Design of frames - structural analysisINITIAL STEPS:1. Prepare of a scheme of the structure, deducing major

dimensions of the building, type of the support structure (walls frames, columns + bracing), chosen materials (basic character). Task of an architect.

2. Finding dimensions of structural member’s cross-section’s (CS) simple, not obligatory formulas are used (e.g. hs as a function of span and character of the support) = PRELIMINARY DESIGN Task of a C.I. (formulas for frames members will be given later + in the practice of Bl005).

3. Loading analysis, esp. determination of all important loads and loads combinations.

Notice: Concerning points 1. and 3. above: The majority of needed information was given in the previous lectures and BL001 subject.

STRUCTURAL ANALYSIS:

It means calculation of internal forces with knowledges of structural mechanics science (other subject). We will learn one or two of simplified procedures for calculation of internal forces suitable for MBMS frames in this subject.

DESIGN ITSELF:

1. Virtual disassemble of designed complex structure into separate members.

2. Finding of the most stressed CSs in these members for all important actions and relevant internal forces. (M+N, M, V, T ).

3. Design of reinforcement for combined or separate action of M+N, M, V, T in these most stressed CSs + special cases. Design at this level= calculation of Asd

based on static equilibrium of forces (minimal sectional area of designed rfcmt.) and applying of detailing‘s rules.

4. Assemblage of separate reinforcement designed in point 3. into the complex reinforcing of actually designed complex structure.

5. Finally making Drawings – the most important output of design for building place.

Procedures for structural analysis of MBMS structure. Which one is more accurate and which one is more useful? Let us judge:

At first we have to keep in mind, the MBMS frames are staticallyindeterminate (hyper-static) structures. It means the staticequilibrium equations are insufficient for determining the internalforces and reactions on that structure. Itis in the time a little bitcomplication for structural analysis, but it is not an obstacle by usageof a Computer.Structural analysis of MSMB frames is usually based on usage ofComputer and FEM (Finite Element Method) software. After fillingthe computer with relevant data, the disigner have to decide whichkind of structural behaviour he will assume: plastic-elastic or justthe elastic behaviour of the structure. (Notice: the most of todaysoftware allows to switch between these two possibilities).The plastic – elastic assumption is very close to real behaviour of the structure, but it is conected with several problems: Follow theslides . . . . . .

1. Internal forces distribution in structures is commonly isinfluenced by loading, geometry and stiffness of CS investigated.By structural analysis we have to find solution where the internaland external forces are in equilibrium (in CSs for design). Easy?No! The force distribution depends on the stiffness of importantCS. By RC structures depends the stiffness of these CS inverse onthe internal forces distribution. E.g. on cracks presence (orabsence) or degree of cracking. You will understand more whenwe learn about cracking of RC CS… In any case, such problemmust be solved with iterative techniques and the convergence ofsuch computing is not guaranteed.

2. By this technique (in case of correct structure analysis) is thecarrying capacity of related structure fully utilized. By primarybeams can occur large deflections due to creation and presenceof so called „plastic hinges“. Will be shown in few moments.

3. The structure (its partial members) will satisfy according to ULSbut probably not according to SLS (deflection, crack width).

Acc. to theory of elasticity is the CS‘s stiffness expressed:B = E. J

Where E is modulus of elasticity of used concrete class.J is moment of inertia of relevant CS.

By assumption of elastic behaviour of a structure, is the stiffness B(i)

of all frame CSs taken as a constant. More accurate: depends on the

multiple „E . J“ only, not on the load.

This concept (linear elastic computation + set of correction

coefficients) is proven as safe by tens of years. Therefore is allowed

by the majority of design codes, including EUORCODE 0 and 2. The

mentioned/possible corrections usually concern redistribution of

bending moments around the frame and reduction of M in

supports.

Case of the Linear solution - frequently used:

More to corrections:

Extreme simple example for redistribution - stress transfer: Simple supp. Beam (SSB) x double fixed beam

Redistribution of internal forces, primarily values of bending moment M.

Reduction of bending moments to a face of the support.

By DFB: if in span acting Med >Mrd

collapse can be prohibited with transfer (redistribution) of excessive part of Med into fixed supports. If the reserve in fixed end DM is greater than redistributed part of Med in span, is collapse prohibited. In the middle of the beam

builds plastic hinge..

By real MSMB frames the excessive parts of M must be redistributed step by x step to the whole structure, as the condition ∑Mi=0 for each joint must be fulfilled. It is very laborious, leads to occurring of plastic hinges over-deflections etc → redistribu�on is not frequently used.

By SSB if Med

overruns MRd the collapse is unavoidable…

|Rmax|= 0,65.1/12.p.L2

REDUCTIONPrinciple of reduction of the bending moment in a primary beam. Reduction to the face of the support. Fig shows case of monolithic connection of beams with support.

With the limit:

B ) The structural analysis (calc. of internal forces) can be done with greater respect to reality, with non – linear and cracked CS.+ no corrections allowed!!. Other flaws: Risk of divergence, over-deflections, creation of plastic hinges.

