STRUCTURAL DESIGN & CONCEPTUALIZATION
(8 Questions, Difficulty Level: MODERATE)
by Raison John J. Bassig
by Raison John J. Bassig
In the building you are designing, you indicated that the overall size of a reinforced concrete beam must be 350mm wide by 600mm deep due to architectural requirements. You also indicated in your material specifications that the main deformed steel reinforcement bars, which are non-prestressed, should be PNS Grade 275 and that the concrete mix must achieve a compressive strength of 4,500psi.
Shown in the structural plans prepared by your structural engineer, the beam was indicated to be reinforced with 4-Φ25mm main rebars arranged in a single layer at the tension side, surrounded by Φ10mm rectangular closed-loop stirrups and provided with a clear concrete cover of 40mm all-around.
Q#1: What is the effective depth of section of the given beam?
a. 560.0mm
b. 537.5mm
c. 550.0mm
d. 547.5mm
e. 590.0mm
SOLUTION:
Effective Depth of Section (d),
Therefore, the correct answer is b. 537.5mm.
MY EXPLANATION AND DERIVATION OF ANSWER:
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP),
The above provision in the Code can be simply illustrated based on our given beam as,
The tension reinforcements in our given problem are the 4-Φ25mm longitudinal bars arranged in a single layer (the other two rebars found at the top of the beam section are in the compression side of the beam - which may or may not handle compression stresses, depending on the beam design). This means that the centroid of such tension bars would be half of its diameter or ( 25mm / 2 ).
As the Effective Depth of Section (d) is measured from the extreme compression fiber (the fiber OPPOSITE of the TENSION SIDE, in our case, at the top of the midspan of the beam) to the centroid of the tension steel (half of the diameter of the tension bars), we simply subtract the given clear concrete cover [40mm], the given stirrup diameter [10mm], and half of the diameter of the main tension bars ( 25mm / 2 ) from the given [600mm] overall height (H) of the beam,
So, calculating the Effective Depth of Section (d), we have,
As the Effective Depth of Section (d) is calculated to be d = 537.5mm, the remaining depth of the beam which is [600mm - 537.5mm = 62.5mm] would be considered the "Effective Concrete Cover" (not to be confused with clear concrete cover).
Therefore, the correct answer is b. 537.5mm.
Effective Depth of Section (d),
d = 600mm - 40mm - 10mm - ( 25mm / 2 )
d = 537.5mm
d = 537.5mm
Therefore, the correct answer is b. 537.5mm.
MY EXPLANATION AND DERIVATION OF ANSWER:
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP),
"EFFECTIVE DEPTH OF SECTION is a distance measured from extreme compression fiber to centroid of tension reinforcement."
(Sec. 402.3 Terminology of the 2015 NSCP)
(Sec. 402.3 Terminology of the 2015 NSCP)
The above provision in the Code can be simply illustrated based on our given beam as,
The tension reinforcements in our given problem are the 4-Φ25mm longitudinal bars arranged in a single layer (the other two rebars found at the top of the beam section are in the compression side of the beam - which may or may not handle compression stresses, depending on the beam design). This means that the centroid of such tension bars would be half of its diameter or ( 25mm / 2 ).
As the Effective Depth of Section (d) is measured from the extreme compression fiber (the fiber OPPOSITE of the TENSION SIDE, in our case, at the top of the midspan of the beam) to the centroid of the tension steel (half of the diameter of the tension bars), we simply subtract the given clear concrete cover [40mm], the given stirrup diameter [10mm], and half of the diameter of the main tension bars ( 25mm / 2 ) from the given [600mm] overall height (H) of the beam,
So, calculating the Effective Depth of Section (d), we have,
d = [ H ] - [ Clear Concrete Cover ] - [ Stirrup Diameter ] - [ Half of Main Bar Diameter ]
d = [600mm] - [40mm] - [10mm] - ( [25mm] / 2 )
d = [600mm] - [40mm] - [10mm] - (12.5mm)
d = 537.5mm
d = [600mm] - [40mm] - [10mm] - ( [25mm] / 2 )
d = [600mm] - [40mm] - [10mm] - (12.5mm)
d = 537.5mm
As the Effective Depth of Section (d) is calculated to be d = 537.5mm, the remaining depth of the beam which is [600mm - 537.5mm = 62.5mm] would be considered the "Effective Concrete Cover" (not to be confused with clear concrete cover).
Therefore, the correct answer is b. 537.5mm.
Q#2: What is the minimum area of steel reinforcement that must occupy the given beam section?
a. 957.73 sq.mm.
b. 1,050.00 sq.mm.
c. 997.82 sq.mm.
d. 952.67 sq.mm.
e. 992.55 sq.mm.
SOLUTION:
Minimum Area of Steel (As,min),
Therefore, the correct answer is a. 957.73 sq.mm.
MY EXPLANATION AND DERIVATION OF ANSWER:
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP),
Note that these formulae are SI-based (metric) units. So, in my question, let's first convert the specified concrete compressive strength (f'c) given at [4,500psi] from English unit to metric unit,
The other relative data to be used in the equations are already given in their correct units,
Now let's determine the minimum steel area (As,min) using the two (2) given equations based on the provisions of the Code,
Since we are looking for the minimum value, and 957.73mm2 is greater than 952.67mm2, then the larger value shall govern as the minimum.
Therefore, the correct answer is a. 957.73 sq.mm.
Minimum Area of Steel (As,min),
As,min = ( 0.25√(f'c) / fy )( bw d )
As,min = (0.25√[31.03MPa] / [275MPa] ) ( [350mm] x [537.5mm] )
As,min = 952.67mm2
As,min = ( 1.4 / fy )( bw d )
As,min = ( 1.4 / [275MPa] ) ( [350mm] x [537.5mm] )
As,min = 957.73mm2
As,min = (0.25√[31.03MPa] / [275MPa] ) ( [350mm] x [537.5mm] )
As,min = 952.67mm2
As,min = ( 1.4 / fy )( bw d )
As,min = ( 1.4 / [275MPa] ) ( [350mm] x [537.5mm] )
As,min = 957.73mm2
Therefore, the correct answer is a. 957.73 sq.mm.
MY EXPLANATION AND DERIVATION OF ANSWER:
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP),
"A minimum area of flexural reinforcement As,min shall be provided at every section where tension reinforcement is required by analysis."
(Sec. 409.6.1.1 of the 2015 NSCP)
"As,min shall be the greater of (a) and (b), except as provided in Section 409.6.1.3. For a statically determinate beam with a flange in tension, the value of bw shall be the lesser of bf and 2bw.
(Sec. 409.6.1.1 of the 2015 NSCP)
"As,min shall be the greater of (a) and (b), except as provided in Section 409.6.1.3. For a statically determinate beam with a flange in tension, the value of bw shall be the lesser of bf and 2bw.
a. ( 0.25√(f'c) / fy )( bw d )
b. ( 1.4 / fy )( bw d )"
(Sec. 409.6.1.2 of the 2015 NSCP)b. ( 1.4 / fy )( bw d )"
Note that these formulae are SI-based (metric) units. So, in my question, let's first convert the specified concrete compressive strength (f'c) given at [4,500psi] from English unit to metric unit,
f'c = 4,500psi = 4,500psi x ( 1MPa / 145psi ) = 31.03MPa
The other relative data to be used in the equations are already given in their correct units,
fy (Specified steel yield strength) = PNS Grade 275 = 275MPa
bw (Beam width) = 350mm
d (Effective depth of section) = 537.5mm (solved in Q#1)
bw (Beam width) = 350mm
d (Effective depth of section) = 537.5mm (solved in Q#1)
Now let's determine the minimum steel area (As,min) using the two (2) given equations based on the provisions of the Code,
From first conditional equation:
From second conditional equation:
As,min = ( 0.25√(f'c) / fy )( bw d )
As,min = ( 0.25 √[31.03MPa] / [275MPa] ) ( [350mm] x [537.5mm] )
As,min = 952.67mm2
As,min = ( 0.25 √[31.03MPa] / [275MPa] ) ( [350mm] x [537.5mm] )
As,min = 952.67mm2
From second conditional equation:
As,min = ( 1.4 / fy )( bw d )
As,min = ( 1.4 / [275MPa] ) ( [350mm] x [537.5mm] )
As,min = 957.73mm2
As,min = ( 1.4 / [275MPa] ) ( [350mm] x [537.5mm] )
As,min = 957.73mm2
Since we are looking for the minimum value, and 957.73mm2 is greater than 952.67mm2, then the larger value shall govern as the minimum.
Therefore, the correct answer is a. 957.73 sq.mm.
Q#3: What is the actual steel ratio of the given beam?
a. 0.015656
b. 0.010246
c. 0.009350
d. 0.010437
e. 0.010018
SOLUTION:
Actual steel ratio (ρ),
Therefore, the correct answer is d. 0.010437.
MY EXPLANATION AND DERIVATION OF ANSWER:
The Steel Ratio (ρ) of a reinforced concrete beam is simply the cross-sectional area of steel reinforcement (As) over the effective cross-sectional area of concrete ( bw x d ),
In my problem, the Area of Steel (As) can be determined by the summation of all areas of each main rebars given as [4-Φ25mm], meaning 4 sets of 25mm-diameter rebars,
The other relative data needed to compute for the Steel Ratio are the beam width (bw) given in the problem as [350mm] and the effective depth of section (d) which we have calculated in Q#1 to be [537.5mm],
Solving for the Steel Ratio (ρ),
This shows that the Area of Tensile Steel is 1.0437% the Effective Cross-Sectional Area of Concrete in our given beam.
Therefore, the correct answer is d. 0.010437.
Actual steel ratio (ρ),
ρ = ( 4 x π (25mm)2 / 4 ) / ( [350mm] x [537.5mm] )
ρ = 0.010437
ρ = 0.010437
Therefore, the correct answer is d. 0.010437.
MY EXPLANATION AND DERIVATION OF ANSWER:
The Steel Ratio (ρ) of a reinforced concrete beam is simply the cross-sectional area of steel reinforcement (As) over the effective cross-sectional area of concrete ( bw x d ),
ρ = As / ( bw d ) ===> Formula for the Steel Ratio
In my problem, the Area of Steel (As) can be determined by the summation of all areas of each main rebars given as [4-Φ25mm], meaning 4 sets of 25mm-diameter rebars,
As = 4 sets x ( π diameter2 / 4 )
As = 4 sets x ( π [25mm]2 / 4 )
As = 4 sets x ( 490.875mm2 )
As = 1,963.49mm2
As = 4 sets x ( π [25mm]2 / 4 )
As = 4 sets x ( 490.875mm2 )
As = 1,963.49mm2
The other relative data needed to compute for the Steel Ratio are the beam width (bw) given in the problem as [350mm] and the effective depth of section (d) which we have calculated in Q#1 to be [537.5mm],
Solving for the Steel Ratio (ρ),
ρ = [ Cross-Sectional Area of Steel ] / [ Effective Cross-Sectional Area of Concrete ]
ρ = As / ( bw d )
ρ = [1,963.49mm2] / ( [350mm] [537.5mm] )
ρ = [1,963.49mm2] / ( 188,125mm2 )
ρ = 0.010437
ρ = As / ( bw d )
ρ = [1,963.49mm2] / ( [350mm] [537.5mm] )
ρ = [1,963.49mm2] / ( 188,125mm2 )
ρ = 0.010437
This shows that the Area of Tensile Steel is 1.0437% the Effective Cross-Sectional Area of Concrete in our given beam.
Therefore, the correct answer is d. 0.010437.
Q#4: What would be the balanced steel ratio of this beam?
a. 0.054455
b. 0.015656
c. 0.055902
d. 0.001210
e. 0.011070
SOLUTION:
Balanced steel ratio (ρb),
Therefore, the correct answer is a. 0.054455.
MY EXPLANATION AND DERIVATION OF ANSWER:
A Balanced Steel Ratio (ρb) is a steel ratio of the reinforced concrete beam that will produce a balanced strain design condition (where strain in concrete and strain in steel will both reach its limit at the same time, and thus, will fail simultaneously).
Recall that in the design of reinforced concrete members, we do NOT want concrete and steel to fail AT THE SAME TIME (i.e., a BALANCED strain design). Likewise, we also do NOT want a scenario where concrete will crush BEFORE steel has yielded (i.e., an OVER-REINFORCED strain design).
Both a BALANCED DESIGN, where actual steel ratio is equal to the Balanced Steel Ratio [ ρ = ρb ], and an OVER-REINFORCED DESIGN, where actual steel ratio is greater than the Balanced Steel Ratio [ ρ > ρb ], will most likely exhibit catastrophic BRITTLE FAILURE, wherein the beam will suddenly crush without any warning.
The IDEAL design is to allow for a DUCTILE FAILURE, where the ductile STEEL will YIELD FIRST before the brittle concrete will crush (i.e., UNDER-REINFORCED strain design). Such ductility of the beam will give ample warning before failing if the reinforced concrete reaches its ultimate strength. Thus, the actual steel ratio (ρ) of a reinforced concrete beam must fall somewhere between the values of the Minimum Steel Ratio (ρmin) and the Balanced Steel Ratio (ρb).
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP), several design assumptions are prescribed in analyzing the sectional strength of a flexural reinforced concrete member (i.e., beams),
The above assumption simply means that [ εcu = 0.003 ], as shown in my illustration below,
Furthermore, the Code states,
The above assumptions in the Code can be illustrated as,
The value of β1 is just a stress distribution factor (due to the changing of the actual parabolic-shaped concrete stress area to its equivalent rectangular-shaped "stress block" for ease of computing). The β1 (beta-factor) is simply dependent on the f'c (the specified compressive strength of concrete), where it states in the Code,
The above table simply means that,
In the structural design of reinforced concrete members, all FORCES (recall that: Force = Stress x Area) should be in EQUILIBRIUM (i.e., Force in Compression Concrete + Force in Tension Steel = 0). So, given the provisions and assumptions of the Code mentioned above regarding the equivalent concrete stress block,
So, from the above equilibrium equation [ fy As ] = [ 0.85 f'c α bw ], the Depth of the Stress Block (α), aside from what is prescribed in the Code as [ α = β1 c ], can also be expressed as,
This means that the distance c, which is the distance of the neutral axis from from the extreme concrete compression fiber, based in the Code as [ c = α / β1 ], can also be expressed by substituting (α) using the above equation, thus,
In a BALANCED DESIGN, the distance c, which is now termed as distance (cb), measured from the extreme concrete compression fiber to the neutral axis, is so located that both the assumed MAXIMUM COMPRESSIVE STRAIN OF CONCRETE [ εcu = 0.003 ] along the extreme compression fiber and the TENSILE STRAIN OF STEEL AT YIELD POINT, where [ εy (steel strain at yield point) = εs (steel strain at service loads) ], along the centroid of steel, will reach such corresponding values and points at the SAME TIME.
The value of εy (the strain in steel at yield point) can be derived in the Code,
Stress in steel BELOW fy (below the steel yield point) will show a strain diagram that is of an OVER-REINFORCED design, wherein the Strain of Steel at Service Loads (εs) has NOT yet reached the Steel Strain at Yield Point (εy), or [ εy > εs ]. This means that the STEEL DID NOT YIELD but the CONCRETE HAS ALREADY CRUSHED.
Such undesirable condition can also be applied to a BALANCED strain design, where both steel and concrete would have failed simultaneously. So, at balanced design, [ εs = εy ].
Recall that [ Modulus of Elasticity (E) = Stress / Strain (ε) ], or [ Stress = E x ε ]. Since the Code permits that the Modulus of Elasticity of Steel (Es) can be taken as 200,000MPa, then,
The distance cb (from extreme compression fiber to neutral axis at balanced strain condition) can be determined by using simple proportion from its balanced strain diagram,
From the above proportion equation, we can substitute the value of εy (the strain in steel at yield point) based on the equation [ εy = fy / 200,000MPa ] considering the Modulus of Elasticity of Steel,
Since the formula for solving distance c is [ c = As fy / ( 0.85 f'c b β1 ) ], such distance c, at balanced design conditions, will be [ c = cb ],
Recall that steel ratio is [ ρ = As / ( bw d ) ]. Such ratio ρ, at balanced conditions, will be [ ρb = As,bal / ( bw d ) ]. So, from the above equation, we can produce [ As,bal / ( bw d ) ] by just mutiplying both sides of the equation with a value of [ 1 / d ],
The resulting formula for the Balanced Steel Ratio (ρb) shows you that in order to calculate ρb, you would just simply need two (2) commonly known values: the Specified Compressive Strength of Concrete (f'c) and the Specified Yield Strength of Steel (fy). Take note that the above formula is in S.I. Units.
In my problem, the f'c was given in psi (English unit). So, we first convert it to metric units,
Since the value of f'c falls between 28MPa and 55MPa, based on the provisions of the Code, the Beta-factor (β1) would be,
Now that all necessary values are known, we can compute the Balanced Steel Ratio (ρb) in our given problem using the formula,
This means that if the Area of Tensile Steel (As) reaches 5.4455% of the Effective Cross-Sectional Area of Concrete ( bw x d ) in a beam specified to have an f'c = 31.03MPa and an fy = 275MPa, the beam will result in a balanced strain design condition where both ductile steel and brittle concrete will fail simultaneously.
Therefore, the correct answer is a. 0.054455.
Balanced steel ratio (ρb),
ρb = 0.85 f'c β1 600MPa / ( fy ( 600MPa + fy ) )
ρb = (0.85) [31.03MPa] (0.828) (600MPa) / ( [275MPa] ( 600MPa + [275MPa] )
ρb = 0.054455
ρb = (0.85) [31.03MPa] (0.828) (600MPa) / ( [275MPa] ( 600MPa + [275MPa] )
ρb = 0.054455
Therefore, the correct answer is a. 0.054455.
MY EXPLANATION AND DERIVATION OF ANSWER:
A Balanced Steel Ratio (ρb) is a steel ratio of the reinforced concrete beam that will produce a balanced strain design condition (where strain in concrete and strain in steel will both reach its limit at the same time, and thus, will fail simultaneously).
