MSMB framesDESIGN AND DETAILING PART 2

Explanation of abbreviations used (upper right corner)

E stands for Essential, which means that the issue needs to be understood in detail.EP means essential in principles, only the meaning and principles need to be known.IF means informative only for information, will not be tested.TT means „Think trough" usually refers to more complex pictures, not to learn, just to try to understand why it is just as it is.

Stirrups in columns

Main rebars in columns

Recomm. f = 12 mm -28mm12 mm is the minimum of used diameter Length of used longitudinalbars is one storey + overlapping

Cross-Sections of columnsRectangular, squares, Circular. Min amount of bars is 4 by rectangle CS, by circular CS min = 6 barsdepth d%250 mm, 200 mm

in an one storey structure

Recomm. f = 6 to 8 (10) mmmin fss = max [0,25 ds, 6 (5 nets)By regular zonesmax ss= min [20 ds,max, b, 400mm] By critical zones - lappingis 0,6 ss used

b

d

E

(Over)lapping

Lapping length lo= a6 . lbd

a6 is coefficient oflappinga6 = 1,0 to 1,5 according

kind of stress in lapped bars (compression –tension) and % of lapped bars.

See educative video ED2 once more!!

burstingbuckling

Stocky memberSlender member

ViolatonsVioationstypestypes of violations cfWays of violations of compressive members

(Over)lapping length lO is commonly multiltiple of anchoriglength Lbd

lo= a6 . Lbd

a6 is coefficient of lapping, with values in the range of 1,0 to 1,5; according to the kind of stress (compression – tension) and % of lapped bars in an actual CS.As the art of stressing is usually not fully clear (depends on each possible loading case) it is usually recommended to take a6 =1,5.

EMBMS frame construction seqence II. summary

1. Construction starts from footing (80% cases=pad)+the rfcmt extends outof the pad for lO=60d of used rfcmt.

Example of bursting due to axial overload and…Which detailing rule is broken? (Eurocode)

. . . . . . . . . . . . . . . . .

Buckling failureBursting failure(Perhaps combined a littlebit with buckling.)

Bursting failure in the foot of a column, probebily due to earthquake

Detail of bursting

As a protection against bursting additional longitudinal bars conected by clips in the middle of the column‘s sides should be lines. See previous figures.

E

Stirrups position in a column according to ACI 11 line

COLUMN’S AND BEAM’S INTERSECTIONS and JUNCTIONS

(JOINTS)

intersection junction

E

Kinds of intersections (joints)

Internal intersectionKnee jointMarginal jointColumn – pad junction

Internal intersection or connection

of two or more members, we can differ:

1. Internal joint without change of column’s cross-section and RF, see right.

2. Internal joint with a marginal change of column’s cross-section.

3. Internal joint with an important change of column’s cross-section.

Ad 1) Usually stirrups go through the intersection in a column, by cross-beam not.Ad 2) Stirrups are closer for0,6 lo in the lapping length lo (Eurocode) + in the space bellow the intersection (ACI, only)

Internal joint with a marginal change of column’s cross-section.

Marginal change = slope less than 1 : 6 (1 : 7, 1 : 8 etc.)

Additional stirrups

E

1

6

b

dInternal intersection or

connection of two or more members

Internal joint without change of column’s cross-section and RF, see right.

Internal joint without change of column’s cross-section and change of RF.

Internal joint with a marginal change of column’s cross-section.

Internal joint with an important change of column’s cross-section.

EP

Knee joint – core analysis

“O“ For opening M

“C“ For closing M

E“O“ “C“

By both kinds of joint stressing we are looking for principal stresses in the core. Direction of tensile stress determines reinforcement position of the joint.Basic rule:Reinforcement should be placed in the direction of principal tension or perpendicularly to the cracks (possible cracks).

Knee joint – centreline has a significant change in direction It is area of discontinuity

E

In the Eurocode are given pre-solved examples of knee joints with proper truss-model, solved solution of truss-model output in form of forces in assumed reinforcement + detailing recom..

Knee joint – centreline has important change in direction area of discontinuity

E

You have to memorize the simple solution for the case with moderate opening M (previous slide) and model with closing M (this slide), only.

IF

FRAME EXTERNAL JOINT (COLUMN x CROSS - BEAM JUNCTION

Alternative shape

EIn the Eurocode it is not taken this case as area of discontinuity, so for the detailing is presumed tension in upper fibres and compression in the bottom ones.

The reinforcement should have the shape as shown.Besides sufficient anchoring space for of the column concreting should be kept.

The reinforcement should be anchored in the connected column. The anchoring length is at least partially in the area of a poor bond, so for the calculation of the ultimate bond stress fbd hold true:

fbd=2.25 ⋅ η1 ⋅ η2 ⋅ fctd

The coefficient η1 which expresses the quality of the bond have to betaken : η1= 0,7.

Knee joint – core analysis

For opening M

For closing M

TT-EP

What is enough acc. to Eurocode (bottom) is not correct accordingto API (right).

FRAME EXTERNAL JOINT (COLUMN x CROSS - BEAM JUNCTION (alternative detailing)

Min. radius is respected

To leave column open for concreting

EP

TT

TT

Junction of a column and a footing pad

From the static point of View

According to technical execution

Fixed connection (99% of cases) With notched

hingeHinged connection

With rolled hinge

Column fixed in a footing padIF

The bent bar should be long enough to cross at least two perpendicular bars.

In this position should be one stabilizing bar to protect the bent bar against overturning.

Consists ofcolumn reinforcement and pad reinforcement in this lecture Will be discussed the reinforcement of the column, only.

Column – pad hinge connectionE

H/(0,5Fd) = 0,25(h-a)/0,5h

H = 0,25 (h-a) Fd

For safetyH = 0,25 (h-a) Fd

Is used

EAxial stress is concentrated in the notch and leads to transversal stress. The magnitude of Horizontal –Transversal force can be deduced from the force and geometrical similarity:1. Step: axial force Fd is divided into two halves, then we can derive:

For horizontal force H, additional stirrups should be designed. Their stress can be presumed as pure (transversal) tension. The distribution of stirrups alongside column foot is on the following slide.

E

IF

IF

I-TT

EP

END OF THE PART 2 AND FRAMES, TOO.

https://www.edisk.cz/stahni/67518/ev3_and_ev4.zip_31.4MB.html/

Two educative videos download from the link below: