2.1 General View. Two philosophies of design for reinforced concrete have long been prevalent. ... ACI 318-99: Limit State at Failure Approach was introduced.
REINFORCED CONCRETE DESIGN
Dr. AHMED AJEL
Design Methods of Reinforced Concrete Structure 2.1 General View Two philosophies of design for reinforced concrete have long been prevalent. Working stress Design (WSD) was the principle method used from the early 1900s until the early 1960s. after the 1963 edition of the ACI code, there was a rapid transition to Ultimate Strength Design , largely because of its more rational approach. Ultimate strength design , referred to in the Code as the strength Design Method (SDM) is conceptually more realistic in its approach to structural safety and reliability at the strength limit state. Working Stress Design Method Stresses are computed in both the concrete and steel using principles of mechanics that include consideration of composite behavior Actual Stresses < Allowable Stresses Strength Design Method The Strength of members is computed at ultimate capacity Load Factors are applied to the loads Internal forces are computed from the factored loads Required Strength < Actual Strength
aci 318
Early 1900s: WSD was mainly used. ACI 318-56: USD was first introduced. ACI 318-63: Treated WSD and USD on equal basis. ACI 318-71: Based entirely on strength approach (USD). WSD was small part called Alternate Design Method (ADM). ACI 318-77: ADM moved to Appendix A. USD was called Strength Design Method. ACI 318-83: ADM moved to Appendix B. ACI 318-89: ADM back to Appendix A. ACI 318-99: Limit State at Failure Approach was introduced ACI 318-02: Change load factor to 1.2DL + 1.6LL.
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Dr. AHMED AJEL
2.2 Elastic Flexural Behavior of Beams Elastic design is considered valid for the homogeneous plain concrete beam as long as the tensile stress does not exceed the modulus of rupture fr . elastic design can also be applied to reinforced concrete beam using the working stress design method (WSD) approach. The beam is a structural member used to support the internal moments and shears. It would be called a beam-column if a compressive force existed.
C=T M = C*(jd) = T*(jd)
The stress in the block is defined as: σ = (M*y) / I 2
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REINFORCED CONCRETE DESIGN
Dr. AHMED AJEL
2.3 Flexure Stress Stages of Beams By consider any reinforced concrete beam carry an incrementally accumulative increase load, the beam will pass through sex stress stages which are:
Stage 1: Elastic Uncracked Stage: The applied load on beam less than the load which cause cracking. Stage 2: Elastic Cracked- threshold Stage: The applied load make the bottom fiber stress equal to modulus of rupture of concrete. Stage 3: Elastic Cracked Stage : The external applied load cause the bottom fibers to equal to modulus of rupture of the concrete. Entire concrete section was effective, steel bar at tension side has same strain as surrounding concrete. Stage 4: Inelastic Cracking Stage : The tensile strength of the concrete exceeds the rupture fr and cracks develop. The neutral axis shifts upward and cracks extend to neutral axis. Concrete loses tensile strength and steel starts working effectively and resists the entire tensile load. Stage 5: Ultimate Strength Stage: The reinforcement yields. Stage 6: Failure Stage : The material stresses will be exceed its corresponding capacity .
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Dr. AHMED AJEL
2.4 Development of Flexural Equations This is important not only as acceptable alternative ACI design method , but also for the later evaluation of crack width under service loads. Under the action of transverse loads on a beam strains, normal stresses and internal forces developed on a cross section are as shown in Figure 2.1. Stage 1: Before Cracking (Uneconomical). Stage 3: After Cracking(Service Stage). Stage 6: Ultimate (Failure).
o Uncracked Section
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REINFORCED CONCRETE DESIGN
Dr. AHMED AJEL
Assuming perfect bond between steel and concrete, we have εs= εc ⇒
⇒
⇒
Where n is the modular ratio
Tensile force in Steel Replace steel by an equivalent area of concrete,
Homogenous section & under bending ⇒
o Section Cracked, Stresses Elastic ́ and If we will assume that the crack goes all the way to the N.A and will use the transformed section,
To locate N.A. , tension force = compressive force (by def. NA) (Note, for linear stress distribution and with ∑ ⇒∫ , thus ∫ neutral axis).
and ∫
̅
, by definition, gives the location of the
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Dr. AHMED AJEL
Note, N.A. location depends only on geometry & n (
)
Tensile and compressive forces are equal to & and neutral axis is determined by equating the moment of the tension area to the moment of compression area: ( )( ) ( ) 2nd degree equation ⇒ ⇒ Where
(
).
2.5 Alternative Design Method Alternative Method of Design for structures and their elements is the same as for Limit state Method except for safety considerations. To be more specific, Codal provision for durability, fire resistance, detailing of reinforcement, serviceability, stability, etc remain the same as for Limit Stress Method. However, the design loads are characteristic loads and the stresses developed due to loads shall be within the corresponding permissible stresses (i.e. characteristic strength divided by Factor of Safety).
The strength design method can be attributed to several factors including: The uniform treatment of all types of loads, i.e. all load factors are equel to unity. The different variability of different types of loads (dead and live load) is not acknowledged. The unknown factor of safety against failure (F.S. is not known explicitly) . The typically more conservative designs, which generally require more reinforcement or larger member sizes for the same design moments when compared to the strength design method.