Fully developed plastic hinge

Max. distance of links Acc. to EN 1992-1-1 ?

As you can see, everything is different → here is an opportunity for simplified techniques

CS

Comparison of assumptions for elastic model solution with assumptions for ULS and SLS design

Design of the support structures - framesNowadays (as was mentioned) in structural analysis in the majority based often on the presumption of an elastic behaviour of the structure. These results are very good balanced with reality due to usage of different and proven correction coefficients. Was discussed. Moreover we can transfer the problem of structural analysis of the whole MSMB frame into problem of much smaller part which is called partial frame, cut-out frame etc.

How to derive a cut-out frame?

We have to analyse structural model of the whole MBMS structure and look for symmetricity, antimetricity, uniformity and repetitions in original structure.

The continuity can be defined as fixed (rotation=0)Or with equality and opposite of rotations – case in practice. Afterwards are solved important cut outs, only.

E.g. from this full frame can be cut out more sub-frames until simple beams crossing. Each of this part can be analysed separately and results assembled backwardsTherefore it is important to define the way of continuity(connection) between them.

Another example of cut out sequence.

Frame model

Cross-Sections (CSs)of MBMS frame membersand their centroids, possible simplifications

The model of the frame is a frame

For structural analysis for MSMB frame is as substitution centroid axis of original new frame model used. The same principle is used by simplified solutions.

There are solved symmetrical 3-bays of x-storeys frames, here. For such structure can be for ver�cal load ↓ presumed zero translations and effective are rotations , only. Moreover certain relations between them (rotations) can be defined and used for simplifying.

Technique used in the BL005 practice

Solved in the practice. Presumption over equality and opposite of rotations will be used. For different loading cases:

Than beams are marked with couple of numerals belonged to relevant end joints.e.g. (6,10)

2. Description of the model= marking of beams or joints it is strongly recommended to mark joints.

For Loading Case 1 Permanent load and addition of LC3 + LC4

For Loading Case 1 Permanent load and addition of LC3 + LC4

For Loading Case 3 and 4 (inverse load distiribution)

Usually a simplified slope and deflection method of solution is used. The simplification is frequently based on these facts:1. By MBMS frames exists important uniformity in

arrangement of structure. It means the properties and loading of similar squares (bays, storeys) are identical or close to be identical. Therefore instead the whole frame can be solved a part of it, only. This part is called cut-out frame or partial frames

2. By symmetricity of the frame can be vertical and lateral loading solved separately with limited deformations effects.

3. Static calculation itself

For lateral loading (wind action) completely different simplification will be presented. This one uses no rotations and translations, It is pure force technique. Contents a set of relative simple steps … will follow.

Used basic quantities and relations:

ki = Ei . Ji /li= beam (flexural (in bending)) stiffness of the beam i of length li (or Li).

Kj = ki = storey stiffness of storey j.

Mjs = Lj . Hj = storey moment of storey j.

Hj = horizontal (lateral) force equal to storey j.

Lateral = horizontal load ak�non →→→

We have to perform Simplified structural analysis on the shown part of MBMS frame.For lateral action of wind.Presumptions:1. All storeys above and below

2. solved one are infinitely stiff.

2. Internal forces are between analysed different member distributed according member‘s stiffness (rigidity).That’s all!Have look at the sequence:

L1

1. We have to start one storey

above and calculate storey moment:

M2S= L2.(H2+H3+H4+H5)

2. The actual storey stiffness:

K2=(k5,9 +k6,10+k7,11+k8,12)

3. The M2S should be

divided in to individual columns according to their stiffness:

e.g. M5,9= M2S . (k5,9/ K2)

etc….

L2

Lt L1

We have divided storey moment M2S in to individual columns, but at each

column must be once more divided in to head and foot of the column. As the storey 2 is the regular (?) storey of solved MBMS frame the moments in

columns will divide 50 : 50 in to heads and foots. See figure above→

1. We have to repeat former

process once more. Storey moment:

M1S= L1.(H1 +H2+H3+H4+H5)

2. The actual storey stiffness:

K1=(k1,5 +k2,6+k3,7+k4,8)

3. The M1S should be

divided in to individual columns according to their stiffness:

e.g. M1,5 = M1S . (k1,5/ K1)

etc….

L1

We have to divide storey moment M1S in to individual columns, and at each M in

column must be once more divided in to head and foot of it.. As the storey 1 is not regular storey (columns foots are fixed into pads) the moments from

columns will be divided in to the ratio 40% to heads and 60% to foots. See

figure above→

It is exhausting, but it is not finished yet. The remaining step is to divide moment difference at each joint to surrounding part off cross beam. For these we have to respect classic mechanical’ sign convince:

Bending moments M are to be plotted to the tensile margin of the beam and the sign is + for bottom and left margin – see figure

By the joint 6 can be defined bending moment difference��� = ��,�� − ��,�

(����+)Then��,� = ���( k6,5 /(k6,5 + k6,7)��,� = ???????

Watch signs!

The end of structural analysis