Transcript
University Of Sharjah
Collage of Higher Education and Scientific Research
M.Sc. in Civil Engineering Program
Assignment № 1 to
Environmental Impact Assessment Analysis
(0401564)
Done By Hussain Osama Qasem
U00028420
Submitted to Prof. Abdullah Shanableh
Hussain Osama Qasem U00028420
1
Q. Analyze the structural design process in terms of applied loads, generated stresses (tension,
compression, shear, flexure and torsion) and material properties available to deal with the
applied loads. Define for each stress type (or combination of stresses) the relationship between
the load and behavior/resistance of material, and then explain the process using risk analysis
and explain how risk is managed in the design process. You can define the risk as the
potential of failure under the load.
Solution:
Loads can be classified (in terms of resistance into the following categories
1. Normal loads to be resisted by reinforcement or framing (gravitational loads, shear,
flexure, torsion, ... etc)
2. Environmental loads to be resisted by reinforcement framing (wind loads and seismic
loads)
In general the loads, the material and the capacity are given into this diagram
Hussain Osama Qasem U00028420
2
The interpretation of this triangle is the following statement on basis of load “for any
given load to be sustained by a certain material. The capacity is the ability of the section of this
material to resist the load”. Also, for any structure we have the ratio of load to capacity (or the
L/C ratio) has to stay less than 1.00 always.
There are many types of load resisting structures such as reinforced and prestressed
concrete structures, steel structures, timber structures and stone structures. Reinforced concrete
design will be taken as an example to analyze its process against risk assessment and
management.
Risk analysis for inelastic bending (flexure) loads
Parameter Value Comment Yield strength of reinforcement fy Material property
Steel Area As Material property
Base of concrete section b Material property
Effective depth of concrete section d Material property
Effective compression depth of the
section A Material property
Span of concrete section l Material property
Compressive strength of concrete fc’ Material property
Young’s modulus of elasticity of
concrete Ec Material property
Moment of inertia of the section I Material Property
The given load (regardless of its type) L Load
General capacity reduction factor ϕ Reduction factor for the whole capacity
(0.9)
Failure Criteria
Criterion Value Comment Tension equals compression in the
section T = C To have equilibrium in the section
Compression strength is reduced C = (0.85) fc’ a b The reduction factor for concrete is 0.85
Load is magnified to the ultimate UL To magnify the loads
The ultimate moment is calculated by
structural analysis from ultimate loads Mu
To determine the maximum moments
applied (depends on type of structure)
Hussain Osama Qasem U00028420
3
The magnified load over reduced
capacity equation
𝑀𝐿
𝑅𝐶=
𝑀𝑢
𝜙𝐴𝑠𝑓𝑦 𝑑 −𝑎2
< 1.00 This number HAS to be less than 1.00
always
Also, deflection is of great concern δ = f (L, l, Ec, I)
The inputs of this function varies a lot
between different cases of members
whether it was loading or material
properties
Risk Assessment
Hazard Identification Loads on slabs or beams
Hazard Quantification (Load) L = M
Section response (Capacity) 𝐶 = 𝐴𝑠𝑓𝑦 𝑑 −𝑎
2
Risk characterization
Probability of failure because one or more of the following
1. Underestimating or miscalculation of loads
2. Miscalculation of capacity
3. Material degradation
4. Excess deflection of the member
Risk management
1. Overestimating loads (magnification of load)
2. Underestimating Material Strength (reduction of capacity)
3. Combination between safety and Economy
Commentary:
Concrete flexural sections are designed based on the fact that the moment is nothing but
tension and compression acting simultaneously but each force dominates an area in the section
where the other force has the opposite are. It has been an axiom that concrete is very week in
tension (about 10% of compressive strength). So, from very long ages, designers have to choose
between one of the following; either making the whole building in pure compression which is
not all the time applicable or to provide some reinforcement (steel here) to take care of the
tension loading in the flexural section.
Hussain Osama Qasem U00028420
4
For steel reinforcement, codes have developed an envelope that the reinforcement has to
fall within its limits. That means steel has a minimum area to guarantee the elastic behavior of
the section under the given flexural loading which is
𝐴𝑠,𝑚𝑖𝑛 =3 𝑓𝑐
′𝑏𝑑
𝑓𝑦 𝐵𝑆.𝑈. 𝑜𝑟 𝐴𝑠,𝑚𝑖𝑛 =
𝑓𝑐′𝑏𝑑
4𝑓𝑦 𝑆𝐼.𝑈.
The same thing for maximum allowed steel area that designers use it to guarantee that the section
will not undergoes a sudden failure mechanism.