Recall that in the design of reinforced concrete members, we do NOT want concrete and steel to fail AT THE SAME TIME (i.e., a BALANCED strain design). Likewise, we also do NOT want a scenario where concrete will crush BEFORE steel has yielded (i.e., an OVER-REINFORCED strain design).
Both a BALANCED DESIGN, where actual steel ratio is equal to the Balanced Steel Ratio [ ρ = ρb ], and an OVER-REINFORCED DESIGN, where actual steel ratio is greater than the Balanced Steel Ratio [ ρ > ρb ], will most likely exhibit catastrophic BRITTLE FAILURE, wherein the beam will suddenly crush without any warning.
The IDEAL design is to allow for a DUCTILE FAILURE, where the ductile STEEL will YIELD FIRST before the brittle concrete will crush (i.e., UNDER-REINFORCED strain design). Such ductility of the beam will give ample warning before failing if the reinforced concrete reaches its ultimate strength. Thus, the actual steel ratio (ρ) of a reinforced concrete beam must fall somewhere between the values of the Minimum Steel Ratio (ρmin) and the Balanced Steel Ratio (ρb).
ρmin ≤ ρ < ρb ==> Condition that must be met where Steel will yield
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP), several design assumptions are prescribed in analyzing the sectional strength of a flexural reinforced concrete member (i.e., beams),
"Maximum strain at the extreme concrete compression fiber shall be assumed equal to 0.003."
(Sec. 422.2.2.1 of the 2015 NSCP)
(Sec. 422.2.2.1 of the 2015 NSCP)
The above assumption simply means that [ εcu = 0.003 ], as shown in my illustration below,
Furthermore, the Code states,
"Concrete stress of 0.85 f'c shall be assumed uniformly distributed over an equivalent compression area bounded by the edges of the cross section and a line parallel to the neutral axis located at a distance α from the fiber of maximum compressive strain, as calculated by:
Distance from the fiber of maximum compressive strain to the neutral axis, c, shall be measured perpendicular to the neutral axis."
(Sec. 422.2.2.4.1 and Sec. 422.2.2.4.2 of the 2015 NSCP)
α = β1 c
Distance from the fiber of maximum compressive strain to the neutral axis, c, shall be measured perpendicular to the neutral axis."
(Sec. 422.2.2.4.1 and Sec. 422.2.2.4.2 of the 2015 NSCP)
The above assumptions in the Code can be illustrated as,
The value of β1 is just a stress distribution factor (due to the changing of the actual parabolic-shaped concrete stress area to its equivalent rectangular-shaped "stress block" for ease of computing). The β1 (beta-factor) is simply dependent on the f'c (the specified compressive strength of concrete), where it states in the Code,
"Values of β1 shall be in accordance with Table 422.2.2.4.3"
(Sec. 422.2.2.4.3 of the 2015 NSCP)
(Sec. 422.2.2.4.3 of the 2015 NSCP)
(Table 422.2.2.4.3 of the 2015 NSCP)
The above table simply means that,
β1 = [ 0.85 ] if the f'c was specified to be 17MPa to 28MPa
β1 = [ 0.85 - 0.05 ( f'c - 28MPa ) / 7 ] if the f'c specified is greater than 28MPa but less than 55MPa
β1 = [ 0.65 ] if the f'c was specified to be 55MPa or greater
β1 = [ 0.85 - 0.05 ( f'c - 28MPa ) / 7 ] if the f'c specified is greater than 28MPa but less than 55MPa
β1 = [ 0.65 ] if the f'c was specified to be 55MPa or greater
In the structural design of reinforced concrete members, all FORCES (recall that: Force = Stress x Area) should be in EQUILIBRIUM (i.e., Force in Compression Concrete + Force in Tension Steel = 0). So, given the provisions and assumptions of the Code mentioned above regarding the equivalent concrete stress block,
[ Force in Tension Steel ] = [ Force in Compression Concrete ]
[ Steel Stress x Area of Steel ] = [ Concrete Stress x Area of Concrete Stress Block ]
[ fy x As ] = [ 0.85 f'c x α bw ]
[ Steel Stress x Area of Steel ] = [ Concrete Stress x Area of Concrete Stress Block ]
[ fy x As ] = [ 0.85 f'c x α bw ]
So, from the above equilibrium equation [ fy As ] = [ 0.85 f'c α bw ], the Depth of the Stress Block (α), aside from what is prescribed in the Code as [ α = β1 c ], can also be expressed as,
fy As = 0.85 f'c [ α ] bw
α = As fy / ( 0.85 f'c bw )
α = As fy / ( 0.85 f'c bw )
This means that the distance c, which is the distance of the neutral axis from from the extreme concrete compression fiber, based in the Code as [ c = α / β1 ], can also be expressed by substituting (α) using the above equation, thus,
c = [ α ] / β1
c = [ As fy / ( 0.85 f'c bw ) ] / β1
c = As fy / ( 0.85 f'c bw β1 ) ===> Formula for determining c
c = [ As fy / ( 0.85 f'c bw ) ] / β1
c = As fy / ( 0.85 f'c bw β1 ) ===> Formula for determining c
In a BALANCED DESIGN, the distance c, which is now termed as distance (cb), measured from the extreme concrete compression fiber to the neutral axis, is so located that both the assumed MAXIMUM COMPRESSIVE STRAIN OF CONCRETE [ εcu = 0.003 ] along the extreme compression fiber and the TENSILE STRAIN OF STEEL AT YIELD POINT, where [ εy (steel strain at yield point) = εs (steel strain at service loads) ], along the centroid of steel, will reach such corresponding values and points at the SAME TIME.
The value of εy (the strain in steel at yield point) can be derived in the Code,
"For non-prestressed bars and wires, the stress below fy shall be Es times steel strain. For strains greater than that corresponding to fy, stress shall be considered independent of strain and equal to fy."
(Sec. 420.2.2.1 of the 2015 NSCP)
"Modulus of elasticity, Es, for nonprestressed bars and wires shall be permitted to be taken as 200,000 MPa."
(Sec. 420.2.2.2 of the 2015 NSCP)
(Sec. 420.2.2.1 of the 2015 NSCP)
"Modulus of elasticity, Es, for nonprestressed bars and wires shall be permitted to be taken as 200,000 MPa."
(Sec. 420.2.2.2 of the 2015 NSCP)
Stress in steel BELOW fy (below the steel yield point) will show a strain diagram that is of an OVER-REINFORCED design, wherein the Strain of Steel at Service Loads (εs) has NOT yet reached the Steel Strain at Yield Point (εy), or [ εy > εs ]. This means that the STEEL DID NOT YIELD but the CONCRETE HAS ALREADY CRUSHED.
Such undesirable condition can also be applied to a BALANCED strain design, where both steel and concrete would have failed simultaneously. So, at balanced design, [ εs = εy ].
Recall that [ Modulus of Elasticity (E) = Stress / Strain (ε) ], or [ Stress = E x ε ]. Since the Code permits that the Modulus of Elasticity of Steel (Es) can be taken as 200,000MPa, then,
[ Stress in Steel ] = [ Modulus of Elasticity of Steel ] x [ Strain in Steel ]
[ fy ] = [ Es ] x [ εy ]
fy = [ 200,000MPa ] x εy
εy = fy / 200,000MPa
[ fy ] = [ Es ] x [ εy ]
fy = [ 200,000MPa ] x εy
εy = fy / 200,000MPa
The distance cb (from extreme compression fiber to neutral axis at balanced strain condition) can be determined by using simple proportion from its balanced strain diagram,
cb / d = εcu / ( εy + εcu )
cb / d = 0.003 / ( εy + 0.003 )
cb / d = 0.003 / ( εy + 0.003 )
From the above proportion equation, we can substitute the value of εy (the strain in steel at yield point) based on the equation [ εy = fy / 200,000MPa ] considering the Modulus of Elasticity of Steel,
cb / d = 0.003 / ( [ εy ] + 0.003 )
cb / d = 0.003 / ( [ fy / 200,000MPa ] + 0.003 )
cb / d = 0.003 / ( ( 600MPa + fy ) / 200,000MPa )
cb / d = 600MPa / ( 600MPa + fy )
cb = 600MPa ( d ) / ( 600MPa + fy ) ==> Formula for determining cb
cb / d = 0.003 / ( [ fy / 200,000MPa ] + 0.003 )
cb / d = 0.003 / ( ( 600MPa + fy ) / 200,000MPa )
cb / d = 600MPa / ( 600MPa + fy )
cb = 600MPa ( d ) / ( 600MPa + fy ) ==> Formula for determining cb
Since the formula for solving distance c is [ c = As fy / ( 0.85 f'c b β1 ) ], such distance c, at balanced design conditions, will be [ c = cb ],
[ c (at balanced design) ] = [ cb ]
[ As,bal fy / ( 0.85 f'c bw β1 ) ] = [ 600MPa ( d ) / ( 600MPa + fy ) ]
As,bal / bw = 0.85 f'c β1 600MPa (d) / ( fy ( 600MPa + fy ) )
[ As,bal fy / ( 0.85 f'c bw β1 ) ] = [ 600MPa ( d ) / ( 600MPa + fy ) ]
As,bal / bw = 0.85 f'c β1 600MPa (d) / ( fy ( 600MPa + fy ) )
Recall that steel ratio is [ ρ = As / ( bw d ) ]. Such ratio ρ, at balanced conditions, will be [ ρb = As,bal / ( bw d ) ]. So, from the above equation, we can produce [ As,bal / ( bw d ) ] by just mutiplying both sides of the equation with a value of [ 1 / d ],
[ As,bal / bw = 0.85 f'c β1 600MPa (d) / ( fy ( 600MPa + fy ) ) ] x [ 1 / d ]
As,bal / ( bw [ d ] ) = 0.85 f'c β1 600MPa (d) / ( [ d ] fy ( 600MPa + fy ) )
[ As,bal / ( bw d ) ] = 0.85 f'c β1 600MPa / ( fy ( 600MPa + fy ) )
ρb = 0.85 f'c β1 600MPa / ( fy ( 600MPa + fy ) ) ===> Formula for the Balanced Steel Ratio
As,bal / ( bw [ d ] ) = 0.85 f'c β1 600MPa (d) / ( [ d ] fy ( 600MPa + fy ) )
[ As,bal / ( bw d ) ] = 0.85 f'c β1 600MPa / ( fy ( 600MPa + fy ) )
ρb = 0.85 f'c β1 600MPa / ( fy ( 600MPa + fy ) ) ===> Formula for the Balanced Steel Ratio
The resulting formula for the Balanced Steel Ratio (ρb) shows you that in order to calculate ρb, you would just simply need two (2) commonly known values: the Specified Compressive Strength of Concrete (f'c) and the Specified Yield Strength of Steel (fy). Take note that the above formula is in S.I. Units.
