flange lateral bending stress ( f) under the wind pressure
TRANSCRIPT
Flange Lateral Bending Stress (fl) Under the Wind Pressure By
Atorod Azizinamini, Ph.D.,P.E.
1- Introduction and Objectives
The purpose of preparing this document is to evaluate the application of AASHTO LRFD
Bridge Design Specification (Third edition) to calculate flange lateral bending stress, fl,
for a specific design example and compare the results to detail finite element analysis.
Specific objectives are as follows:
a) calculate the flange lateral bending stress using two and three dimensional finite
element analyses
b) calculate the flange lateral bending stress using code recommendations
c) calculate the magnification factor using detail finite element analysis approach
and that recommended by the code
d) Incorporate flange lateral bending stress term in constructability limit state check
for a three span continuous bridge designed using High performance steel
e) Provide preliminary conclusions with respect to advantage and shortcomings of
the procedures suggested by AASHTO LRFD Bridge Design Specification to
calculate flange lateral bending stress
2- Brief Summary of the Bridge Configuration
Bridge Considered is a three span continuous steel plate girders. Following is brief summary of the specific design information: No. of Spans = 3 Length Span No. (ft) _________________________________________________________________ 1 135 2 175
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
3 135 No. of Lanes = 3 No. of Girders = 4 Skew Angle = 0 Dist. Curbline To Exterior Girder = 3 DECK DATA AND MATERIAL PROPERTIES _________________________________ Slab Thickness = 8.5 in Haunch Thickness = 3 in Sacrificial Wearing Surface = 0.5 in Concrete Compressive Strength (f'c) = 3000 psi COMPOSITE TYPE FOR ANALYSIS: Deck is Considered Composite Throughout LOADING _______ The Live Load Vehicle is the AASHTO HL-93 Loading ADDITIONAL DEAD LOADS _____________________ Superimposed Dead Load = 175 plf Future Wearing Surface = 20 psf Additional Girder Dead Weight = 10 % of Girder Dead Load
3- Wind Load Analysis
a- Elastic Three Dimensional Finite Element Analysis- Complete Bridge Model
Three dimensional model of the bridge was developed using both SAP2000 and
ANSYS5.7. The purpose of this exercise was to ensure the accuracy of the three
dimensional modeling. The modeling techniques used in developing three
dimensional model of the bridge is based on the past experiences gained from full
scale testing and modeling of the steel bridges in both laboratory and field. Both
model were subjected to same loading and produced approximately same results.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
ANSYS5.7 model was selected to carryout all three dimensional analysis as one of
the objective of the work was to conduct geometrical non-linear analysis.
The code specified lateral load (non-factored) due to wind is 0.3 kips per linear foot. The
depth of the girder used is 54 inches. In the three dimensional finite element analysis a
pressure loading of 65 lb/ft2 was applied to outside girder (perpendicular to the web face),
which is approximately equivalent to 0.3 kips per linear ft wind load.
The stress contours and deflected shape of the bridge under the applied wind pressure is
shown in Figure 1. The close up of the model for the first span is shown in figure 2.
Deflected shape (plan view) Stress in x-direction
Figure 1 the deflection and stress of the bridge under the wind pressure
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Figure 2 Flange stress under the wind pressure, a closer view- First Span
The maximum lateral deflection of the girder is approximately 10 inches (due to
unfactored wind load of 65 lb/ft2).
Summary of the flange lateral bending stresses obtained from three dimensional finite
element analysis of the entire bridge is shown in Table 1 (see column no. 7) for three
different locations along outside girder. Section 1 in table 1 is located in the first span. As
indicated in table 1, location of maximum flange lateral bending stresses is different for
various analyses types. Section 2 in table 1 is located over the pier and section 3 is
located at mid-span of middle span.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Table 1 Bottom flange stress under wind pressure from a 3-span bridge
Sec. Sy
(in3 ) M (kips-in)
Maximum Flange Lateral Bending
Stresses, (ksi)
Sec.
Mod.
