Previous developments have subjected to fixed loads. loads. Structures are also subjected to live loads whose position may vary on the structure. structure. This chapter focuses on such loads for . 1
Influ nf lue ence nc e Lines Li nes ons er e r ge n g. . s the car moves across the bridge, change with the position of the car and the maximum force in each member will be at a different car location. The design of each mem er mus e ase on e maximum probable load each .
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gure . r ge russ ruc ure Subjected to a Variable Position Load
er e o r e, t e t r u s s an a y s s forr ea fo each me memb mbe er wou w ould ld posi tion positio n tha thatt ca cause uses s the t he rea re ate test st fo forc rce e or st stre ress ss in eac h me m em b er.
3
If a structure is to be safely designed, members must be proportioned such that the and live loads is less than the availa lab ble secti tio on ca acit . Structural analysis for variable 1.Determining the positions of response function is maximum and 2.Computing the maximum . 4
In flue fl uenc nce e L ine in e Definitions u u reaction, axial force, shear force, or . Inf nflu lue enc nce e Li Line ne ≡ graph of a a function of the position of a dow downward ard uni unitt loa load d mov movin in acro across ss the structure. st ati sta tica call lly y de d ete termi rmina nate te st stru ruct ctures ures are al al ways pi pie ecewi cewise se linea li nearr. 5
Once an influence line is constructed: • Determine where to lace live load on a structure to maximize the drawn response function; and • Evaluate the maximum magnitude of the response function based on the loading.
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Calculating Response (Equilibrium Method)
ILD for A
1
0
L 1
ILD for Cy
0
L
7
1 x
MB a
0
VB
A
∑ Fy = 0
⇒ V = Ay − 1 B
Ma = 0 ⇒ M
MB
a
A y
B
= A y a −1(a − x)
a
VB
∑ Fy = 0
⇒ V = Ay B
Ma = 0 ⇒ M
B
= Ay a
8
1 – a/L
VB 0 a -a/L
ILD for VB
–
B
0
ILD for MB 9
Beam Example 1
Calculate and draw the su ort reaction response functions.
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Beam Example 2
Calculate and draw the response functions for R A, M A, RC and VB.
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BD: Link Member
Calculate and draw the response functions for Ax, Ay, AB and VB . NOTE: Unit load . 12
Muller-Breslau Principle influence line for a response function is iven b the deflected shape of the released structure due to a unit displacement (or rotation) at the location and in the direction of the response . A released structure is obtained constraint corresponding to the the original structure.
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CAUTION: Principle is only valid . Releases: Support reaction - remove translational support restraint. Internal shear - introduce an internal glide support to allow differential displacement movement. Bending moment - introduce an internal hinge to allow differential ro a on movemen .
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©
The McGraw-Hill Companies, Inc. Permission required for reproduction or display.
Influence Line for Shear
15
Influence Line for Bending Moment 16
Application of MullerBreslau Principle
17
18
19
y = (L – x) (a/L)
1
2
=
20
Qualitative Influence Lines In many practical applications, it is necessary to determine only the general shape of the influence lines but not the numerical values o e or na es. uc an influence line diagram is known as gram. numerical values of its ordinates is known as a uantitative influence line diagram. 21
NOTE: An advanta e of constructing influence lines using the Muller-Breslau Principle is that the response function of interest can be determined . determining the influence lines for other functions as was the case with the equilibrium method.
22
Influence Lines for Trusses In a gable-truss frame building, to the top chord joints through roof purlins as shown in Fig. T.1. Similarly, highway and railway brid e truss-structures transmit floor or deck loads via stringers to floor beams to the truss joints as shown schematically in Fig. T.2.
Fig. T.1. Gable Roof Truss
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Fig. T.2. Bridge Truss 24
These load paths to the truss joints provide a reasonable assurance that the primary resistance in the axial force. Consequently, influence lines for axial member forces are developed by placing a unit load on the truss and making judicious use of free body diagrams and the equations of .
25
Due to the load transfer process in truss systems, no discontinuity will exist in the diagrams. Furthermore, since we are restrictin our attention to statically determinate structures, the influence line diagrams will be piecewise linear .
26
Example Truss Structure
u w functions for Ax, Ay, FCI and FCD.
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Use of Influence Lines
Single Moving Concentrated Load Each ordinate of an influence response function due to a sin le concentrated load of unit magnitude placed on the structure at the location of that or na e. us,
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A
B
C
D
x B
D A
B
C
B
(M B )max ⇒ place P at B
−
B max
⇒ 29
1. The value of a response concentrated load can be obtained by multiplying the magnitude of the load by the ordinate of the response position of the load. . ax mum pos ve va ue o the response function is point load by the maximum positive ordinate. Similarly, the maximum negative value is obtained by multiplying the po n oa y e max mum 30 negative ordinate.
