CE 331, Fall 2010 Influence Lines for Trusses 1 / 7 An influence line shows how the force in a particular member changes as a concentrated load is moved along the structure. For example, say we are designing the seven‐panel parallel‐chord roof truss shown below. We’ve determined that the dead load causes the maximum compressive force in the top chord at midspan (member “M11”). Part of the live load consists of a 5000 lb air conditioner unit. Unfortunately, the location of the air conditioner is not specified at the time we are designing the roof truss. We need to design the truss to support the air conditioner at any location (nodes N10 through N15). 8 ft 10 ft Typ Figure 1. 7‐panel parallel‐chord roof truss A brute‐force approach would be to apply the 1500 lb load at Node N10, calculate the bar force in Member M11, move the load to Node N11, calculate the bar force in Member M11, and so on for all six possible load locations. This process is illustrated in Figure 2. The structural analysis program RISA was used to calculate the bar forces. In RISA, compression is positive, which is opposite to the sign convention used in your Mechanics of Materials (MoM) class where compression is negative. We will use the MoM sign convention in our calculations this semester. Note that the maximum force in Member M11 (10,714 lb) occurs when the weight of the air conditioner is applied to Node N13. CE 331, Fall 2010 Influence Lines for Trusses 2 / 7 Figure 2. Member axial forces (lbs) due to unit loads at six different joints. Compression = +’ve. Influence Lines for Trusses CE 331, Fall 2010 3 / 7 The axial forces in Member M11 due to the 5000 lb air conditioner are summarized in the 2nd column of Table 1 below. The forces in M11 (zero) when a load is applied at the truss ends (N9 and N16) were added. In the third column, the axial forces in Member M11 due to a unit load are listed. These forces, when plotted, form an influence line for the axial force in Member M11, shown in Figure 3. Table 1. Forces in Member M11 Figure 3. Influence line for Member M11 (MoM sign convention) f_11, lb 0 ‐2679 ‐5357 ‐8036 ‐10714 ‐7143 ‐3571 0 f_11/‐5000 lb = I 0.000 0.536 1.071 1.607 2.143 1.429 0.714 0.000 Influence Value
2.50
Load at Joint N9 N10 N11 N12 N13 N14 N15 N16 2.00
1.50
1.00
0.50
0.00
N9 N10 N11 N12 N13 N14 N15 N16
Location of Concentrated Load
Each value of the influence line shown in Figure 3 required a separate structural analysis. This same influence line can be calculated with one analysis using the equation shown below (derived elsewhere from Betti’s Law of Reciprocal Deflections.) Δδxi
δi
=
f
Px
i
Px
(1) Where Δδxi = the deflection at Joint “x” due to the deformation of member “i” δ i = the deformation of member “i” f i Px = the force in member “i" due to a concentrated load at Joint “x” Px = the concentrated load at Joint “x” For this example, “x” = Joints N9 through N16 and “i" = Member M11. We can calculate the ratio on the right‐hand side of Equation 1 by calculating the ratio on the left‐hand side. The ratio on the left‐hand side can be calculated for “x” = N9 through N16 in one step using RISA. In theory, δi could be any number, but it’s easiest if we let δi be 1”. We cause Member M11 to elongate by 1” in RISA by applying a temperature change to the member. The magnitude of the temperature is calculated as follows. CE 331, Fall 2010 Influence Lines for Trusses 4 / 7 δ = α ΔT L
1" = (6.5 × 10 −6 / o F ) (ΔT ) (10 ft × 12 in / ft ) ΔT = 1282 o F
This temperature “load” is applied to Member M11 in RISA and the resulting node deflections are calculated. Figure 4 shows the load applied to Member M11, the resulting deflected shape of the truss, and a table of joint deflections. Note that the Y deflections for Joints N9 through N16 equal the influence values, I, shown in Column 3 of Table 1. (a) (b) (c) Figure 4. (a) Temperature load to cause 1” elongation of Member M11, (b) deflected shape of truss due to 1” elongation of Member M11, and (c) joint deflections due to 1” elongation of Member M11. Another way to elongate a truss member one inch is to apply a pair of loads to the joints on either end of the member. The magnitude of the loads is calculated below for this example. Influence Lines for Trusses CE 331, Fall 2010 5 / 7 δ=
PL
AE
L =10 ft × 12 in /
A = 2.96 in
ft
=120 in
2
E = 29,000 ksi
2
P=
(1in )(2.96 in )(29,000 ksi )
in
= 715.3 k
120
Figure 5 below shows the loads applied to the joints on either end of Member 11, the resulting deflected shape, and a table of joint deflections. The deflected shape and joint deflections are identical with those shown in Figure 4. CE 331, Fall 2010 Influence Lines for Trusses 6 / 7 (a) (b) (c) Figure 5. (a) Loads to cause 1” elongation of Member M11, (b) deflected shape of truss due to 1” elongation of Member M11, and (c) joint deflections due to 1” elongation of Member M11. Influence Lines for Trusses CE 331, Fall 2010 7 / 7 The procedure to calculate the maximum compressive chord force in a truss is illustrated below. 1. Calculate the maximum compressive chord force due to dead loads and note the member in which this occurs. max. chord force due to dead load = 50k and occurs in Member 11. f11D = 50k C
2. Draw the influence line for Member A using a structural analysis program. Apply a pair of 715.3k loads to the joints on either end of Member 11. The resulting
vertical deflections are the influence values, I.
Joint N9 N10 N11 N12 N13 N14 N15 N16 I 0.000 0.536 1.071 1.607 2.143 1.429 0.714 0.000 3. Apply the live load to cause the maximum compressive force in Member A. Equation 1 from previous page:
For this example:
P
f1113
= P13
Δδxi
δi
Δδxi
δi
f
=
Px
i
Px
= (−5 )
k
, ∴ fi
2.143in
in
1
Px
= Px
Δδxi
δi
= − 10.72 k I
The sign convention for Eqn 1 is shown below: Px and Δx :
= +’ve fi and δi : tension/elongation = +’ve Therefore, the live load causes a compressive force in Member 11. f11L =10.70k C Note, we use the influence diagram to decide where to place the live load to cause maximum total force in the member of interest. Since the bar force due to dead load was compressive, we want to place the live load to cause a compressive bar force due to live load. 4. Calculate the total force due to factored loads and check the strength of the member. Pu = 1.2 f11D + 1.6 f11L = 1.2 (50k C) + 1.6 (10.7k C) = 77.1k C. Pu = 77.1k < 81k = φ Pn , OK A similar procedure can be used to check the maximum diagonal force.