2D† Sap,
Constant
properties
2D† Sap,
specified
Properties
2D† Sap,
Specified
Properties
3D† Ansys, One
Girder Model
3D†† Ansys,
Entire Bridge
(1) (2) (3) (4) (5) (6) (7)
1 53 1092
(x=49.5ft) 832.5
15.7
(x=66.3 ft)
12.1
(x=43.5ft)
9.4
(x=45.1ft)
2
232 2213(x=135ft) 2941 12.7
(x=135ft) 8.4 (x=135ft) 8.4 (x=135ft)
3
49 1232
(x=222.5 ft) 535
10.9
(x=222.5ft)
9.5
(x=222.5ft)
8.2
(x=222.5ft)
† 0.3 k/ft applied to one girder and resulting stress divided by number of girders
†† 65 lb/ft2 which is equivalent of 0.3 k/ft applied to outside girder only. Analysis takes care of number of
girders
The maximum and minimum flange lateral bending stresses shown in figure 1 and 2 are
for first span and are approximately 9.4 ksi. The maximum and minimum flange lateral
bending stresses over the pier is 8.4 ksi. The maximum and minimum flange lateral
bending stresses in middle of the second span is 8.2 ksi.
b- Elastic Three Dimensional Finite Element Analysis- One Girder Model
A three dimensional model of one outside girder was also constructed using Ansys 5.7. In
this model the shell elements were used to model both the web and flanges using the
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
actual web and flange dimensions. The applied load was in the form of pressure and was
specified at the 65 pounds per square ft, which is equivalent to 0.3 kips per linear ft. From
this three dimensional model the maximum flange lateral bending stresses at the same
locations as that reported for complete model of the bridge were extracted and are shown
in column 6 of table 1. It should be noted that the stresses obtained from this three
dimensional analysis were divided by four (number of girders in the cross section).
c- Two Dimensional Analysis
One of the objectives of the work reported here was to determine the best approach to
calculate the flange lateral bending stresses. As a result, series of two dimensional
analyses were carried out.
Two different two dimensional analyses were carried out using Sap2000. In each analysis
one outside girder was modeled using two dimensional beam elements. One of the
analysis used constant section properties (section sizes at the middle of the first span)
over the entire length of the bridge. The other analysis used actual section properties
along the girder. Results of the Sap2000 in terms of moment for uniform cross section are
shown in figure 3. Section 2 is over the pier and section 3 is at middle of the middle span.
Figure 3 the moment diagram from the 2-D analysis with uniform section properties along the girder
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
The moment obtained from two dimensional analyses were converted to flange lateral
bending stresses using equation (1) shown below.
yf S
M=σ (1)
Where, Sy is the section modulus about the minor axis of one flange and M is the wind
induced moment per each girder (the moment shown in Figure 3 is for one girder. It is
assumed that total wind load is resisted by all four girders, which is in accordance with
results obtained from three dimensional finite element analyses).
Table 1 shows summary of the results obtained from two dimensional analyses in terms
of moment and stresses at three locations along one girder.
d- Calculating the Flange Lateral Bending Stresses using AASHTO
Recommendation
Section 4.6.2.7 of AASHTO LRFD in its commentary provides a recommendation on
how to calculate the resulting moments in the girders due to wind pressure. The
recommendation are as follows:
b
bw N
WLWLM
810
22
+= (2)
Using equation (2), the maximum moment for first span could be calculated as follows:
fkftklfftklf
M w −=+= 3.187)4(8
)135)(3.0(
10
)4.23)(3.0( 22
(3)
Using equation (2) the maximum moment for middle span could be calculated as follows:
fkftklfftklf
M w −=+= 1.303)4(8
)175)(3.0(
10
)1.23)(3.0( 22
(4)
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
The flange lateral bending stresses could then be obtained by dividing the resulting
moments by section modules. Table 2 provides summary of flange lateral bending
stresses using the AASHTO LRFD recommendation.
Table 2 Lateral flange bending stress under the wind pressure based on the AASHTO recommendations
Sec. Sy (in3 ) M (kips-in) Flange Lateral Bending Stresses, (ksi)
Section
Modules
AASHTO LRFD
Recommendation AASHTO LRFD Recommendation
1 53 2247.6 42.4
2 232 N.A. N.A.