In my problem, the f'c was given in psi (English unit). So, we first convert it to metric units,
f'c = 4,500psi x ( 1MPa / 145psi ) = 31.03MPa
Since the value of f'c falls between 28MPa and 55MPa, based on the provisions of the Code, the Beta-factor (β1) would be,
β1 = [ 0.85 - 0.05 x ( f'c - 28MPa ) / 7 ]
β1 = [ 0.85 - 0.05 x ( [31.03MPa] - 28MPa ) / 7 ]
β1 = 0.828
β1 = [ 0.85 - 0.05 x ( [31.03MPa] - 28MPa ) / 7 ]
β1 = 0.828
Now that all necessary values are known, we can compute the Balanced Steel Ratio (ρb) in our given problem using the formula,
ρb = 0.85 f'c β1 600MPa / ( fy ( 600MPa + fy ) )
ρb = 0.85 [31.03MPa] [0.828] 600MPa / ( [275MPa] ( 600MPa + [275MPa] ) )
ρb = 13,103.3484MPa2 / 240,625MPa2
ρb = 0.054455
ρb = 0.85 [31.03MPa] [0.828] 600MPa / ( [275MPa] ( 600MPa + [275MPa] ) )
ρb = 13,103.3484MPa2 / 240,625MPa2
ρb = 0.054455
This means that if the Area of Tensile Steel (As) reaches 5.4455% of the Effective Cross-Sectional Area of Concrete ( bw x d ) in a beam specified to have an f'c = 31.03MPa and an fy = 275MPa, the beam will result in a balanced strain design condition where both ductile steel and brittle concrete will fail simultaneously.
Therefore, the correct answer is a. 0.054455.
Q#5: What would be the maximum area of steel reinforcement that could occupy the given beam if the section must be designed as tension-controlled only?
a. Approx. 5,751.28 sq.mm.
b. Approx. 7,683.26 sq.mm.
c. Approx. 5,602.42 sq.mm.
d. Approx. 4,802.08 sq.mm.
e. Approx. 7,875.00 sq.mm.
SOLUTION:
Calculate distance (c) from extreme compression fiber to neutral axis at tension-controlled steel strain (εT) of 0.005,
Solving for maximum area of steel reinforcement (As,max),
Therefore, the correct answer is c. Approx. 5,602.42 sq.mm.
MY EXPLANATION AND DERIVATION OF ANSWER:
In older versions of the American Concrete Institute (ACI) Code, where the National Structural Code of the Philippines (NSCP) is also based on, a Maximum Steel Ratio (ρmax) equal to 75% of the Balanced Steel Ratio (ρb) was prescribed to ensure that the reinforced concrete section is ductile.
However, instead of using the old equation [ ρmax = 0.75 ρb ] which, as you can see, would still result in having a HIGH AMOUNT OF STEEL AREA, the updated editions of the ACI and NSCP now controls the ductility of the beam by applying a CERTAIN LIMIT on the net tensile strain (εT) located along the extreme layer of tension steel.
As the Code assumes a maximum compression strain in concrete of 0.003 [εcu = 0.003], the Code also states that, for a beam section to be classified as TENSION-CONTROLLED (where steel would only need to handle tensile stresses of the beam and will NOT require additional steel reinforcement in compression), the said limit for the Tension Strain in Steel (εT) must be greater than or equal to 0.005 [εT ≥ 0.005].
My illustration above, showing the strain diagrams of a compression-controlled, transition, and tension-controlled beam section, is similar in concept to the strain diagrams produced by an over-reinforced design, balanced design, and under-reinforced design, respectively.
An OVER-REINFORCED design has a very large amount of steel area (As) that would result in a small amount of deformation or strain in steel (εT), thus will produce a COMPRESSION-CONTROLLED beam section (brittle section) where the beam, if not properly reinforced with additional compression steel, could be catastrophic when it fails as the BRITTLE CONCRETE WILL CRUSH FIRST without any warning.
Consequently, an UNDER-REINFORCED design has a smaller, but ample, amount of steel area (As) that would result in a larger deformation or strain in steel (εT), thus will exhibit a TENSION-CONTROLLED beam section (ductile section) where the beam, without nay need for additional compression reinforcement, would be safe in the event of failure due to the DUCTILE STEEL YIELDING FIRST instead of the brittle concrete crushing instantly.
As you have noticed, there is an inverse proportion between the steel area and the resulting steel strain. The LARGER the STEEL AREA on a given beam section, the SMALLER the value of the STRAIN produced in the tension steel - which is UNDESIRABLE (steel did NOT yield, as such, additional steel will be required at the compression side of the beam to help with the concrete stress).
On the other hand, the SMALLER the STEEL AREA on the same given beam section, the LARGER the value of the STRAIN in tension steel - which is DESIRABLE (steel yielding first).
In my problem, if the beam section is required to be tension-controlled, then, the deformation or strain at the extreme tension steel εT should NOT be less than 0.005. So, if the steel strain εT is at its minimum [ εT = 0.005 ], then, the corresponding steel area As would be at its maximum As,max for a tension-controlled section.
Let's first compute the distance c of the neutral axis, with [ εcu = 0.003 maximum ] and [ εT = 0.005 minimum ].
We could use the formula [ c = As fy / ( 0.85 f'c bw β1 ) ] to determine c, but in this case, both values for c and As,max are unknown. So instead, we'll compute for distance c using the proportion equation of the strain diagram,
Since we have already solved d (the effective depth of section) in Q#1 as [537.5mm], then,
Knowing the value of c, we can now use the formula [ c = As fy / ( 0.85 f'c bw β1 ) ] to solve for the unknown area of steel (As) which is now considered our maximum area of steel (As,max) for a tension-controlled section.
The other values required in the formula, [ bw = 350mm ], [ fy = 275MPa ], and [ f'c = 31.03MPa ] were all given in the problem, while [ β1 = 0.828 ], dependent on the f'c, was solved in Q#4.
Since my question calls for a tension-controlled section, then the As,max = 5,602.42mm2 must prevail, as such maximum steel area, the steel strain will be at its tension-controlled limit of εT = 0.005 minimum.
Therefore, the correct answer is c. Approx. 5,602.42 sq.mm.
Calculate distance (c) from extreme compression fiber to neutral axis at tension-controlled steel strain (εT) of 0.005,
c (at εT = 0.005) = d ( 3 / 8 )
c (at εT = 0.005) = [537.5mm] ( 3 / 8 )
c (at εT = 0.005) = 201.5625mm
c (at εT = 0.005) = [537.5mm] ( 3 / 8 )
c (at εT = 0.005) = 201.5625mm
Solving for maximum area of steel reinforcement (As,max),
As,max = c ( 0.85 f'c bw β1 ) / fy
As,max = 201.5625mm ( [0.85] [31.03MPa] [350mm] [0.828] ) / [275MPa]
As,max = 5,602.42mm2
As,max = 201.5625mm ( [0.85] [31.03MPa] [350mm] [0.828] ) / [275MPa]
As,max = 5,602.42mm2
Therefore, the correct answer is c. Approx. 5,602.42 sq.mm.
MY EXPLANATION AND DERIVATION OF ANSWER:
In older versions of the American Concrete Institute (ACI) Code, where the National Structural Code of the Philippines (NSCP) is also based on, a Maximum Steel Ratio (ρmax) equal to 75% of the Balanced Steel Ratio (ρb) was prescribed to ensure that the reinforced concrete section is ductile.
ρmax = 0.75 ρb ===> based on old provision in earlier editions of the Code.
