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Unformatted text preview: 2.3.3 Limitation of neutral axis depth x z = la d Fst εs • The relationship among the depth of neutral axis and the maximum
concrete strain and steel strain has been given by Eq. (2.1) ε s = ε cu d−x
x The depth of the neutral axis is then x= d
1 + ε s / ε cu (2.27) The ultimate strain of concrete εcu = 0.0035; and for the high-yield
steel (fy = 460 N/mm2), the yield strain εy = 0.00219. Substituting
these values into Eq. (2.27) gives x= d
= 0.615 d
0.0035 (2.28) Hence, to ensure yielding of the tension steel at the ultimate limit
state: x ≤ 0.615 d 53 • At the ultimate limit state, it is important that
o sections in flexure should be ductile.
o failure should occur with the gradual yielding of the tension steel reinforcement and not by a sudden catastrophic
compression failure of the concrete.
• BS 8110 limits x ≤ 0.5d
o To ensure yielding of the tension steel at the ultimate limit state.
o To prevent a brittle failure of the cross-section. 2.3.4 Types of failure εc = εcu εc = εcu εc = εcu εs = εy εs > εy εs < εy Type 1 Type 2 Type 3 Fig. 2.3-3 Strain distributions of sections in different types of failure • Sections in different failure types
ο Type 1: Balanced section (x = xb) εc = εcu , εs = εy
ο Type 2: Under-reinforced section (x < xb) εc = εcu , εs > εy
ο Type 3: Over-reinforced section (x > xb) εc = εcu , εs < εy
54 • Failure behaviour
ο Balanced section Tension steel yielding (εs = εy) and concrete crushing (εc = εcu)
at failure at the same time
Crashing of concrete Yielding of steel reinforcement ο Under-reinforced section Steel strain reaches εy before concrete strain reaches εcu . The failure of an under-reinforced beam is characterised
by large steel strains, and hence by extensive cracking of the
concrete and by substantial deflection. The ductility of such a
beam provides ample warning of impending failure; for this
reason, and for economy, designers usually aim at underreinforcement.
ο Over-reinforced section Concrete strain reaches εcu before steel strain reaches εy . 55 The failure of an over-reinforced beam is initiated by the
crushing of the concrete, while the steel strain is still relatively
low. The failure is therefore characterised by a small deflection
and by the absence of extensive cracking in the tension zone.
The failure, often explosive, occurs with little warning.
ο It is not allowed to design a beam with a very low steel ratio. Failure of the beam with a very low reinforcement ratio:
main crack → quickly expand → beam broken (brittle failure) Main crack (critical crack) Reinforcement ratio should be controlled by ρmin .
2.3.5 Moment redistribution (refer to Section 3.6 in the text)
(1) Concept of moment redistribution of a continuous beam made of
an ideally elastic-plastic material
• Stress-strain relationship and moment curvature characteristics
f M fy (a) MP ε 1/r
(b) Fig. 2.3-4 An ideally elastio-plastic material. (a) Stress-strain
relationship; (b) moment curvature characteristics
56 • For a section of the beam subjected to an increasing moment M, the
curvature 1/r at that section increases linearly with M until the value
MP (plastic moment of resistance) is reached; the curvature then
increases indefinitely, as shown in Fig. 2.3-4(b).
• Example 2.3.5-1: Fig. 2.3-5(a) shows a two-span uniform beam made
of an ideal elastic-plastic material, subjected to mid-span loads, P.
The value of the plastic moment of resistance of cross-sections is MP . Fig. 2.3-5 (a) Two-span uniform beam made of an ideal elastic-plastic
material; (b) elastic moments; (c) plastic moments of the uniform beam; (d)
collapse mechanism; (e) plastic moments of the non-uniform beam in
Example 2.3.5-2 (p.60).