3 49 3637.4 74.2
As noted from table 2 the flange lateral bending stresses, using AASHTO LRFD
recommendations, are not calculated for section near pier. It is believed that the
AASHTO LRFD recommendations are only applicable for middle of each span.
4- Magnification Factor
It is assumed that the compression flange, when subjected to lateral loads, acts as a beam
column. The lateral load causes the flange to displace in lateral direction. As a result the
maximum moment in the compression flange will increase due to secondary effects.
AASHTO LRFD treats this magnification in a similar way that AISC building code treats
design of beam columns. Appendix A provides derivation of the moment magnification
factor as used in AISC building code.
As indicated in appendix A the general from of the magnification factor is as follows:
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
ek
m
P
PC
−1
Where P is the applied axial load to the beam column and Pek is the critical column
buckling load. Cm is a factor that accounts for different loading cases. Appendix A
provides more detail discussion of this magnification factor.
As noted from Appendix A the derivation of the moment magnification factor is based on
assuming a beam column being supported at both ends and subjected to end axial load
and some type of lateral load.
The magnification factor used in Equation 6.10.1.6-4 of the AASHTO LRFD is similar to
that shown above and is as follows:
cr
bm
F
f−1
85.0
According to AASHTO LRFD code the lateral flange bending stress should be magnified
only if the unbraced length between the cross frames exceeds that shown below:
yc
bm
bbpb
F
fRC
LL 2.1≤ (5)
Where
yctP F
ErL = (6)
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
For the design example under consideration the Lp is as follows:
ftinLP 6.650
290003.3 == (7)
Equation (5) is function of the compressive stress in the flange due to gravity load as
indicated in fbm term.
The Strength Limit State III is where wind load on structure WS influences the loading
combination. The Strength Limit State III could be written as follows:
1.25 fDC1 + 1.4 fWS
During the construction phase the fbm in equation (5) could be due to dead weight of the
steel girders or weight of the steel girders plus the wet concrete.
For the sake of discussion, the calculation of magnification factor is shown for middle
span.
a) magnification factor using self weight of concrete
This is a very unlikely event where concrete is placed and high wind loads are applied to
the girders before concrete has had a chance to harden. Therefore this scenario is a safe
guard for approximately 24 hour time period where concrete is cast but is not hardened
yet. For this scenario, for the middle span the maximum factored compressive stress, fbm
due to gravity load only is 25.8 ksi. This includes the weight of the girder plus the
concrete. The limiting unbraced length between cross frames would then become:
ftftLb 0.11
50
8.25)0.1(0.1
)6.6(2.11.23 =>= (8)
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Since actual unbraced length between the cross frames is 23.1 ft and the limiting value is
11.0 ft, we must then magnify the fl term.
Substituting the numerical values, magnification factor becomes:
29.2
41
8.251
85.0 =−
=MF (10)
b) Magnification factor using only the weight of the girder
If concrete is not cast during severe wind condition, the fbm can be computed based on the
self weight of the steel girders. The moment induced in mid-span of the middle span, due
to weight of the girder only is 2021 ft-kips. The resulting compressive stress in the
compression flange could then be obtained as follows:
ksiS
Mf
xbm 0.3
833
)2021(25.1 === (11)
The limiting value for the unbraced length is calculated using Equation (5).
ftftLb 3.32
50
0.3)0.1(0.1
)6.6(2.11.23 =<= (12)
Therefore, for this scenario, there is no need to magnify the flange lateral bending stress
due to wind loads.
c- Magnification factor using Nonlinear Geometric Analysis
Nonlinear finite element analyses were carried out to account for second order effect
directly. ANSYS5.7 was used. Complete three dimensional model of the bridge was used
in the analysis.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Several scenarios were simulated in the nonlinear geometrical finite element analysis
using the full three dimensional model of the bridge. A) scenario where the dead weight
consisted of the weight of the girders only, b) scenario where dead weight consisted of
weight of the girders plus the weight of the wet concrete before it is hardened and c)
scenario where dead weight consisted of weight of the girder plus weight of the concrete
after concrete is hardened.