For informational purposes only, let's try to use the above old equation in solving for the maximum area of steel (As,max) for our given problem. Since the balanced steel ratio (ρb) was already computed in our solution for Q#4 as [ ρb = 0.054455 ], then,
If steel ratio (ρ) is [ ρ = As / ( bw d ) ], then, maximum steel ratio ρmax would be [ ρmax = As,max / ( bw d ) ]. Thus,
ρmax = 0.75 [ ρb ]
ρmax = 0.75 [ 0.054455 ]
ρmax = 0.04084125
ρmax = 0.75 [ 0.054455 ]
ρmax = 0.04084125
If steel ratio (ρ) is [ ρ = As / ( bw d ) ], then, maximum steel ratio ρmax would be [ ρmax = As,max / ( bw d ) ]. Thus,
ρmax = [ As,max ] / ( bw d )
As,max = ρmax bw d
As,max = 0.04084125 [350mm] [537.5mm]
As,max = 7,683.26mm2 ==> based on old provision in earlier editions of the Code.
As,max = 0.04084125 [350mm] [537.5mm]
As,max = 7,683.26mm2 ==> based on old provision in earlier editions of the Code.
However, instead of using the old equation [ ρmax = 0.75 ρb ] which, as you can see, would still result in having a HIGH AMOUNT OF STEEL AREA, the updated editions of the ACI and NSCP now controls the ductility of the beam by applying a CERTAIN LIMIT on the net tensile strain (εT) located along the extreme layer of tension steel.
As the Code assumes a maximum compression strain in concrete of 0.003 [εcu = 0.003], the Code also states that, for a beam section to be classified as TENSION-CONTROLLED (where steel would only need to handle tensile stresses of the beam and will NOT require additional steel reinforcement in compression), the said limit for the Tension Strain in Steel (εT) must be greater than or equal to 0.005 [εT ≥ 0.005].
A tension-controlled section is when [ εT ≥ 0.005 ]
A transition section is when [ 0.002 < εT < 0.005 ]
A compression-controlled section is when [ εT ≤ 0.002 ]
A transition section is when [ 0.002 < εT < 0.005 ]
A compression-controlled section is when [ εT ≤ 0.002 ]
My illustration above, showing the strain diagrams of a compression-controlled, transition, and tension-controlled beam section, is similar in concept to the strain diagrams produced by an over-reinforced design, balanced design, and under-reinforced design, respectively.
An OVER-REINFORCED design has a very large amount of steel area (As) that would result in a small amount of deformation or strain in steel (εT), thus will produce a COMPRESSION-CONTROLLED beam section (brittle section) where the beam, if not properly reinforced with additional compression steel, could be catastrophic when it fails as the BRITTLE CONCRETE WILL CRUSH FIRST without any warning.
Consequently, an UNDER-REINFORCED design has a smaller, but ample, amount of steel area (As) that would result in a larger deformation or strain in steel (εT), thus will exhibit a TENSION-CONTROLLED beam section (ductile section) where the beam, without nay need for additional compression reinforcement, would be safe in the event of failure due to the DUCTILE STEEL YIELDING FIRST instead of the brittle concrete crushing instantly.
As you have noticed, there is an inverse proportion between the steel area and the resulting steel strain. The LARGER the STEEL AREA on a given beam section, the SMALLER the value of the STRAIN produced in the tension steel - which is UNDESIRABLE (steel did NOT yield, as such, additional steel will be required at the compression side of the beam to help with the concrete stress).
On the other hand, the SMALLER the STEEL AREA on the same given beam section, the LARGER the value of the STRAIN in tension steel - which is DESIRABLE (steel yielding first).
In my problem, if the beam section is required to be tension-controlled, then, the deformation or strain at the extreme tension steel εT should NOT be less than 0.005. So, if the steel strain εT is at its minimum [ εT = 0.005 ], then, the corresponding steel area As would be at its maximum As,max for a tension-controlled section.
Let's first compute the distance c of the neutral axis, with [ εcu = 0.003 maximum ] and [ εT = 0.005 minimum ].
We could use the formula [ c = As fy / ( 0.85 f'c bw β1 ) ] to determine c, but in this case, both values for c and As,max are unknown. So instead, we'll compute for distance c using the proportion equation of the strain diagram,
c / 0.003 = d / ( 0.005 + 0.003 )
c = d ( 0.003 / 0.008 )
c = d ( 3 / 8 ) ===> Formula for c, where εT is at 0.005 minimum (tension-controlled section)
c = d ( 0.003 / 0.008 )
c = d ( 3 / 8 ) ===> Formula for c, where εT is at 0.005 minimum (tension-controlled section)
Since we have already solved d (the effective depth of section) in Q#1 as [537.5mm], then,
c (at εT = 0.005) = [537.5mm] ( 3 / 8 )
c (at εT = 0.005) = 201.5625mm
c (at εT = 0.005) = 201.5625mm
Knowing the value of c, we can now use the formula [ c = As fy / ( 0.85 f'c bw β1 ) ] to solve for the unknown area of steel (As) which is now considered our maximum area of steel (As,max) for a tension-controlled section.
The other values required in the formula, [ bw = 350mm ], [ fy = 275MPa ], and [ f'c = 31.03MPa ] were all given in the problem, while [ β1 = 0.828 ], dependent on the f'c, was solved in Q#4.
c = As fy / ( 0.85 f'c bw β1 )
[ c (at εT = 0.005) ] = [ As,max (at εT = 0.005) ] fy / ( 0.85 f'c bw β1 )
As,max (at εT = 0.005) = [ c (at εT = 0.005) ] ( 0.85 f'c bw β1 ) / fy
As,max (at εT = 0.005) = [201.5625mm] ( [0.85] [31.03MPa] [350mm] [0.828] ) / [275MPa]
As,max (at εT = 0.005) = 5,602.42mm2
[ c (at εT = 0.005) ] = [ As,max (at εT = 0.005) ] fy / ( 0.85 f'c bw β1 )
As,max (at εT = 0.005) = [ c (at εT = 0.005) ] ( 0.85 f'c bw β1 ) / fy
As,max (at εT = 0.005) = [201.5625mm] ( [0.85] [31.03MPa] [350mm] [0.828] ) / [275MPa]
As,max (at εT = 0.005) = 5,602.42mm2
Informational Note: If we would have used the previously-computed As,max = 7,683.26mm2, which was based on the old Code provision of [ ρmax = 0.75 ρb ], the steel strain εT will result in a value less than 0.005,
As you can see, using the old provision of [ ρmax = 0.75 ρb ] results in our beam being classified as a transition section (NOT tension-controlled), and is almost close to being a compression-controlled section (where steel may NOT yield without additional compression reinforcement).
c = As fy / ( 0.85 f'c bw β1 )
c = 7,683.26mm2 [275MPa] / ( [0.85] [31.03MPa] [350mm] [0.828] )
c = 276.426mm
εT = εcu (d - c) / c
εT = 0.003 (537.5mm - 276.426mm) / 276.426mm
εT = 0.0028333
c = 7,683.26mm2 [275MPa] / ( [0.85] [31.03MPa] [350mm] [0.828] )
c = 276.426mm
εT = εcu (d - c) / c
εT = 0.003 (537.5mm - 276.426mm) / 276.426mm
εT = 0.0028333
As you can see, using the old provision of [ ρmax = 0.75 ρb ] results in our beam being classified as a transition section (NOT tension-controlled), and is almost close to being a compression-controlled section (where steel may NOT yield without additional compression reinforcement).
Since my question calls for a tension-controlled section, then the As,max = 5,602.42mm2 must prevail, as such maximum steel area, the steel strain will be at its tension-controlled limit of εT = 0.005 minimum.
Therefore, the correct answer is c. Approx. 5,602.42 sq.mm.
Q#6: Calculate the nominal bending strength of the given beam.
a. Approx. 283kN-m
b. Approx. 1,084kN-m
c. Approx. 184kN-m
d. Approx. 328kN-m
e. Approx. 274kN-m
SOLUTION:
Nominal moment or bending strength (Mn) with respect to steel (Mn.steel):
Nominal moment or bending strength (Mn) with respect to concrete (Mn.concrete):
Therefore, the correct answer is e. Approx. 274kN-m.
MY EXPLANATION AND DERIVATION OF ANSWER:
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP),
Note that the said assumptions (of Sec. 422.2 of the 2015 NSCP) have already been outlined in Q#4 and will still be in consonance to what we will be outlining here.
Flexural (or bending) strength refers to the MOMENT STRENGTH (recall that: Moment = Force x Distance). Similar to the equilibrium of forces (recall that: Force = Stress x Area) that I have previously mentioned in our solution for Q#4 (where: Forces in Compression Concrete + Forces in Tension Steel = 0), so is the EQUILIBRIUM of MOMENTS another condition in the structural design of reinforced concrete beams. As such,
In any given reinforced concrete beam section, their exists a PAIR of different forces: one is the TENSILE FORCE (T) in Steel [ T = fy x As ] and the other is the COMPRESSIVE FORCE (C) in Concrete [ C = 0.85 f'c x α bw ].
These two forces run parallel to each other, are oppositely directed, and are equal in magnitude [ C = T ]. Thus, it forms a FORCE COUPLE where the distances between the two forces (i.e., their Moment Arm) are the same. Such Moment Arm, as seen in my illustration above, is simply the Effective Depth of Section ( d ) minus Half the Depth of the Rectangular Stress Block ( α / 2 ), or ( d - α / 2 ).
