → R2 = 8.47 k ΣM 1 = 0, → R1 = 3.53k ΣFv = 0, 3.53k + 8.57k − (0.5klf )( 24 ft ) = 0, OK

3.50 3.53 12.46k-ft V, k

12.5k-ft

9.94ft

7.06ft

=0k,

OK

k-ft

24.70

-4.97

12.46

=0k-ft,

M, k-ft

OK

-12.24

max MD = 12.46k-ft in "middle" of Span 1

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ML :

Bend beam at middle of Span 1 to cause positive bending (since MD = +12.46k-ft)

Live load Span 1 to cause the deflected shape above.

wL = 1.0klf

Therefore, ML = wL L2 / 8 = (1.0klf)(17ft)2 / 8 = 36.1k-ft ML = +36.1k-ft MD+L:

MD+L = MD + ML = 12.46k-ft + 36.1k-ft

MD+L = 48.6k-ft

Check: Envelope of maximum moments due to MD and ML from computer-aided analysis:

From the computer-aided analysis: •

the max MD = 12.46k-ft at 7.06ft from the left support

•

the max ML = 36.1k-ft at 8.50ft from the left support

•

the max MD+L = 48.1k-ft at 7.97ft from the left support