For the three cases described above, Table 3 gives the magnification factors to be used in
conjunction with flange lateral bending stresses at mid-span of the middle span (section 3
in table 2). The magnification factors reported in Table 3 is simply the ratios between
flange lateral stresses obtained from nonlinear finite element analysis divided by the
corresponding value from linear finite element analysis.
Table 3 Maginification factor for mid-span of middle span
Dead Load Considered fbm, ksi Magnification Factor
2D Analysis AASHTO Ansys, 3D nonlinear
Self weight of Girders only 3.0 1.0 1.01
Wet concrete & Girder
weight
25.8 2.29 1.31
Composite girder 1.0 1.03
Figures 4 through 5, gives maximum flange lateral stresses any where along the bridge as
obtained from nonlinear and corresponding linear finite element analysis for the three
construction scenarios described above. The magnification factors reported in Table 4 is
the maximum flange lateral bending stresses anywhere along the flanges obtained from
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
nonlinear finite element analysis divided by maximum flange lateral bending stress
anywhere along the bridge obtained from linear finite element analysis.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Linear Analysis
Nonlinear Analysis
Figure 4 The longitudinal stress contours in x-direction under wind and steel girder weight
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Linear Analysis
Nonlinear Analysis
Figure 5 The longitudinal stress contours in x-direction under wind, steel girder weight and slab weight
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Linear Analysis
Nonlinear Analysis
Figure 6 The longitudinal stress contours in x-direction under wind, steel girder weight and slab weight in composite condition
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Table 4 Maginification factor for various construction scenarios
Dead Load Considered Magnification Factor
AASHTO Ansys, 3D nonlinear
Self weight of Girders only 1.0 1.06
Wet concrete & Girder
weight
2.29 1.24
Composite girder 1.0 1.03
5- Summary
Previous sections provided different approaches for calculating the flange lateral bending
stresses. Methods used included
a) Three dimensional model of the entire bridge
b) Three dimensional model of one girder
c) Two dimensional model of girder using uniform sectional properties
d) Two dimensional model of the girder using specified section properties
e) AASHTO LRFD code recommendations
Table 5 provides summary of the flange lateral bending stresses for mid-span of the
middle span
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Table 5- Unfactored flange lateral bending stresses due to 0.3 k/ft. wind load
Method Used Flange Lateral Bending Stresses, ksi
3-D Entire bridge 8.2
3-D One girder 9.5
2-D Uniform Section Properties 25.1
2-D Specified Section Properties 10.9
AASHTO Recommendations 74.2
Further, the magnification factors to be used in conjunction with flange lateral bending
stresses due to wind loads were calculated using different approaches. Methods used
included:
a) magnification factors calculated using dead weight of the girder only
b) magnification factor calculated using dead weight of wet concrete and steel girder
c) magnification factor calculated using weight of hardened concrete and steel girder
Table 6 provides summary of the magnification factor for flange lateral bending stresses
for mid-span of the middle span
Table 6 Maginification factor for mid-span of middle span
Dead Load Considered fbm, ksi Magnification Factor
2D Analysis AASHTO
Requirement
Nonlinear Analysis
Self weight of Girders only 3.0 1.0 1.01
Wet concrete & Girder
weight
25.8 2.29 1.31
1.0 1.03
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
6- Constructibility Flexural Checks (AASHTO 6.10.3.2)
The Strength Limit State III is where wind load on structure appear. This load
combination is as follows:
1.25 fDC1 + 1.4 fWS
For the sake of discussion the calculations are shown for mid-span of middle span.
For this example, during construction, the maximum compressive stress in the
compression flange, fbu , due to factored (1.25 factor) gravity load is given by table 7
Table 7 Factored Stress in Compression Flange Due to Gravity Load
Dead Load Considered fbu, ksi
2D Analysis
Self weight of Girders only 3.0
Wet concrete & Girder weight 25.8
According to AASHTO equation 6.10.1.6-1 the flange lateral bending stress should be
less than 0.6Fyc.