The formula for solving the Nominal Moment with respect to Steel (Mn.steel) is straight-forward,
The formula for solving the Nominal Moment with respect to Concrete (Mn.concrete) would entail a bit more calculations as to how it was derived,
Recall that [ α = As fy / ( 0.85 f'c bw ) ], and if we are to multiply the equation by the value [ d / d ], it will produce an expression [ As / ( bw d ) ] which is equal to the Steel Ratio ρ, since [ ρ = As / ( bw d ) ]. So,
Since Mn.concrete = [ 0.85 f'c α bw ] x [ d - α / 2 ], we can substitute the values of ( α ) within such equation with the above equation [ α = ω d / 0.85 ],
At this point, we have already determined the following relevant data to solve the nominal bending strength in my given problem,
Now with the two formulae, [ Mn.steel = fy As ( d - α/2 ) ] and [ Mn.concrete = R bw d2 ], we can solve the nominal moment or bending strength of the given beam using either of the two,
Nominal moment with respect to Steel (Mn.steel):
Nominal moment with respect to Concrete (Mn.concrete):
Therefore, the correct answer is e. Approx. 274kN-m.
Nominal moment or bending strength (Mn) with respect to steel (Mn.steel):
Mn.steel = fy As ( d - α / 2 )
Mn.steel = [275MPa] [1,963.49mm2] ( [537.5mm] - α / 2 )
Mn.steel = [275MPa] [1,963.49mm2] ( [537.5mm] - [58.49mm] / 2 )
Mn.steel = 274,437,242.736N-mm or ~274kN-m
Mn.steel = [275MPa] [1,963.49mm2] ( [537.5mm] - α / 2 )
α = As fy / ( 0.85 f'c bw )
α = [1,963.49mm2] [275MPa] / ( 0.85 [31.03MPa] [350mm] )
α = 58.49mm
α = [1,963.49mm2] [275MPa] / ( 0.85 [31.03MPa] [350mm] )
α = 58.49mm
Mn.steel = [275MPa] [1,963.49mm2] ( [537.5mm] - [58.49mm] / 2 )
Mn.steel = 274,437,242.736N-mm or ~274kN-m
Nominal moment or bending strength (Mn) with respect to concrete (Mn.concrete):
Mn.concrete = R bw d2
Mn.concrete = R [350mm] [537.5mm]2
Mn.concrete = [2.71353852MPa] [350mm] [537.5mm]2
Mn.concrete = 274,385,383.315N-mm or ~274kN-m
Mn.concrete = R [350mm] [537.5mm]2
R = f'c ω ( 1 - 0.59 ω )
R = [31.03MPa] ω ( 1 - 0.59 ω )
R = [31.03MPa] [0.0924967] ( 1 - 0.59 [0.0924967] )
R = 2.71353852MPa
R = [31.03MPa] ω ( 1 - 0.59 ω )
ω = ρ fy / f'c
ω = [0.010437] [275MPa] / [31.03MPa]
ω = 0.0924967
ω = [0.010437] [275MPa] / [31.03MPa]
ω = 0.0924967
R = [31.03MPa] [0.0924967] ( 1 - 0.59 [0.0924967] )
R = 2.71353852MPa
Mn.concrete = [2.71353852MPa] [350mm] [537.5mm]2
Mn.concrete = 274,385,383.315N-mm or ~274kN-m
Therefore, the correct answer is e. Approx. 274kN-m.
MY EXPLANATION AND DERIVATION OF ANSWER:
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP),
"Nominal flexural strength Mn shall be calculated in accordance with the assumptions of Section 422.2."
Sec. 422.3.1.1 of the 2015 NSCP)
Sec. 422.3.1.1 of the 2015 NSCP)
Note that the said assumptions (of Sec. 422.2 of the 2015 NSCP) have already been outlined in Q#4 and will still be in consonance to what we will be outlining here.
Flexural (or bending) strength refers to the MOMENT STRENGTH (recall that: Moment = Force x Distance). Similar to the equilibrium of forces (recall that: Force = Stress x Area) that I have previously mentioned in our solution for Q#4 (where: Forces in Compression Concrete + Forces in Tension Steel = 0), so is the EQUILIBRIUM of MOMENTS another condition in the structural design of reinforced concrete beams. As such,
[ Moment from Steel Tensile Force ] = [ Moment from Concrete Compressive Force ]
In any given reinforced concrete beam section, their exists a PAIR of different forces: one is the TENSILE FORCE (T) in Steel [ T = fy x As ] and the other is the COMPRESSIVE FORCE (C) in Concrete [ C = 0.85 f'c x α bw ].
These two forces run parallel to each other, are oppositely directed, and are equal in magnitude [ C = T ]. Thus, it forms a FORCE COUPLE where the distances between the two forces (i.e., their Moment Arm) are the same. Such Moment Arm, as seen in my illustration above, is simply the Effective Depth of Section ( d ) minus Half the Depth of the Rectangular Stress Block ( α / 2 ), or ( d - α / 2 ).
The formula for solving the Nominal Moment with respect to Steel (Mn.steel) is straight-forward,
Mn.steel = [ Force in Steel ] x [ Distance from the Force in Concrete ]
Mn.steel = [ Tensile Stress x Area of Steel ] x [ Effective Depth - Half of Stress Block Depth ]
Mn.steel = [ fy As ] x [ d - α / 2 ] ===> Formula for Mn (with respect to Steel)
Mn.steel = [ Tensile Stress x Area of Steel ] x [ Effective Depth - Half of Stress Block Depth ]
Mn.steel = [ fy As ] x [ d - α / 2 ] ===> Formula for Mn (with respect to Steel)
The formula for solving the Nominal Moment with respect to Concrete (Mn.concrete) would entail a bit more calculations as to how it was derived,
Mn.concrete = [ Force in Concrete ] x [ Distance from Force in Steel ]
Mn.concrete = [ Compressive Stress x Area of Stress Block ] x [ Effective Depth - Half of Stress Block Depth ]
Mn.concrete = [ 0.85 f'c α bw ] x [ d - α / 2 ]
Mn.concrete = [ Compressive Stress x Area of Stress Block ] x [ Effective Depth - Half of Stress Block Depth ]
Mn.concrete = [ 0.85 f'c α bw ] x [ d - α / 2 ]
Recall that [ α = As fy / ( 0.85 f'c bw ) ], and if we are to multiply the equation by the value [ d / d ], it will produce an expression [ As / ( bw d ) ] which is equal to the Steel Ratio ρ, since [ ρ = As / ( bw d ) ]. So,
α = As fy / 0.85 f'c bw x [ d / d ]
α = As fy d / 0.85 f'c bw d
α = [ As / ( bw d ) ] ( fy d / 0.85 f'c )
α = [ ρ ] ( fy d / 0.85 f'c )
α = [ ρ fy / f'c ] ( d / 0.85 )
α = [ ω ] d / 0.85
α = As fy d / 0.85 f'c bw d
α = [ As / ( bw d ) ] ( fy d / 0.85 f'c )
α = [ ρ ] ( fy d / 0.85 f'c )
α = [ ρ fy / f'c ] ( d / 0.85 )
where:
[ ρ fy / f'c ] ===> Formula for the Reinforcement Index (ω)
[ ρ fy / f'c ] ===> Formula for the Reinforcement Index (ω)
α = [ ω ] d / 0.85
Since Mn.concrete = [ 0.85 f'c α bw ] x [ d - α / 2 ], we can substitute the values of ( α ) within such equation with the above equation [ α = ω d / 0.85 ],
Mn.concrete = [ 0.85 f'c [ α ] bw ] x ( d - [ α ] / 2 )
Mn.concrete = ( 0.85 f'c [ ω d / 0.85 ] bw ) x ( d - [ ω d / 0.85 ] / 2 )
Mn.concrete = ( f'c ω d bw ) ( d - ( ω d / 1.7 ) )
Mn.concrete = f'c ω d bw d ( 1 - ω / 1.7 )
Mn.concrete = f'c ω bw d2 ( 1 - 0.588235294 ω )
Mn.concrete = [ f'c ω ( 1 - 0.59 ω ) ] bw d2
Mn.concrete = R bw d2 ===> Formula for Mn (with respect to Concrete)
Mn.concrete = ( 0.85 f'c [ ω d / 0.85 ] bw ) x ( d - [ ω d / 0.85 ] / 2 )
Mn.concrete = ( f'c ω d bw ) ( d - ( ω d / 1.7 ) )
Mn.concrete = f'c ω d bw d ( 1 - ω / 1.7 )
Mn.concrete = f'c ω bw d2 ( 1 - 0.588235294 ω )
Mn.concrete = [ f'c ω ( 1 - 0.59 ω ) ] bw d2
where:
[ f'c ω ( 1 - 0.59 ω ) ] ===> Formula for the Coefficient of Resistance (R)