The following figure explains some parameters
Figure acquired from Reinforced Concrete Design Book by McCormac and Brown 8th
ed.
Now for columns and compression reinforced concrete members
Risk analysis for compressive loads
Parameter Value Comment Yield strength of
reinforcement fy Material property
Steel Area As Material property
Gross sectional area of the
member Ag Material property
Compressive strength of fc’ Material property
Hussain Osama Qasem U00028420
5
concrete
The given load (regardless of
its type) L Load
General capacity reduction
factor ϕ
For tied reinforcement (0.65)
For Spiral reinforcement (0.75)
Failure Criteria
Criterion Value Comment Compression is resisted by
both concrete and steel C’ = Cc + Cs By statics
The ultimate load is UL Magnify the load
The capacity equation C = ϕ (0.8)[0.85 fc’(Ag – As) + As fy] Reduce the capacity
The magnified load over
reduced capacity equation
𝑀𝐿
𝑅𝐶=
𝑈𝐿
𝜙 0.8 0.85 𝑓𝑐′ 𝐴𝑔 − 𝐴𝑠 + 𝑓𝑦𝐴𝑠
< 1.00 This number HAS to be less
than 1.00 always
Risk Assessment
Hazard Identification Loads on compression members
Hazard Quantification (Load) L = P
Section response (Capacity) C = [0.85 fc’(Ag – As) + As fy]
Risk characterization
Probability of failure because one or more of the following
1. Underestimating or miscalculation of loads
2. Miscalculation of capacity
3. Material degradation
Risk management
1. Overestimating loads (magnification of load)
2. Underestimating Material Strength (reduction of capacity)
3. Combination between safety and Economy
Commentary:
In compressive loads the failure might be because of the previously discussed factors, the
deflection here will not be that important as the deflection is resisted by compressive
characteristics of concrete. So, no risk is there regarding the deflection of columns unlike the
deflection of beams where tension in the beam plays an important role in the falure of the beam
because of deflection.
Hussain Osama Qasem U00028420
6
Risk analysis for shear loads
Parameter Value Comment Yield strength of reinforcement fy Material property
Steel Area As Material property
Base of concrete section b Material property
Effective depth of concrete section d Material property
Span of concrete section l Material property
Compressive strength of concrete fc’ Material property
Shear strength of concrete Vc Material Property
Spacing between stirrups s Material Property
The given load (regardless of its type) L Load
The ultimate shear applied Vu Translation from loads into shear
General capacity reduction factor ϕ Reduction factor for the whole capacity
(0.75)
Failure Criteria
Criterion Value Comment The maximum load is calculated UL Magnify the load
The Maximum moment is calculated Vu = 0.5 UL l To determine the maximum moments
applied (depends on type of structure)
Stirrups are needed if the ultimate shear is
greater than half of the reduced capacity Vu > 0.5 ϕ Vc
Since stirrups are repeated two times
per section
Shear is resisted by both steel and
concrete ϕ (Vs + Vc) = Vu By statics
After selecting desired steel diameter,
spacing is calculated 𝑠 =𝐴𝑠𝑓𝑦𝑑
𝑉𝑠
Calculate Vs from the given equation
in the previous line
The magnified load over reduced capacity
equation
𝑀𝐿
𝑅𝐶=
𝑉𝑢
𝜙 𝑉𝑐 +𝐴𝑠𝑓𝑦𝑑𝜙𝑠
< 1.00 This number HAS to be less than 1.00
always
Risk Assessment
Hazard Identification Shear loads
Hazard Quantification (Load) L = V
Section response (Capacity) 𝐶 = 𝑉𝑐 +𝐴𝑠𝑓𝑦𝑑
𝜙𝑠
Risk characterization
Probability of failure because one or more of the following
1. Underestimating or miscalculation of loads
2. Miscalculation of capacity
3. Material degradation
Risk management
1. Overestimating loads (magnification of load)
2. Underestimating Material Strength (reduction of capacity)
3. Combination between safety and Economy
Hussain Osama Qasem U00028420
7
Risk analysis for torsional loads
Parameter Value Comment Yield strength of reinforcement fy Material property
Shear reinforcement Av Material property
Torsional reinforcment At Material property
Area enclosed by the outside