57 ο Fig. 2.3-5(b) shows the elastic beading moments. At mid-span: M B = M D = 5
Pl ; at support: M C = Pl
16 ο Suppose P is just large enough for the moment at C to reach MP , i.e. MC = MP. A further increase in P, to P′ say, will not increase
MC . A plastic hinge is said to have developed at C. The beam
behaves as though it is hinged there (section C deforms as a plastic
Under the increased load P’, the moment at B is
MB = 5
Pl + ( P ′ − P ) l
4 This is because the increased loads cause an increase in the
overall moment. Since the moment at C cannot be increased,
more moments will go to the elastic sections (B and D).
As P is increased, both MB and MD increase and will eventually
reach the value MP [see Fig. 2.3-5(c)]. The beam behaves as
though it is also hinged at B and D.
ο The beam will collapse in the mode shown in Fig. 2.3-5(d), where the beam is no longer a structure, but a mechanism (the collapse
ο Let Pu be the value of P at collapse. From Fig. 2.3-5(c), MB = Pu l M C
2 where both MB and MC are now equal to MP ; hence
58 MP = Pu l M P
or M P = Pu l
6 (2.29) MP
l (2.30) Then,
Pu = 6 Therefore, at collapse the moment at section C is
MC = MP = 1
6 If the beam had remained elastic, the bending moment diagram at
collapse would have been that of Fig.2.3-5(b) with Pu substituted
for P, and the (hypothetical) elastic moment at C could have been
( M elastic ) C = 3
16 ο Comparing M C = Pu l / 6 with ( M elastic ) C = 3Pu l / 16 shows that the rotation of the plastic hinge at C has resulted in a moment
redistribution: After the formation of the plastic hinge at C, the bending
moment there is smaller (and that B greater) than what it would
have been if the beam had remained elastic.
The bending moment at C after the moment distribution is
Pu l , but not
Pu l ( Pu l <
Pu l ).
ο Moment redistribution ratio βb (BS 8110: Part 1: Clause 22.214.171.124): βb = moment at a section after redistribu tion
≤ 1 (2.31)
moment at the section before redistribu tion
59 The moment redistribution ratio at section C is βb = Pu l / 6
3Pu l / 16 • Theoretically, the moment of resistance at the support can be made to
any desired value, and the span moment can be calculated to be in
equilibrium with the support moment and external loads.
The two-span beam shown in Fig. 2.3-5(a) is now a non-uniform beam
made of an ideal elastic-plastic material.
ο The moment of resistance (the plastic moment of resistance) at C is mPC . Suppose, by design, the value of m PC = 0.5M P , where MP is
the plastic moment of resistance of the cross-sections in Example
2.3.5-1 (p. 57).
ο Suppose the beam is designed to have the same load-carrying capacity as that in Example 2.3.5-1, i.e. Pu = 6 MP
. The value of
l moment resistance at B can be calculated from the equilibrium
between internal and external forces,
MB = Pu l M C
2 Notice that MC = mPC = 0.5MP and Pu = 6 M P / l . Then, the corresponding plastic moments of resistance at cross-sections B
and D are determined, as shown in Fig. 2.3-5(e),
60 mPB = mPD = 1.25M P
ο Thus, by considering moment redistributions the designer has a choice – the loads Pu may be resisted by the beam with plastic
moments of resistance 0.5MP at C and 1.25MP at B and D,
(Compared to Example 2.3.5-1, where the loads Pu are resisted by
the beam with plastic moment of resistance MP at C, B and D.)
ο The moment redistribution at C is βb = where 0.5M P = P l / 12
0. 5 M P
3Pu l / 16 3Pu l / 16 3
is the actual ultimate moment at C, and
12 is the elastic moment.
ο Therefore, different combinations of MC and MB can be made, provided that rotation of a plastic hinge does not cause failure of
ο In general, structural design which takes account of moment redistributions is called plastic design or limit design.
(The term ‘limit design’ should not be confused with limit state
design, because limit design refers only to the design that takes
account of moment redistribution.) 61 ...
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This note was uploaded on 12/24/2011 for the course CIVL 232 taught by Professor Jskuang during the Spring '06 term at HKUST.
- Spring '06