Beton Subiecte Examen

8/11/2019 Beton Subiecte Examen

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1. What shape has the computed setion of the slab

Rectangular shape

2. What are the minimum and maximum no. of bars in a slab section?

-

at least 5 bars/meter for slabs with mm300hp

-

at leats 4 bars/meter for slabs with mm400h300 p

-

at least 3 bars/meter for mm400hp

-

maximum 12 bars/meter

3. The slab can be with resistant reinforcement on one or both directions?

- Both direction

a)

Oneway spanning slabs )0,2l( oyox

The main reinforcement is placed parallel with the small span of the slab. Perpendicular is

placed constructive reinforcement,

b)

Two way spanning slabs

The reinforcement is placed respecting on every direction the provisions

4. What type of reinforcing can be used for slab?

Wired mesh:

a)

One way spanning slabs )0,2l( oyox -The main reinforcement is placed parallel with the

small span of the slab. Perpendicular is placed constructive reinforcement. For taking over the

local bending moments on support from the big span, there are placed riders, that are m/65 (PC 52, PC 60) or m/85 (OB 37); they are prolonged outer the support with ox25,0 .

b) Twoway spanning slabs-The reinforcement is placed respecting on every direction the

provisions.

Slab reinforcing with welded meshes

a) One-way spanning slab b) Two-way spanning slab


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5. What members of a structure are subjected to bending and shear force?

Beams girders

6. What types of sections the beams have ?

-

T sections-

Rectangular sections

7.

When the rectangular section of a beam is singly reinforced?

If lim sd the section is singly reinforced, on contrary being necessary the double reinforcing.

ck

u

sdfbd

M2

lim is function of lim where

ydd

x

0035.0

0035.0limlim and is given in

Table in Guides for designing.

8. Where are placed the reinforcements in case of doubly reinforced section?

9. What represent X ?


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X= the compressed zone;

10. In which stage of working the design of steel is made ?

11.

What types of stress(produced by what?) are taking over the longitudinal reinforcement in a

beam ?

The bending moment produces tension of the beam and the stress is taken by the long reinf.

12. In case of T section when the neutral axis passes through the flange?

If b/bw5 the neutral axis passes through the flange,xhf, or if the condition is satisfied:

d

h

d

h

b

b

fdb

M ff

wcdw

Eds 5.012

13.How is determined the width of the flange (beff) ?

Detrminarea limii efective a plcii n calculul la toate strile limit se face pe baza distanei l 0 dintre

punctele de efort nul din diagrama de momente ncovoietoare.

Limea efectiv rezult din relaia:

bbbbb weffeffeff 2,1

Unde:

b

bieff

eff bbb

bb

b 21

,

min

Unde : 002,12,1 2,01.02.0 llbbeff

14.The T section can be doubly reinforced?

Yes, when lim1 s .


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15.What type of reinforcement can be provided for shear ?

For shear we use stirrups for reinforcement and bend up bars.

16.Types of shear failure?

shear bond failure, dowel failure, shear compression failure, diagonal tension failure, flexural failure.

P P

45

dowel action

Shear bond failure Dowel failure

P P

crushing concrete

P P

Shear compression failure Shear compression failure

P P

Flexural failure

17.Mechanism of shear transfer? For member without shear reinforcement? For

members with shear reinforcement(stirrups)?

Consider a typical member inn bending and shear, which is reinforced with longitudinal

steel against bending. Under its applied loads, the member will crack in one of the characteristic

manners, such as that illustrated in Fig. 5.24

The shear force is transmitted through the cracked member by a combination of three

mechanisms. The first is dowel action of the reinforcement (ordinary reinforcements). This result

from the resistance of the reinforcement to local bending as illustrated in Fig. 5.25a and the

resistance of the concrete to localized crushing near the reinforcement. The second mechanism

by which cracked members resist shear is known as aggregate interlock and results from the

forces transmitted across the crack by interlocking pieces of aggregate Fig. 5.25b. The third

mechanism by which cracked members resist shear is through the resistance of the concrete in

the uncracked portion of the beam where the axial stress is compressive. The free body diagram


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for the portion of the member between the left-hand support and the first inclined crack is

illustrated in Fig. 5.26 the total capacity to resist shear force, Q is the result of the combined

effects of the three mechanisms illustrated inn that figure.

It has been found that dowel action is generally first to reach its capacity, followed by

failure of the aggregate interlock mechanism followed by shear failure of the concrete in the

compression zone. However, the precise proportion of the total shear force carried by each

mechanism is difficult to establish and, consequently, the shear strength of the concrete is most

often represented by a single expression which accounts for all three mechanisms acting

together.