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Dr. AHMED AJEL
2.6 BASIC ASSUMPTIONS AND PERMISSIBLE STRESSES Basic Assumptions Basic assumptions for design applicable to flexural and compression members are as follows: (a) Plane section before bending remains plane after bending. (b) The tensile stress of concrete is neglected unless otherwise mentioned. (c) The strain-stress relation for concrete as well as for steel reinforcement is linear. (d) Perfect bond between steel and concrete. Loading Stages – Uncracked section _Cracked section Permissible Stresses Load factors for all types of loads are taken to be unity for this design method. Permissible stresses are defined as characteristic strength divided by factor of safety. The factor of safety is not unique values either for concrete or for steel; therefore, the permissible stresses at service load must not exceed the following :
Flexure extreme fiber stress in compression: 0.45 f’c. Tensile stress in reinforcement Grade 40 and 50 (fy=280 or 300 MPa) : 138 MPa (20,000 psi). Grade 60 (fy=420) : 165.5 MPa (24,000 psi). Modulus of elasticity for steel reinforcement 200 GPa (29,000 psi). Modular Ratio n = Es /Ec . Transformed Section : Substitute steel area with (nAs) of fictitious concrete.
Location of neutral axis depends on whether we are analyzing or designing a section. REVIEW (ANALYSIS): We seek to locate the N.A by taking the first moments:
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Design : Objective is to have & preset & determine As , and we thus seek the optimal value of k in such a way that concrete and steel reach their respective limits simultaneously.
Balanced design in terms of : What is the value of such that the steel and concrete will both reach their maximum allowable stress values simultaneously. Governing equations:
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Dr. AHMED AJEL
Review (Analysis): Start by determining
,
If
steel reaches max. allowable value before concrete, and
If
concrete reaches max. allowable value before steel , and
Or
Where √
(
)
Design : We define
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REINFORCED CONCRETE DESIGN
Dr. AHMED AJEL
Example1: The Fig. below show a section in reinforced concrete beam of width b = 300 mm and effective depth d = 500 mm , area of steel reinforcement is As = 1500 mm2 , modular ratio is n = 8.Compute the stress in the steel and concrete if the applied bending moment M= 70 kN.
511mm
311mm
Solution: Solution by (Internal Couple Method): √
(
) √
(
)
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REINFORCED CONCRETE DESIGN
Dr. AHMED AJEL
(
)
Solution by (Transformed Section Method):as previous method we find k then compute I , (
(
)
)
(
)
Example2: for the same beam in example1, b = 250 mm and effective depth d = 400 mm , area of steel reinforcement is As = 1000 mm2 , modular ratio is n = 8.Compute the maximum service moment can be applied on the beam , if the allowable stress in the steel and concrete was , . Solution: Solution by (Internal Couple Method):
As in previous example
If assumed that the concrete stress reaches the allowable value.
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Dr. AHMED AJEL
If assumed that the steel stress reaches the allowable value.
The applied moment Solution by (Transformed Section Method): From above , (
)
If assumed that the concrete stress reaches the allowable value.
If assumed that the steel stress reaches the allowable value. (
)
(
)
M
Example3: Determine the moment capacity of the section if `
584 mm
650 mm
Solution:
15`00 mm2
250 mm
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√
(
Dr. AHMED AJEL
) √
(
)
(
)(
(
)
)
⇒ Steel reaches elastic
limit.
Note , had we used the alternate equation for moment (wrong) we would have overestimated the design moment:
Example4: Design a beam to carry LL=27.72 kN/m , DL = 14.59 kN/m, L= 9.75 m. Solution:
́
√
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(
Dr. AHMED AJEL
)
(
)(
)
Estimate beam weight at 7.3 kN/m (
(
))
Take b=450 mm & d= 803 mm and h=920 mm Check beam weight =0.92 x 0.45x24.0=9.94….o.k
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Example5: For the simply supported beam shown reinforced by 4ϕ 25 bars (fy =420 MPa ), the concrete strength ( ́ ) evaluate the following : 1. If the span of beam =4 m and dead load = 8 kN/m , live load =10 kN/m check the actual flexural stress in concrete and steel . 2. The length of beam span that make the concrete in tension face start to crack. 3. The actual stress in concrete and steel if the span of beam =7m .
Solution: 1. Total load W= wd +wl =8 +10 = 18 k N/m M= n= As =4
√
Ab = 4 491 =1964 mm2
Assume Transformed section area At = (Ac-As) + n As = Ac + (n-1)As=500
+(9.28 -1)1964
=191261.92 mm2
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Dr. AHMED AJEL
(
) (
(
(
)
)
)
(
)
( (
(
)
Concrete allowable compression stress:
́
Actual bending stress (
)
)
)
For Compression Fiber: Actual compression concrete stress :
For Tension Fiber: (
Actual tension concrete stress :
)
√́
The concrete stress that make initial crack: √
Actual steel stress :
(
)
Steel allowable stress: 16
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REINFORCED CONCRETE DESIGN
Dr. AHMED AJEL
2. To make the concrete start to crack put the concrete tension stress at the extreme fiber equal to concrete stress at rupture ( ) √ ( (
)
)
⇒
Any moments greater than (49.12 kN.m) will cause concrete cracks. For maximum beam span length that make the concrete section cracked:
⇒ 3.
Since the moments The concrete section cracked. √
(
) √
(
(
)
)
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Concrete allowable compression stress:
Dr. AHMED AJEL
́
⇒
Steel allowable stress:
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Dr. AHMED AJEL
Assignment (1) Q1 : For the rectangular beam shown in fig. find the following : fy =420 MPa ́ . 1- Bending stresses at the top, bottom and at reinforcement when M=35 kN.m . 2- Bending moment to make initial crack. 3- The bending stresses when the applied bending moment equal to 80 kN.
Q2: Calculate the bending stresses in concrete and steel if fy =345 MPa and the concrete strength ́ , n=8 .
Q3: Evaluate the steel area required for the a simply supported beam with the section shown below. use bars with fy =420 MPa and concrete compressive strength strength ́ , the beam span = 5m and carry loads wd=7 kN/m ,wl=10 kN/m . use transformed section method and neglect beam self-weight.
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