examples: rigid frames example 1 from estructures v1.1...
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ARCH 631 Note Set 7.3 F2012abn
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Examples:
Rigid Frames
Example 1 From eStructures v1.1, Schodek and Pollalis, 2000 Harvard College
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Example 1 (continued)
ARCH 631 Note Set 7.3 F2012abn
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Example 1 (continued)
ARCH 631 Note Set 7.3 F2012abn
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Example 1 (continued)
ARCH 631 Note Set 7.3 F2012abn
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Example 1 (continued)
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Example 1 (continued)
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Example 1 (continued)
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Example 2
The rigid frame shown at the right has the loading and supports as show. Using
superpositioning from approximate analysis methods, draw the shear and bending
moment diagrams.
Solution:
Reactions The two loading situations for which approximate reaction values
are available are shown below. These values must be calculated and added together (allowed by superpositioning).
RAH = -0.907wh + 0.0551Ph/L = m
mkNmmkN
5
)6)(50(0551.0)6)(10(907.0 / = -51.11 kN
RAV = -0.197wh2/L + 0.484P = )50(484.05
)6)(10(197.0 2/
kNm
mmkN
= 10.02 kN
MRA = -0.303wh2 + 0.0112Ph
2/L =
m
mkNmmkN
5
)6)(50(0112.0)6)(10(303.0
22/ = -105.05 kN-m
RDH = -0.093wh - 0.0551Ph/L = m
mkNmmkN
5
)6)(50(0551.0)6)(10(093.0 / = -8.89 kN
RDV = 0.197wh2/L + 0.516P = )50(516.05
)6)(10(197.0 2/
kNm
mmkN
= 39.98 kN
Member End Forces The free-body diagrams of all the members and joints of the frame are shown below.
The unknowns on the members are drawn as anticipated, and the opposite directions are drawn on the joint. We
can begin the computation of internal forces with either member AB or CD, both of which have only three
unknowns.
50 kN
B C
A
6 m
5 m
D
10 k
N/m
h w
P
h
L L/2
A D A D
L/2
RAH = 0.907wh
RAv = 0.197wh2/L
MRA = 0.303wh2
RAH = 0.0551Ph/L
RAv = 0.484P
MRA = 0.0112Ph2/L
RDH = 0.093wh
RDV = 0.197wh2/L
RDH = 0.0551Ph/L
RDV = 0.516P
ARCH 631 Note Set 7.3 F2012abn
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Member AB With the magnitudes of reaction forces at A know, the unknowns are at end B of BAx, BAy,
and MBA, which can get determined by applying 0 xF , 0 yF , and 0 BM . Thus,
0)6(1011.51 xx BAmkNkNF BAx = 8.89 kN, 002.10 yy BAkNF BAy = 10.02 kN
0)6(89.8)3)(6(1005.105 / BA
mkNmkNA MmkNmmM MBA = 21.16kN-m
Joint B Because the forces and moments must be equal and opposite, BCx = 8.89 kN, BCy = 10.02 kN and
MBC = 21.16 kNm
Member CD With the magnitudes of reaction forces at D know, the unknowns are at end C of CDx, CDy,
and MCD, which can get determined by applying 0 xF , 0 yF , and 0 BM . Thus,
089.8 xx CDkNF CDx = 8.89 kN, 098.39 yy CDkNF CDy = 39.98 kN
0)6(89.8 CDD MmkNM MDC = 53.34 kN-m
Joint C Because the forces and moments must be equal and opposite, CBx = 8.89 kN, CBy = 39.98 kN and
MCB = 53.34 kN-m
Member BC All forces are known, so equilibrium can be checked.
(Remember: To find the point of zero shear with a distributed load, divide the peak {triangle} shear by the
distributed load; ex. 51.11kN/(10kN/m) = 5.11 m)
51.1
1 kN
-8.8
9 kN
+ V
deflected shape (based on +/- M)
+ V
+ M
+ M 10.02 kN
(5.1
1 m
)
-39.98 kN 25.5
3 kN m
-53.34 kNm
-53.34 kNm
21.16 kNm
8.89 kN
105.
05 k
N m
0
43.21 kNm
21.1
6 kN m
A
B
6 m
10.02 kN
51.11 kN
105.05 kNm
BAx
BAy
MBA
BCx
MBC
BAy
B
BAx MBA
BCx
BCy
MBC
B
50 kN/m
C
5 m
BCy
10 k
N/m
CBy
CBx
MCB
CDy
C
CDx
MCD
CBx
CBy
MCB
D
C
6 m
39.98 kN
8.89 kN
CDx
CDy
MCD
ARCH 631 Note Set 7.3 F2012abn
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Example 3
Using Multiframe4D, verify the bending moment diagram for the example in Figure 9.9:
Assuming steel (E = 29,000 ksi)
y
x
5538.615 kip-f t
0.000 kip-f t
Mz'
5538.615 kip-f t
0.000 kip-f t
Mz'
5055 k-ft
9.9
y
x
Load Case 1
3600
Sectionsmies-slender
mies-stif f
P = 216 k
1425 k-ft
ARCH 631 Note Set 7.3 F2012abn
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Example 3 (continued)
Displacement:
y
x
Maximum Actions for all members (column-1, beam-2, column-3):
(axes orientation reference)
Maximum Stresses for all members (column-1, beam-2, column-3):
Beam-Column stress verification (combined stresses) when d = 24 in, A = 28 in2. Ix = 2380 in
4:
ksiksiksift
in
in
in
in
k
I
Mc
A
P
S
M
A
Pf
ftk
93.9322.8671.712
2380
)2
24(1425
28
21642max