Sample Problem 4: Analysis of SRRB Strength Capacity
A rectangular beam with b = 250 mm and d = 460 mm is reinforced for tension only with 3-25 mm
bars. The beam is simply supported over a span of 6 m and carries a uniform dead load of 680 N/m
including its own weight. Calculate the uniform live load that the beam can carry. Assume fy = 276.5
MPa and f ’c = 32 MPa. Use NSCP 2015 Provisions.
Solution:
Solve for α from C = T, Assume steel yields first
which mCheck if fs ≥ fy
0.85f ′ cαb = AS fy
Check if fs ≥ fy .
600 d − c
fs =
c
Determine type of failure to know ∅.
Solve for ∅Mn
′
α
∅M = ∅0.85f cαb � d −
α
2
∅M = ∅As fy � d − (YIELDING)
2
α
∅M = ∅As fs � d − (NON-YIELDING)
2
Calculate the uniform live load that the beam can
carry
Prepared by: Engr. John Mark G. Payawal, MSCE 158
Solution to Problem 4: Analysis of SRRB Strength Capacity
α
0.85f ′ c εc = 0.003 0.05 f ′ 𝑐𝑐 − 28
2 β1 = 0.85 −
x α 𝐂𝐂
c 7
= 𝟎𝟎. 𝟖𝟖𝟖𝟖𝐟𝐟 ′ 𝐜𝐜𝛂𝛂𝐛𝐛
N.A 0.05 32 − 28
α β1 = 0.85 −
d = 460 d−
2 d−c
7
mm
3 − 25 𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝐓𝐓 = 𝐀𝐀𝐬𝐬 𝐟𝐟𝐲𝐲
fs
𝛃𝛃𝟏𝟏 = 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 (𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬)
or T = As fs εt =
STRESS BLOCK (non-yielding) Es
b = 250
α
STRAIN DIAGRAM
Step 1: Solve for 𝛂𝛂 from 𝐂𝐂 = 𝐓𝐓, Assume steel 600 d −
β1
yields first, 𝐟𝐟𝐬𝐬 ≥ 𝐟𝐟𝐲𝐲 fs = α
0.85f ′ cαb = AS fy β1
AS fy 𝐟𝐟𝐬𝐬 = 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑. 𝟏𝟏𝟏𝟏 𝐌𝐌𝐌𝐌𝐌𝐌
α=
0.85f ′ cb Since fs > fy , assumption in step 1 is correct.
𝛂𝛂 = 𝟓𝟓𝟓𝟓. 𝟖𝟖𝟖𝟖𝟖𝟖 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
Step 2: Check if 𝐟𝐟𝐬𝐬 ≥ 𝐟𝐟𝐲𝐲 Step 3: Determine type of failure to know ∅.
600 d − c 0.003 d − c
fs = εs =
c c
α = cβ1 𝛆𝛆𝐬𝐬 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
Since 28 < f′c = 32 MPa < 55, Since 𝛆𝛆𝐬𝐬 > 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎, 𝐓𝐓. 𝐂𝐂. 𝐅𝐅, ∅ = 𝟎𝟎. 𝟗𝟗. (see page 145 of lecture)
𝛃𝛃𝟏𝟏 =?
Prepared by: Engr. John Mark G. Payawal, MSCE 159
Solution to Problem 4: Analysis of SRRB Strength Capacity
STEP 4: Solve for ∅𝐌𝐌𝐧𝐧
α ′
∅M = ∅0.85f cαb � d −
2
59.879
∅M = (0.9)0.85(32)(59.879)(250) � 460 −
2
∅𝐌𝐌 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔 𝐤𝐤𝐤𝐤 − 𝐦𝐦
α
∅M = ∅As fy � d − (YIELDING)
2
α
∅M = ∅As fs � d − (NON-YIELDING)
2
Since fs > fy , we’ll use
α
∅M = ∅As fy � d − (YIELDING)
2
π 2
59.879
∅M = 0.9 (3 � � 25 )(276.5) � 460 −
4 2
∅𝐌𝐌 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔 𝐤𝐤𝐤𝐤 − 𝐦𝐦
∴, 𝐭𝐭𝐭𝐭𝐭𝐭 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬, ∅𝐌𝐌𝐧𝐧 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟔𝟔𝟔𝟔 𝐤𝐤𝐤𝐤 − 𝐦𝐦
Prepared by: Engr. John Mark G. Payawal, MSCE 160
Solution to Problem 4: Analysis of SRRB Strength Capacity
STEP 5: Calculate the uniform live load that the beam can carry
Wu
6
Design strength, Mu = ∅M = 157.60 kN − m = Mmax
Wul2
From the simply support beam, Mmax =
8
2
Wu(6)
157.60 kN − m =
8
kN
Wu = 35.02 (store)
m
From NSCP 2015, Dead load and Live load combination is given as,
Wu = 1.2D + 1.6L
Wu = 1.2 068kN/m + 1.6L
kN
35.02 = 1.2 068kN/m + 1.6L
m
𝐋𝐋 = 𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑/𝐦𝐦
∴, 𝐭𝐭𝐭𝐭𝐭𝐭uniform live load that the beam can carry 𝐢𝐢𝐢𝐢 𝐋𝐋 = 𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑𝟑𝟑𝟑𝟑/𝐦𝐦
Prepared by: Engr. John Mark G. Payawal, MSCE 161
Sample Problem 5(Board Exam ): Analysis of SRRB Strength Capacity
A 350 mm x 500 mm rectangular beam is reinforced for tension only with 5-28 mm bars. The beam
has an effective depth of 446 mm. The beam carries a uniform dead load of 4.5 kN/m (including its
own weight), a uniform live load of 3 kN/m, and a concentrated load P and 2P as shown. Assume fy
= 420 MPa and f ’c = 34.5 MPa. Calculate:
1. The ultimate moment capacity of the section in kN-m, and
2. the maximum value of P in kN.