perimeter of concrete cross section Acp Material property
Gross sectional area Ag Material Property
Perimeter of the studied cross
section pcp Material property
Cross sectional area enclosed withen
the outside hoop Aoh Material property
Gross area enclosed by shear path
(0.85 Aoh) Ao Material Property
Perimeter of torsion reinforcing ph Material property
Effective depth of concrete section d Material property
base of concrete section b Material property
Compressive strength of concrete fc’ Material property
Shear strength of concrete Vc Material Property
Cracking torque of concrete Tcr Material property
Spacing between stirrups s Material Property
The given load (regardless of its
type) L Load
Factored axial load Nu Load
The ultimate shear applied Vu Translation from loads into shear
The ultimate torsion applied Tu By direct structural analysis
General capacity reduction factor ϕ Reduction factor for the whole capacity (0.75)
Crack angle θ used to be taken as 45⁰
System of units factor ψn to distinguish between customary units and SI
units
Failure Criteria
Criterion Value Comment The maximum load is calculated UL Magnify the load
The Maximum moment is calculated Vu = 0.5 UL l To determine the maximum
moments applied (depends on
type of structure)
Stirrups are needed if the ultimate
shear is greater than half of the
reduced capacity Vu > 0.5 ϕ Vc
Since stirrups are repeated
two times per section
Shear is resisted by both steel and
concrete ϕ (Vs + Vc) = Vu By statics
After selecting desired steel
diameter, spacing is calculated
𝐴𝑣𝑠
=𝑉𝑠𝑓𝑦𝑑
Calculate Vs from the given
equation in the previous line
The magnified load over reduced
capacity equation
𝑀𝐿
𝑅𝐶=
𝑉𝑢
𝜙 𝑉𝑐 +𝐴𝑠𝑓𝑦𝑑𝜙𝑠
< 1.00 This number HAS to be less
than 1.00 always
This is only for shear
Hussain Osama Qasem U00028420
8
Torsional reinforcement is needed in
normal section 𝑇𝑢 > 𝜙 𝜓1 𝑓𝑐′ 𝐴𝑐𝑝
2
𝑝𝑐𝑝 =
1
4𝑇𝑐𝑟
The ultimate torque applied
exceeds 25% of cracking
torque of concrete
ψ1 = 1 in customary units
ψ1 = 1/12 in SI units
or in members subjected to axial
tension or compression loading 𝑇𝑢 > 𝜙 𝜓1 𝑓𝑐
′ 𝐴𝑐𝑝
2
𝑝𝑐𝑝 1 +
𝜓2𝑁𝑢
𝐴𝑔 𝑓𝑐′ ψ2 = 1/4 for customary units
ψ2 = 3 for SI units
Checking the ability of the Solid
section to take the given toesional
loading
𝑉𝑢𝑏𝑑
2
+ 𝑇𝑢𝑝
1.7𝐴𝑜2
2
≤ 𝜙 𝑉𝑐𝑏𝑑
+ 𝜓3 𝑓𝑐′ ψ3 = 8 for customary units
ψ3 = 2/3 for SI units
Here the magnified load has
to be less than the reduced
capacity for the section
ML ≤ RC Checking the ability of the Hollow
section to take the given toesional
loading 𝑉𝑢𝑏𝑑
+ 𝑇𝑢𝑝
1.7𝐴𝑜2 ≤ 𝜙
𝑉𝑐𝑏𝑑
+ 𝜓3 𝑓𝑐′
Calculate needed torsional
reinforcement
𝐴𝑡𝑠
=𝑇𝑛
𝜙 2 𝐴𝑜𝑓𝑦 cot 𝜃 Calculates steel needed per
unit spacing
In any way, stirrups area must not be
below the given limit 𝐴𝑣 + 2𝐴𝑡 = 𝜓4 𝑓𝑐
′ 𝑏𝑠
𝑓𝑦≥𝜓5𝑏𝑠
𝑓𝑦
ψ4 = 3/4 for customary units
ψ4 = 1/16 for SI units
ψ5 = 50 for customary units
ψ5 = 1/3 for SI units
Also, longitudinal torsional
reinforcing has to be calculated 𝐴𝑙 =𝐴𝑡𝑠𝑝 cot2 𝜃
This area should not fall
below a given limit
Longitudinal torsional reinforcing
must not fall below this limit 𝐴𝑙 ,𝑚𝑖𝑛 =5 𝑓𝑐
′𝐴𝑐𝑝
𝜓6𝑓𝑦−𝐴𝑡𝑝𝑠
ψ6 = 1 for customary units
ψ6 = 12 for SI units
Hussain Osama Qasem U00028420
9
Risk Assessment
Hazard Identification Shear loads
Hazard Quantification (Load) L = V
Section response (Capacity) 𝐶 = 𝑉𝑐 +𝐴𝑠𝑓𝑦𝑑
𝜙𝑠
Risk characterization
Probability of failure because one or more of the following
1. Underestimating or miscalculation of loads
2. Miscalculation of capacity
3. Material degradation
Risk management
1. Overestimating loads (magnification of load)
2. Underestimating Material Strength (reduction of capacity)
3. Combination between safety and Economy
Commentary:
For Shear loads, the failure in the section because of the principal tensile stresses that the
section is to carry. Since the shear loads means a principal tensile stress that is inclined at 45⁰,
so, by default the reinforcement is to be inclined at that angle, but because of difficulties in
framing, codes specify some equations (like the previously stated ones) to overcome this issue.