[ f'c ω ( 1 - 0.59 ω ) ] ===> Formula for the Coefficient of Resistance (R)
Mn.concrete = R bw d2 ===> Formula for Mn (with respect to Concrete)
At this point, we have already determined the following relevant data to solve the nominal bending strength in my given problem,
bw = 350mm (given in the problem)
f'c = 4,500psi = 31.03MPa (given in the problem)
fy = PNS Grade 275 = 275MPa (given in the problem)
d = 537.5mm (solved in Q#1)
As = 1,963.49mm2 (solved in Q#3)
ρ = 0.010437 (solved in Q#3)
f'c = 4,500psi = 31.03MPa (given in the problem)
fy = PNS Grade 275 = 275MPa (given in the problem)
d = 537.5mm (solved in Q#1)
As = 1,963.49mm2 (solved in Q#3)
ρ = 0.010437 (solved in Q#3)
Now with the two formulae, [ Mn.steel = fy As ( d - α/2 ) ] and [ Mn.concrete = R bw d2 ], we can solve the nominal moment or bending strength of the given beam using either of the two,
Nominal moment with respect to Steel (Mn.steel):
Calculate depth of concrete stress block (α) to solve moment arm ( d - α / 2 ),
Calculate Nominal Moment with respect to Steel (Mn.steel),
α = As fy / ( 0.85 f'c bw )
α = [1,963.49mm2] [275MPa] / ( 0.85 [31.03MPa] [350mm] )
α = 58.49mm
α = [1,963.49mm2] [275MPa] / ( 0.85 [31.03MPa] [350mm] )
α = 58.49mm
Calculate Nominal Moment with respect to Steel (Mn.steel),
Mn.steel = fy As ( d - α/2 )
Mn.steel = [275MPa] [1,963.49mm2] ( [537.5mm] - [58.49mm] / 2 )
Mn.steel = 539,959.75MPa-mm2 ( [537.5mm] - 29.245mm )
Mn.steel = 539,959.75N (508.255mm)
Mn.steel = 274,437,242.736N-mm
Mn.steel = ~274kN-m
Mn.steel = [275MPa] [1,963.49mm2] ( [537.5mm] - [58.49mm] / 2 )
Mn.steel = 539,959.75MPa-mm2 ( [537.5mm] - 29.245mm )
Mn.steel = 539,959.75N (508.255mm)
Mn.steel = 274,437,242.736N-mm
Mn.steel = ~274kN-m
Nominal moment with respect to Concrete (Mn.concrete):
Calculate Reinforcement Index (ω) to solve Coefficient of Resistance (R),
Calculate Coefficient of Resistance (R) to solve Nominal Moment with respect to Concrete (Mn.concrete),
Calculate Nominal Moment with respect to Concrete (Mn.concrete),
ω = ρ fy / f'c
ω = [0.010437] [275MPa] / [31.03MPa]
ω = 0.0924967
ω = [0.010437] [275MPa] / [31.03MPa]
ω = 0.0924967
Calculate Coefficient of Resistance (R) to solve Nominal Moment with respect to Concrete (Mn.concrete),
R = f'c ω ( 1 - 0.59 ω )
R = [31.03MPa] [0.0924967] ( 1 - 0.59 [0.0924967] )
R = 2.870172601MPa ( 1 - 0.054573053 )
R = 2.870172601MPa (0.945426947)
R = 2.71353852MPa
R = [31.03MPa] [0.0924967] ( 1 - 0.59 [0.0924967] )
R = 2.870172601MPa ( 1 - 0.054573053 )
R = 2.870172601MPa (0.945426947)
R = 2.71353852MPa
Calculate Nominal Moment with respect to Concrete (Mn.concrete),
Mn.concrete = R bw d2
Mn.concrete = [2.71353852MPa] [350mm] [537.5mm]2
Mn.concrete = 274,385,383.315MPa-mm3
Mn.concrete = 274,385,383.315N-mm3/mm2
Mn.concrete = 274,385,383.315N-mm
Mn.concrete = ~274kN-m
Mn.concrete = [2.71353852MPa] [350mm] [537.5mm]2
Mn.concrete = 274,385,383.315MPa-mm3
Mn.concrete = 274,385,383.315N-mm3/mm2
Mn.concrete = 274,385,383.315N-mm
Mn.concrete = ~274kN-m
Therefore, the correct answer is e. Approx. 274kN-m.
Q#7: What is the ultimate moment capacity of the given beam section?
a. Approx. 254.7kN-m
b. Approx. 246.6kN-m
c. Approx. 178.1kN-m
d. Approx. 1,084.0kN-m
e. Approx. 263.8kN-m
SOLUTION:
Therefore, the correct answer is b. Approx. 246.6kN-m.
MY EXPLANATION AND DERIVATION OF ANSWER:
The Ultimate Moment Capacity (Mu) of a beam section is simply the nominal moment or bending strength (Mn) multiplied by a certain Strength Reduction Factor (Φ),
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP),
As we are concerned with the MOMENT action on the given beam, then, as per Table 421.2.1, the corresponding Strength Reduction Factor (Φ) to be used would be,
Referring to Sec. 421.2.2 in the Code takes you to a more specific table, Table 421.2.2, where,
Note that the εT we have computed in Q#5 (where εT = 0.005) only pertains to the MINIMUM εT in solving the amount of maximum steel area As,max for the beam to still be considered as a tension-controlled section. At this point, we have yet to determine the ACTUAL εT and the ACTUAL distance c based on the given area of steel As (the 4 sets of 25mm-diameter rebars, or As = 1,963.49mm2) in the problem.
So, let's first check our actual distance c, which we can then use to find out the actual net tensile strain (εT),
Next, let's determine our actual net steel strain (εT), by using a proportion equation, to know which Strength-Reduction Factor (Φ) to use,
Now that we know, from our above calculations, that the beam in our given problem has indeed a tension-controlled section (since εT = 0.019827 > 0.005), and the beam's transverse reinforcement are closed-loop stirrups (NOT spirals), then the Strength-Reduction Factor (Φ) we must use, according to Table 421.2.2 of the Code, would be,
So, as we have already solved the nominal bending strength (Mn) of the beam in Q#6 as [~274kN-m], then,
Therefore, the correct answer is b. Approx. 246.6kN-m.
Mu = Φ Mn
Mu = Φ [~274kN-m]
Mu = [0.90] [~274kN-m]
Mu = ~246.6kN-m
Mu = Φ [~274kN-m]
εT = εcu ( d - c ) / c
εT = 0.003 ( [537.5mm ] - c ) / c
εT = 0.003 ( [537.5mm] - [70.64mm] ) / [70.64mm]
εT = 0.019827
εT = 0.003 ( [537.5mm ] - c ) / c
c = As fy / ( 0.85 f'c bw β1 )
c = [1,963.49mm2] [275MPa] / ( [0.85] [31.03MPa] [350mm] [0.828] )
c = 70.64mm
c = [1,963.49mm2] [275MPa] / ( [0.85] [31.03MPa] [350mm] [0.828] )
c = 70.64mm
εT = 0.003 ( [537.5mm] - [70.64mm] ) / [70.64mm]
εT = 0.019827
εT > 0.005 ∴ tension-controlled section, use Φ = 0.90
Mu = [0.90] [~274kN-m]
Mu = ~246.6kN-m
Therefore, the correct answer is b. Approx. 246.6kN-m.
MY EXPLANATION AND DERIVATION OF ANSWER:
The Ultimate Moment Capacity (Mu) of a beam section is simply the nominal moment or bending strength (Mn) multiplied by a certain Strength Reduction Factor (Φ),
Mu = Φ Mn
According to the National Structural Code of the Philippines, 7th Ed. (2015 NSCP),
As we are concerned with the MOMENT action on the given beam, then, as per Table 421.2.1, the corresponding Strength Reduction Factor (Φ) to be used would be,
"Φ = 0.65 to 0.90 in accordance with Sec. 421.2.2" (Moment action)
(Table 421.2.1.a of the 2015 NSCP)
(Table 421.2.1.a of the 2015 NSCP)
Referring to Sec. 421.2.2 in the Code takes you to a more specific table, Table 421.2.2, where,
Note that the εT we have computed in Q#5 (where εT = 0.005) only pertains to the MINIMUM εT in solving the amount of maximum steel area As,max for the beam to still be considered as a tension-controlled section. At this point, we have yet to determine the ACTUAL εT and the ACTUAL distance c based on the given area of steel As (the 4 sets of 25mm-diameter rebars, or As = 1,963.49mm2) in the problem.
