examples: rigid frames example 1 from estructures v1.1...

ARCH 631 Note Set 7.3 F2012abn

1

Examples:

Rigid Frames

Example 1 From eStructures v1.1, Schodek and Pollalis, 2000 Harvard College


2

Example 1 (continued)


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4



5



6



7



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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


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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


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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


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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

examples: rigid frames example 1 from estructures v1.1...

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