Point Res onse Due to a Uniformly Distributed Live Load Influence lines can also be emp oye o e erm ne e values of response functions of loads. This follows directly from point forces by treating the uniform load over a differential segment as a differential point orce, .e., = w l x. us, a response function R at a point 31
dR = dP
= w dx
where y is the influence line , application of dP. To determine the total response function value at a point for a u w x= x = b, simply integrate: b
b
wl ydx = wl ydx
R= a
a
32
in which the last inte ral ex ression represents the area under the segment of the influence line, which corresponds to the loaded portion of the beam.
SUMMARY 1. The value of a response function due to a uniformly portion of the structure can be intensity by the net area under the corresponding portion of the response function influence 33 line.
2. To determine the maximum positive (or negative) value of a response unc on ue o a uniformly distributed live load, those portions of the structure where the ordinates of the response function influence line are positive (or negative). Points 1 and 2 are schematically demonstrated on the next slide for moment MB considered in the point . 34
35
36
Where should a CLL (Concentrated Live Load), a ULL (Uniform Live Load) and UDL on the typical ILD’s shown below to maximize the res onse functions?
Typical End Shear (Reaction) ILD 37
Beam Shear ILD
Typical Interior Bending Moment ILD
Possible Truss Member ILD 38
Live Loads for Railroad Bridges Live loads due to vehicular traffic on highway and railway bridges are represen e y a ser es o moving concentrated loads with loads. In this section, we discuss the use of influence lines to determine: (1) the value of the response function for a given loads and (2) the maximum value series of moving concentrated loads.
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To calculate the response unc on or a g ven pos on o the concentrated load series, series load Pi by the magnitude of the influence line diagram ordinate yi at the position of Pi , i.e. = i yi i
calculated from the slope of the influence line diagram (m) via yi = m x i 40
where x i is the distance to point i measure rom e zero y-ax s intercept, as shown in the . m 1
yb
ya x
b
a
a
=
b
⇒ similar triangles
y b yb ∴ ya = a ; m = b b
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xi For example, consider the ILD s own on e nex s e subjected to the given wheel
Load Position 1: 1 30
=
1 30
1 30
1 30
1 ( )(8(20) + 10(16) + 15(13) + 5(8)) 30
= m∑Pi xi = 18.5k i
42
2/3
10 ft. 20 ft.
-1/3
43
Position 1
Position 2 44
Load Position 2:
VB2 =
30
(−8(6) + 10(20) + 15(17) + 5(12))
= Thus, load position 1 results in the . NOTE: If the arrangement of loads is such that all or most of the heavier loads are located near one o e en s o e ser es, en e analysis can be expedited by for the series so that the heavier loads will reach the maximum 45
lighter loads in the series. In such a case, it may not be necessary to examine all the loading positions. Instead, the analysis can be w v u response function begins to . ., the response function is less than the preceding load position. This process is known as the “ Increase-Decrease Method” .
46
CAUTION: This criterion is not va or any genera ser es o loads. In general, depending on , , and shape of the influence line, the value of the response function, after declining for some loading positions, may start ncreas ng aga n or su sequen loading positions and may attain a .
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Zero Ordinate Location Linear Influence Line b+ 1
xm1 -
+
L
48
− b x+ =
; m+ = m+
b − − b L
b + − b − − b − x− = ; m− = m− : are obtained from
u
y = mx + with y = 0.
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Example Truss Problem: Application of Loads to Maximize Response
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Place UDL = 1.0 k/ft; ULL = 4.0 k/ft CLL = 20 kips compression axial forces in members CM and ML. Calculate the magnitudes of the .
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Force and Moment Envelopes response function as a function of the location of the response function is referred to as the envelope of the maximum values of a response function for the considered.
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For a single concentrated force for a simply supported beam:
+ (V)
max
−
max
⎛ a⎞ = P ⎜1− ⎟ a
−
L
M max = P a ⎜ 1−
⎝
⎟ L⎠
Plot is obtained by treating “a” as a var a e. 53
For a uniformly distributed load on for a simply supported beam:
+ (V)
max
− V
=
wl 2L
=−
M max =
l
2
(L −a )
wl a
2
2
2L
(L −a)
Plot is obtained by treating “a” as a var a e. 54
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