In addition the girder section must satisfy Equations 1 to 3 in Article 6.10.3.2.1 for the
compression flange and Equation 6.10.3.2.2 for the tension flange. In the current example
the web is non-compact according to the AASHTO Article 6.10.6.2.2.3, thus the
following equations should be checked.
ychflbu FRff φ≤+ (13)
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
ncflbu Fff φ≤+3
1 (14)
a) Calculations based on values obtained from code provisions
The flange lateral bending stress is fl is 74.2 ksi which exceeds the Fy and therefore
code requirement is violated regardless of which dead loads are considered in the
calculations
fl = 74.6 ksi x 1.4 = 104.4 ksi
fbu = 3 ksi or 25.8 ksi
b) Calculations based on values obtained from 2-D Analysis and magnification
factor from AASHTO – Dead Weight of Wet Concrete and Steel Girder only
fl = 10.9 ksi x 1.4 = 15.3 ksi ( Using specified section properties)
fbu = 25.8 ksi
Magnification factor = 2.29
Fnc = 36.2 ksi Compressive capacity of compression flange
ychflbu FRff φ≤+
25.8+ 15.3 (2.29) = 60.9 > 50 ksi N.G.
ncflbu Fff φ≤+3
1
25.8+ (1/3) 15.3 x 2.29 = 37.4 > 36.2 ksi N.G.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
c) Calculations based on values obtained from 2-D Analysis and magnification
factor from AASHTO – Dead Weight of Steel Girder only
fl = 10.9 ksi x 1.4 = 15.3 ksi ( Using specified section properties)
fbu = 3 ksi
Magnification factor = 1.0
Fnc = 36.2 ksi Compressive capacity of compression flange
ychflbu FRff φ≤+
3 + 15.3 = 18.3 < 50 ksi O.K.
ncflbu Fff φ≤+3
1
3+ (1/3) 15.3 = 8.1 <36.2 ksi O.K.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Appendix A
Derivation of the Moment Magnification factor Used in
AISC Building Code
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
BEAM-COLUMNS
A beam-column is a member that is subjected to both axial force and bending moment. To begin our discussion, let us consider a simply supported beam-column subjected to an axial force P and uniformly distributed lateral load.
00 =∑M
022
2
=−−+ xwl
Pywx
M
w (intensity)
P x
y
EI = constant
P
P
w
V 2
wl
x
y M
o
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
xwlwx
PyM22
2
+−=
But,
''EIyM −=
→ xwlwx
PyEIy22
2'' +−=−
xwlwx
PyEIy22
2" −=+
Letting EI
Pk =2
xEI
wlx
EI
wyky
22" 22 −=+→
ph yyy +=
xKkxAy xxh cossin β+=
To find yp assume the following formula: CxCxCy p ++= 2
21
This results in:
42
22 22 Elk
wx
EIk
wlx
Elk
wy p −−=
Thus, the general solution is:
y=A sin Kx +Bcos Kx +K42
22 22 EIk
wx
EIk
wlx
EIk
w −−
The constant A and B are obtained using the following B.C’s.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
40)0(
EIk
wBy =⇒=
Al
y ⇒= 0)2
('2
tan4
hl
EIk
w=
Introducing the notation 2
klu =
the general solution could be written as follows:
)(8
12
cos2
sintan16 2
2
4
4
xlxEIu
wl
l
ux
l
uxu
EIu
wly −−
−+=
from which the moment distribution along the length of the member is:
−+=−= 11
2cos
1
2sintan
4 2
22 uxux
uu
wlEIyM
For this problem, the maximum deflection occurs at the midspan.
ymax =
2
ly
ymax = 2
4
4
4
32cos
cos1
16 EIu
wl
u
u
EIu
wl −
−
Ymax can be written as:
Ymax =
−−4
24
5
)2sec2(12
384
5
u
uu
EI
wl
=
−−4
2
5
)2sec2(12
u
uuyo
where EI
wlyo 384
5 4
= is the deflection if only uniform lateral load had existed.
Several observations could be made from this equation.