Shear reinforcement, like flexural reinforcement, does not prevent cracks from

forming in a member. Its purpose is to ensure that the member will not undergo shear failure

before the full bending capacity of the member is reached. When inclined cracks form in a

member with shear reinforcement, the bars which cross the cracks contribute to the shear

resistance of the member, as illustrated in Fig.11.28. The total shear capacity provided by the

section can be considered as a combination of the capacity of the reinforcement and that of the

concrete as illustrated in Fig. 11.28.

Fig.11.28. Shear resistance of member with stirrups:

(a)elevation showing; (b) reinforcement; (c) transfer of shear force through member

If the applied shear force is sufficiently large, the shear reinforcement will reach its yield

strength. Beyond this point, the reinforcement behaves plastically and the cracks open more


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rapidly. As the cracks widen, the proportion of the shear resisted by aggregate interlock is

reduced forcing an increase in dowel action and shear stress in the uncracked portion of the

section. Failure finally occurs by dowel splitting or by crushing of the concrete in the

compression zone.

a)

b)

c)

Fig....Mechanism of shear transfer:(a)dowel action;(b)aggregate interlock;(c)shear stresses in

uncracked concrete

18.What method is used in order to quantify the behavior and strength of members

with shear reinforcement?

In order to quantify the behaviour and strength of members with shear reinforcement, an

equivalent truss model is frequently used to represent the behaviour of members in shear.

When subjected to combined moment and shear, a member with shear reinforcement,

such as the illustrated in Fig.11.29a, develops inclined cracks in the same way as a member

without shear reinforcement. For the cracked member illustrated, compressive stress develops in

the concrete at the top of the section and tensile stress develops in the longitudinal reinforcement

at the bottom of the section as illustrated in Fig.11.29b in addition, compressive forces are

developed along inclined paths between the cracks and tensile forces are developed in each of


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the stirrups crossing the cracks. All of these internal forces are illustrated in Fig.11.29b, and

together they are similar in pattern to the forces generated in a truss such as that illustrated in

Fig.11.29c.

Fig.11.29. Truss model for reinforced beam:

(a)cracking in reinforced beam; (b) zones of compression between cracks;

(c) truss model.

19.

How can be obtained the lengths of zones on which the distance among stirrups iscomputed?

20.How many types of torsion do you know?Equilibrium torsionis that which is required to maintain equilibrium of the member. In such situation,

the external load has no other option but to be carried by torsion. Equilibrium torsion is of primary

interest in design because failure of the member is inevitable if it has insufficient torsional strength.


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Compatibility torsionoccurs in structures having rigidly connected members. It results from the

compatibility of deformations of members meeting at a joint. Owing to the monolithic nature of their

construction, most members in concrete structures undergo a certain degree of compatibility

torsion.The compatibility torsion can be ignored if sufficient steel is provided to ensure ductile behavior

and the spacing of stirrups is sufficiently small to control cracking at the serviceability limit state.

21.How can be increased the torsion strength?

The torsional strength of a concrete member can be significantly increased by providing suitable torsion

reinforcement across the cracks. This is usually provided in the form of closed four-sided stirrups,in

combination with longitudinal bars distributed around the periphery of the section. This reinforcement

controls the propagation of cracks and ensures that when failure occurs due to yielding of the

reinforcement, it is not sudden.

22. In which stage of working the limit state of cracking is verified?

Design to shear force

23.The cracks widths are limited by norms?

24.

Depending on the force eccentricity the failure of columns is how ?

For eccentrically compression, the dominant loading influences the behavior of the members. Forinstance the limit shortening of compressed concrete for an axial force is 0,2%, but for bending it is

0,35%. For eccentrically compression, the deformations will be smaller if the bending will be secondary

load reported to the axial force, or will be near the maximum deformation (0,35%) if bending is

principal. The deformation of the tension steel will be in plastic domain, if bending moment is dominant

and the steel percentage does not exceed the maximum value or they can be under the yielding level, if

bending moment is secondary.


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If the failure takes place with plastic deformation in the tension reinforcement, followed by the

crushing of compressed concrete one considers that is the 1stcase of eccentrically compression.

If the reinforcement has only compression stresses or, being tensioned it is elastic domain, the

failure takes place by the crushing of compressed concrete that is the IInd

case.

25.The columns are subjected to ?

The columns are subjected to compression and bending.

26.The columns are reinforced with what type of reinformcement?

(1) - longitudinal rods and spiral hooping;

(2)longitudinal rods and lateral ties;

(3)structural steel.

27. What is slenderness ratio?

Is function of the parameters which determine the lateral deflection of the column. The slenderness

ratio is defined by:

r

lfo

fl - is the effective length of the member

ris the radius of gyration

28.Depending on the slenderness ratio how the columns are classified?

-

for 10 (short columns), the second-order moments can be ignored

-

for 3010 (slender columns), it is necessary to consider the second-order moments-

for 30 (very slender columns) the failure of columns, under an continuous increasing ofthe axial force will take place by loosing the stability (buckling).

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