So, let's first check our actual distance c, which we can then use to find out the actual net tensile strain (εT),
c = As fy / ( 0.85 f'c bw β1 )
c = [1,963.49mm2] [275MPa] / ( [0.85] [31.03MPa] [350mm] [0.828] )
c = 70.64mm
c = [1,963.49mm2] [275MPa] / ( [0.85] [31.03MPa] [350mm] [0.828] )
c = 70.64mm
Next, let's determine our actual net steel strain (εT), by using a proportion equation, to know which Strength-Reduction Factor (Φ) to use,
εT = εcu ( d - c ) / c
εT = 0.003 ( [537.5mm] - [70.64mm] ) / [70.64mm]
εT = 0.019827
εT = 0.003 ( [537.5mm] - [70.64mm] ) / [70.64mm]
εT = 0.019827
εT > 0.005 ===> OK! (Tension-Controlled section achieved)
Now that we know, from our above calculations, that the beam in our given problem has indeed a tension-controlled section (since εT = 0.019827 > 0.005), and the beam's transverse reinforcement are closed-loop stirrups (NOT spirals), then the Strength-Reduction Factor (Φ) we must use, according to Table 421.2.2 of the Code, would be,
"Φ = 0.90" (for Net tensile strain, εT ≥ 0.005, and Type of transverse reinforcement Other than Spirals)
(Table 421.2.2.f of the 2015 NSCP)
(Table 421.2.2.f of the 2015 NSCP)
So, as we have already solved the nominal bending strength (Mn) of the beam in Q#6 as [~274kN-m], then,
Mu = Φ Mn
Mu = [0.90] [~274kN-m]
Mu = ~246.6kN-m
Mu = [0.90] [~274kN-m]
Mu = ~246.6kN-m
Therefore, the correct answer is b. Approx. 246.6kN-m.
Q#8: Based on the given design of the beam, which of the following conditions holds true?
a. Beam section is between tension- and compression-controlled (i.e., Transition).
b. Steel does not yield.
c. The beam has a balanced strain design condition.
d. Beam is only singly-reinforced.
e. None of the above.
Beam section is between tension- and compression-controlled (i.e., Transition)?
Recall that in Q#5, a beam section is considered to be tension-controlled, transition, or compression-controlled, based on the net tensile strain (εT), where,
As we have calculated in Q#7, the actual net tensile strain (εT) of our given beam was,
Given that our beam was calculated to have a net tensile strain of εT = 0.019827 means that our εT ≥ 0.005. As such, the given beam section is tension-controlled. So, choice a.) Beam section is between tension- and compression-controlled (i.e., Transition) is FALSE.
A tension-controlled section is when [ εT ≥ 0.005 ]
A transition section is when [ 0.002 < εT < 0.005 ]
A compression-controlled section is when [ εT < 0.002 ]
A transition section is when [ 0.002 < εT < 0.005 ]
A compression-controlled section is when [ εT < 0.002 ]
As we have calculated in Q#7, the actual net tensile strain (εT) of our given beam was,
εT = εcu ( d - c ) / c
εT = 0.003 ( [537.5mm] - [70.64mm] ) / [70.64mm]
εT = 0.019827
εT = 0.003 ( [537.5mm] - [70.64mm] ) / [70.64mm]
εT = 0.019827
Given that our beam was calculated to have a net tensile strain of εT = 0.019827 means that our εT ≥ 0.005. As such, the given beam section is tension-controlled. So, choice a.) Beam section is between tension- and compression-controlled (i.e., Transition) is FALSE.
Steel does not yield?
Recall that in Q#4, the ideal design is to allow for a ductile failure (under-reinforced strain design), where the steel must yield first before concrete fails. Thus, the Actual Steel Ratio (ρ) must be between the values of the Minimum Steel Ratio (ρmin) and the Balanced Steel Ratio (ρb).
As determined in Q#2, the minimum steel area (As,min) was calculated as [957.73mm2]. Since steel ratio (ρ), as mentioned in Q#3, is [ ρ = As / ( bw d ) ], then the Minimum Steel Ratio (ρmin) would consequently be,
Also in Q#3, we have determined that our Actual Steel Ratio (ρ) was,
By Q#4, we have computed for the Balanced Steel Ratio (ρb), where,
Now, checking if the values we calculated above satisfies the condition ρmin ≤ ρ < ρb, where steel will yield,
So, choice b.) Steel does not yield is FALSE.
ρmin ≤ ρ < ρb ==> Condition that must be met where Steel will yield
As determined in Q#2, the minimum steel area (As,min) was calculated as [957.73mm2]. Since steel ratio (ρ), as mentioned in Q#3, is [ ρ = As / ( bw d ) ], then the Minimum Steel Ratio (ρmin) would consequently be,
ρmin = As,min / ( bw d )
ρmin = [957.73mm2] / ( [350mm] [537.5mm] )
ρmin = 0.005091
ρmin = [957.73mm2] / ( [350mm] [537.5mm] )
ρmin = 0.005091
Also in Q#3, we have determined that our Actual Steel Ratio (ρ) was,
ρ = As / ( bw d )
ρ = [1,963.49mm2] / ( [350mm] [537.5mm] )
ρ = 0.010437
ρ = [1,963.49mm2] / ( [350mm] [537.5mm] )
ρ = 0.010437
By Q#4, we have computed for the Balanced Steel Ratio (ρb), where,
ρb = 0.85 f'c β1 600MPa / ( fy ( 600MPa + fy ) )
ρb = 0.85 [31.03MPa] [0.828] 600MPa / ( [275MPa] ( 600MPa + [275MPa] ) )
ρb = 0.054455
ρb = 0.85 [31.03MPa] [0.828] 600MPa / ( [275MPa] ( 600MPa + [275MPa] ) )
ρb = 0.054455
Now, checking if the values we calculated above satisfies the condition ρmin ≤ ρ < ρb, where steel will yield,
[ ρmin = 0.005091 ] ≤ [ ρ = 0.010437 ] < [ ρb = 0.054455 ] ==> OK! (Steel DOES yield)
So, choice b.) Steel does not yield is FALSE.
The beam has a balanced strain design condition?
Just as have been mentioned above, our Actual Steel Ratio ρ = 0.010437 which is less than the Balanced Steel Ratio ρb = 0.054455 signifies that the given beam is below balanced strain conditions.
Recall also in Q#4 that the strain of steel at yield point is [ εy = fy / 200,000MPa ], so since our given beam's fy is [275MPa], then,
A balanced strain design condition [ εs = εy ] is where both steel and concrete would have failed simultaneously. Since our net tensile strain was calculated in Q#7 as εT = 0.019827 (which serves as our service load strain εs), such tensile strain falls within the tension-controlled zone (where εT > 0.005) and at a point greater than the strain at yield point of steel (εy = 0.001375 or 0.002), where our [ εs > εy ].
So, choice c.) The beam has a balanced strain design condition is FALSE.
Recall also in Q#4 that the strain of steel at yield point is [ εy = fy / 200,000MPa ], so since our given beam's fy is [275MPa], then,
εy = [275MPa] / 200,000MPa
εy = 0.001375 (which by Code, permitted to be taken as 0.002, for a Grade 280 deformed reinforcement)
εy = 0.001375 (which by Code, permitted to be taken as 0.002, for a Grade 280 deformed reinforcement)
A balanced strain design condition [ εs = εy ] is where both steel and concrete would have failed simultaneously. Since our net tensile strain was calculated in Q#7 as εT = 0.019827 (which serves as our service load strain εs), such tensile strain falls within the tension-controlled zone (where εT > 0.005) and at a point greater than the strain at yield point of steel (εy = 0.001375 or 0.002), where our [ εs > εy ].
So, choice c.) The beam has a balanced strain design condition is FALSE.
Beam is only singly-reinforced?
The term SINGLY-REINFORCED means that the steel in the beam ONLY handles tensile stresses (on the tension side) of such beam. On the other hand, the term DOUBLY-REINFORCED means that the steel in the beam handles BOTH tensile stresses (on the tension side) AND compressive stresses (on the compression side) of such beam.
When a beam is OVER-REINFORCED (where the area of tensile steel is too large), the brittle concrete will crush first when its maximum concrete strain is reached ( at εcu = 0.003 ) while the ductile tensile steel has NOT YET reached its yield point ( at εy ) due to its large steel area. As such, one solution (among others) is to provide ADDITIONAL STEEL REINFORCEMENT AT THE COMPRESSION SIDE of the beam to prevent the brittle concrete from failing immaturely and allowing the steel at the tension side of the beam to yield first. This makes the beam a DOUBLY-REINFORCED BEAM.
But, as we have already determined in our analysis of the other choices above which were proven to be false, our given beam section is already within the tensile-controlled region, where tensile steel will yield first. Thus, our beam would have NO need for any additional compression steel reinforcement (since the tensile steel is more than enough to produce a ductile beam) which classifies it as a SINGLY-REINFORCED BEAM.
So, choice d.) Beam is only singly-reinforced is TRUE.
When a beam is OVER-REINFORCED (where the area of tensile steel is too large), the brittle concrete will crush first when its maximum concrete strain is reached ( at εcu = 0.003 ) while the ductile tensile steel has NOT YET reached its yield point ( at εy ) due to its large steel area. As such, one solution (among others) is to provide ADDITIONAL STEEL REINFORCEMENT AT THE COMPRESSION SIDE of the beam to prevent the brittle concrete from failing immaturely and allowing the steel at the tension side of the beam to yield first. This makes the beam a DOUBLY-REINFORCED BEAM.
But, as we have already determined in our analysis of the other choices above which were proven to be false, our given beam section is already within the tensile-controlled region, where tensile steel will yield first. Thus, our beam would have NO need for any additional compression steel reinforcement (since the tensile steel is more than enough to produce a ductile beam) which classifies it as a SINGLY-REINFORCED BEAM.
So, choice d.) Beam is only singly-reinforced is TRUE.
Therefore, the correct answer is d. Beam is only singly-reinforced.
The given beam in my problem, showing its strain and stress diagrams, with all values as calculated.
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