1. The term in the brackets could be viewed as amplification factor due to axial load.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
2. As axial load increases, the amplification factor increases as shown in figure below.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
When P=0 A.F. (amplification factor), reduced to 1. When u = 2/π
).(2
2
l
EIP
π= , the value of A.F. approaches infinity. In other words, as
P approaches the Euler load, the lateral deflection of the member increases without bound.
3. If P remains constant, ymax is linear function of w. Thus for several
beam with constant axial load and varying lateral load, the law of superposition is held. However, if axial load is varied superposition does not hold. For the design purpose, the term in the brackets (A.F.) could be simplified. Expanding sec u in a power series:
....8064
277
720
61
24
5
2
11sec 8642 +++++= uuuuu
Substituting this in the expression for ymax yields: [ ]....1649.4067.01 42 +++= uouyy omzx Noting that:
eP
P
EI
Plklu
222
π===
= 2
2
l
EIPe
π
ymax =
+++ ...)(004.1)(003.11 2
eeo P
P
P
Py
Or approximately
ymax
+
++≈ ...1
2
eeo P
P
P
Py
Also from a power series (Maclaurin Series):
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
...11
1 2
0
+++==− ∑
=xxx
x
oo
m
m
Therefore:
−≈
eP
Pyy
1
10max
The maximum moment also occurs at midspan.
)1(sec4
)2
(2
2
max −== uu
wlLMM
(from expression on page __)
or
−=2
2
max
)1(sec2
8 u
uwlM
−=20max
)1(sec2
u
uMM
M0 is the moment, that would exist if only lateral load had been applied. Therefore, the term in brackets represents amplification factor due to axial load. Again using the power series, expansion for sec u, ,the expression for Mmax could be simplified as follows:
[ ]...0687.01694.04167.01 6420max ++++= uuuMM
Using the expression for u
22
l
EI
Pklu ==
eP
Pu
2
π=
= 2
2
l
EIPe
π
= 2
2
πlP
EI e
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Mmax= M0
+
+
+
+ ...032.1031.1028.11
32
eee P
P
P
P
P
P
or
+
+
+
+= ...004.1003.11028.11
2
0maxeee P
P
P
P
P
PMM
or approximately,
Mmax =Mo
+
+
+
+ ...1028.11
2
Pe
P
P
P
P
P
ee
−
+=
e
e
P
PP
PMM
1
1028.110max
−
+=
e
e
P
PP
P
M1
028.01
0
or
−≈
eP
PMM
1
10max
Where the term in brackets is the design moment amplification factor.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
BEAM COLUMN WITH A CONCENTRATED LATERAL LOAD
For this case the governing D.E.’s are:
Pyxl
alQEIy +−=− )(" for ax ≤≤0
Pyl
xlQaEIy +
−=− " for lxa ≤≤
For the case of 2
la = it can be shown that:
P
P
( )L
aLQ −
x
y M
P x
y
EI = constant
P
Q
L
a
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
=u
uMM o
tanmax (H.W.)
Where 4
QlM o = (Moment if only concentrated load had
existed)
and 2
klu =
To simplify the expression for maximum moment, we will use the power series expansion for tan u.
....315
17
15
2
3
1tan 753 ++++= uuuuu
++++= ...315
17
15
2
3
11 642
max uuuMM o
eP
Pklu
22
π==
+
+
+
+= ...811.0812.0823.01
32
maxeee
o P
P
P
P
P
PMM
+
+
+
+= ...985.0987.01823.01
2
0eee P
P
P
P
P
PM
+
+
+
+≈ ...1823.01
2
eeeo P
P
P
P
P
PM
−
+≈
e
eo
P
PP
PM
1
1823.01
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
−
+−≈
e
eeo
P
PP
P
P
P
MM1
823.01
max
[ ]
−
−≈
−
−≈
e
eo
e
eo
P
PP
P
M
P
PP
P
MM1
2.01
1
18.01
max
For convenience, let’s write this expression in the following form:
−
+
=
e
e
P
P
P
P
MM1
1
0max
ψ
Where andQL
M40 = ψ =0.2
Notice that for the previous case, maxM has the same form except that ψ =0.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
So in summary:
Note that in each case, the amplification factor =
ek
m
P
PC
−1
Table below gives the summary of mC for different loading cases.