For torsion loads, designers used to depend on the previous load resisting designs. But
now when safety factors have reduced relatively after 2005, designers shall now have concern of
failure because of torsion. Even though being lengthy, the approach of torsion reinforcement
design is like shear design approach in principal. They both care about principal tensile stresses
that develop in the section because of various loads that are encountered by the structure.
Hussain Osama Qasem U00028420
10
It is important to mention that this model of risk analysis is for reinforced concrete
structures. For other types of structures the equations will be different but the concept will be the
same.
It is also important to mention that all types of loading that any reinforced concrete
structure faces (dead, live, hydraulic, snow, seismic … etc) will be translated into the ultimate
magnified load (UL or U) that will have one of three possible effects on structure (flexure,
compression and shear). The magnification of loads can be done using magnification factors
(based on the used code of design) for the loads as following:
1. U = 1.4 (D + F) ACI equation 9-1
2. U = 1.2 (D + F + T) + 1.6 (L + H) + 0.5 (Lr or S or R) ACI equation 9-2
3. U = 1.2 D + 1.6 (Lr or S or R) + (1.0 L or 0.8 W) ACI equation 9-3
4. U = 1.2 D + 1.6 W + 1.0 L + 0.5 (Lr or S or R) ACI equation 9-4
5. U = 1.2 D + 1.0 E + 1.0 L + 0.2 S ACI equation 9-5
6. U = 0.9 D + 1.6 W + 1.6 H ACI equation 9-6
7. U = 0.9 D + 1.0 E + 1.6 H ACI equation 9-7
Where the used abbreviations are as follows:
1. U: the ultimate (magnified) design load
2. D: dead load
3. F: loads due to weight and pressure of fluids
4. T: total effects of temperature, creep, differential settlement and shrinkage-
compensating concrete
5. L: live load
6. H: loads due to weight and lateral earth pressure of soils, groundwater pressure or
pressure of bulk materials
7. Lr: roof live loads
8. S: snow loads
9. R: rain loads
10. W: wind loads
11. E: earthquake or seismic loads.
These load magnification combination conforms to the IBC code and ASCE code requirements.
Hussain Osama Qasem U00028420
11
Risk analysis for service time of the structure
Parameter Value Comment Water Cement Ratio W/C Material property
Clear cover to reinforcement S Material property
Service life time T Material Property
Environmental Condition Ea Load
Failure Criteria
Criterion Value Comment
Service life time is a function of clear
cover to reinforcement, the severity of
the environmental condition and water
cement ratio (in some cases)
T = f (S, Ea, (W/C)*)
This function is widely variable since
we have three unknowns of which one
of them (Environmental condition) is by
itself a very complicated (unstable)
function
Risk Assessment
Hazard Identification Environmental conditions are unbearable enough by the structure
Hazard Quantification (Load) L = Ea
Section response (Capacity) T = f (S, Ea, (W/C)*)
Risk characterization
Probability of failure because one or more of the following
1. Underestimating or miscalculation of cover to reinforcement
2. Underestimating of environmental conditions
3. Material degradation
Risk management
1. Overestimating of environmental effects
2. Underestimating Material Strength
3. Provide good quality -impermeable concrete
4. Use of suitable admixtures
5. Protection of reinforcement
6. Use of corrosion inhibitors
7. Choice of non-corroding reinforcement
8. Combination between safety and Economy
Commentary:
Environmental impacts that are given here are not the normal loading that the structure is
designed to carry. It is the environmental condition that the structure will face; such as humidity,
high concentrations of hard salts (such as, sulfurs, chlorides or carbonates) that will cause the
concrete to degrade and therefore, on the final run, corrosion of reinforcement that will make the
capacity of the section to be reduced
Hussain Osama Qasem U00028420
12
References
1. McCormac, Jack C. and Brown, Russell H., Design of Reinforced Concrete, eighth
edition, John Wiley and sons. (2009)
2. Building Code Requirements for Structural Concrete (ACI 318M – 08) and Commentary.
American concrete institute (2008)
3. Al-Toubat, Salah. Lecture notes for Reinforced Concrete Design I course. Spring
semester 2010. University of Sharjah. UAE.
4. Al-Samarai, Mufid. Lecture notes for Advanced Concrete Technology course. Fall
semester 2011. University of Sharjah. UAE.
5. Own work experience.
top related