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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


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)


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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.


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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


Gross sectional area of the

member Ag Material property

Compressive strength of fc’ Material property


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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 management




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.


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Risk analysis for shear loads



Base of concrete section b Material property


Span of concrete section l 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 management





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Risk analysis for torsional loads


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


base of concrete section b 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


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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𝑕

1.7𝐴𝑜𝑕2

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𝑕

1.7𝐴𝑜𝑕2 ≤ 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


ψ5 = 50 for customary 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−𝐴𝑡𝑝𝑕𝑠

ψ6 = 1 for customary units

ψ6 = 12 for SI units


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Risk Assessment

Hazard Identification Shear loads

Hazard Quantification (Load) L = V

Section response (Capacity) 𝐶 = 𝑉𝑐 +𝐴𝑠𝑓𝑦𝑑

𝜙𝑠






Risk management




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.


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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.


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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)*)



1. Underestimating or miscalculation of cover to reinforcement

2. Underestimating of environmental conditions


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


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


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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.

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Documents

steel area material

behaviorresistance of

certain material

material degradation

section compression

environmental loads

miscalculation of loads

given load