P
P
P
P
P
P
MB MA
2l
2l
1=mC
2.0 where
1
−=
+=
ψ
ψek
m P
PC
Form) d(Simplifie
4.04.06.0 ≥
−=
B
Am M
MC
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
BEAM-COLUMN WITH UNEQUAL END MOMENTS
AB MM >
Assume AB MM ≥ The D.E. for this case is as follows:
ABA Mx
l
MMPyEIy −+=+"
it can be shown that the solution to this D.E. is as follows:
( )2222
cossin
cos
EIk
M
lEIk
MMkx
EIk
M
klEIk
MklMy ABAABA −++++−=
=EI
Pk 2
P
P
l
MM BA +
x
y
M
MA
MA
MB
P P x
y
l
EI = Constant
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
In the previous cases, we knew the location of the maximum moment. However, in this case, the location is not too obvious. Remember that
"EIyM −= and maxM is found by evaluating "y at maximum location and substituting it in "EIyM −= .
The moment is the maximum where shear is zero, therefore to calculate the location where the moment is maximum, we could equate:
'"EIydx
dmV −== to zero.
For this problem, it can be shown that:
( )
++
−=kl
klM
MMM
MM B
ABA
B 2
2
max sin
1cos)(2/
If the member had been bent in a single curvature, maxM could be simply found by BB MM −= or
( ) ( )
+−=
kl
klMMMMMM BABA
B 2
2
maxsin
1cos/2/
for the case: MMM BA =−=
( )kl
klMM
2maxsin
cos12 −=
And its location is at midspan as shown below:
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
When a beam is subjected to axial load and unequal end moment, calculation of maxM could be simplified using the concept of equivalent moment. The figure below shows schematically the concept of equivalent moment.
M M
P P
l
M M
+
=
2l
2l
Mmax
Primary Moment
Secondary Moment
Total Moment
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
eqM could be found by equating the expressions for maxM for two cases
derived:
( ) ( ) ( )kl
klM
kl
klMMMMM eq
BABAB 22
2
sin
cos12
sin
1cos/2/ −=
+−
or
( ) ( )
( )kl
klMMMMM BABA
eq cos12
1cos/2/ 2
−+−
= ( BM )
or ( )Bmeq MCM =
where ( ) ( )
( )Coskl
klMMMMAC BAB
m −+−
=12
1cos/2/ 2α
Austin has shown that an expression approximating mC is as follows:
MA MB Meq Meq
P P P P
MB>MA
MA
MB
Meq Meq
Mmax Mmax
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
( ) 4.0/4.06.0 ≥−= BAm MMC This expression has been adopted by both AISC/ASD and LRFD. Thus the maxM for a beam subjected to axial load and unequal end moment could be calculated as follows:
( )kl
klMM eq 2max
sin
cos12 −=
( )kl
klMCM Bm 2max
sin
cos12 −= (Assuming AB MM ≥ )
The above expression could be simplified noting that
2sin2cos1 2 kl
kl =−
and
=2
cos2
sin4sin 222 klklkl
=2
secmax
klMCM Bm
Letting ukl =2
...720
61
24
5
2
11sec 642 ++++= uuuu
eP
Pklu
22
π==
...27.1268.123.11sec32
+
+
++=→
eee P
P
P
P
P
Pu
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
+
+++≈ ...123.11
2
eee P
P
P
P
P
P
−+≈
e
e
P
PP
P
1
123.11
e
ee
P
PP
P
P
P
−
+−≈
1
23.11
e
e
P
PP
P
−
+≈
1
23.01
eP
P−≈
1
1
Therefore:
e
Bm
P
PMC
M−
=1
max
Note that
e
m
P
PC
−1 is an amplification factor.
In all the cases considered, beams were simply supported. In the event that end conditions are different than simply supported it could be shown that
2
2
l
EIPe
π= would be replaced by: ( )2
2
kl
EIP ke
π= where k is the effective length
factor if only axial load had existed.
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com
Therefore:
B
ke
m M
P
PC
M−
=1
max .
PDF Created with deskPDF PDF Writer - Trial :: http://www.docudesk.com