2P
P
2m 2m 2m
Prepared by: Engr. John Mark G. Payawal, MSCE 162
Solution to Problem 5(Board Exam ): Analysis of SRRB Strength Capacity
α
0.85f ′ c εc = 0.003 0.05 f ′ c − 28
2 β1 = 0.85 −
x α 𝐂𝐂
c 7
= 𝟎𝟎. 𝟖𝟖𝟖𝟖𝐟𝐟 ′ 𝐜𝐜𝛂𝛂𝐛𝐛
N.A 0.05 34.5 − 28
α β1 = 0.85 −
d = 446 d−
2 d−c
7
mm
5 − 28 𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝐓𝐓 = 𝐀𝐀𝐬𝐬 𝐟𝐟𝐲𝐲
fs
𝛃𝛃𝟏𝟏 = 𝟎𝟎. 𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖𝟖 (𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬)
or T = As fs εt =
STRESS BLOCK (non-yielding) Es
b = 350
α
STRAIN DIAGRAM
Step 1: Solve for 𝛂𝛂 from 𝐂𝐂 = 𝐓𝐓, Assume steel 600 d −
β1
yields first, 𝐟𝐟𝐬𝐬 ≥ 𝐟𝐟𝐲𝐲 fs = α
0.85f ′ cαb = AS fy β1
AS fy 𝐟𝐟𝐬𝐬 = 𝟏𝟏𝟏𝟏𝟏𝟏𝟏𝟏. 𝟖𝟖𝟖𝟖 𝐌𝐌𝐌𝐌𝐌𝐌
α=
0.85f ′ cb Since fs > fy , assumption in step 1 is correct.
𝛂𝛂 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟗𝟗𝟗𝟗 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
Step 2: Check if 𝐟𝐟𝐬𝐬 ≥ 𝐟𝐟𝐲𝐲 Step 3: Determine type of failure to know ∅.
600 d − c 0.003 d − c
fs = εs =
c c
α = cβ1 𝛆𝛆𝐬𝐬 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎
Since 28 < f′c = 34.5 MPa < 55, Since 𝛆𝛆𝐬𝐬 > 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎, 𝐓𝐓. 𝐂𝐂. 𝐅𝐅, ∅ = 𝟎𝟎. 𝟗𝟗. (see page 145 of lecture)
𝛃𝛃𝟏𝟏 =?
Prepared by: Engr. John Mark G. Payawal, MSCE 163
Solution to Problem 5(Board Exam ): Analysis of SRRB Strength Capacity
STEP 4: Solve for ∅𝐌𝐌𝐧𝐧
α ′
∅M = ∅0.85f cαb � d −
2
125.99
∅M = (0.9)0.85(34.5)(125.99)(350) � 446 −
2
∅𝐌𝐌 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟕𝟕𝟕𝟕 𝐤𝐤𝐤𝐤 − 𝐦𝐦
α
∅M = ∅As fy � d − (YIELDING)
2
α
∅M = ∅As fs � d − (NON-YIELDING)
2
Since fs > fy , we’ll use
α
∅M = ∅As fy � d − (YIELDING)
2
π 2
125.99
∅M = 0.9 (5 � � 28 )(420) � 446 −
4 2
∅𝐌𝐌 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟕𝟕𝟕𝟕 𝐤𝐤𝐤𝐤 − 𝐦𝐦
∴, 𝐭𝐭𝐭𝐭𝐭𝐭 𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝𝐝 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬, ∅𝐌𝐌𝐧𝐧 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟕𝟕𝟕𝟕 𝐤𝐤𝐤𝐤 − 𝐦𝐦
Prepared by: Engr. John Mark G. Payawal, MSCE 164
Solution to Problem 5(Board Exam ): Analysis of SRRB Strength Capacity
STEP 5: Calculate the maximum value of P At point B: Isolating left side of B
2P
in kN
P
A Wu C
2m B 2m 2m D MuB
Maximum moment can occur at B and C.
Wu = 1.2D + 1.6L MuB = ∅M = 445.73 kN − m
Wu = 1.2(4.5) + 1.6(3) MuB = RA 2 − 10.2 2 1
Wu = 10.2 kN/m Solving for RA,
At point C: Isolating right side of C Taking moment at C (use original beam),
RA 4 − 10.2 6 1 − 1.2 2P 2 + 1.2 P 2 = 0
RA = 15.3 + 0.6P
MuC
MuB = RA 2 − 10.2 2 1
MuC = ∅M = 445.73 kN − m 445.73 = RA 2 − 10.2 2 1
MuC = 1.2P 2 + 10.2 2 1 445.73 = 15.3 + 0.6P 2 − 10.2 2 1
445.73 = 1.2P 2 + 10.2 2 1 𝐏𝐏 = 𝟑𝟑𝟑𝟑𝟑𝟑. 𝟗𝟗𝟗𝟗 𝐤𝐤𝐤𝐤
𝐏𝐏 = 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐 𝐤𝐤𝐤𝐤
∴, the maximum value of P in kN 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 𝐭𝐭𝐭𝐭𝐭𝐭𝐭𝐭 𝐌𝐌𝐮𝐮 𝐰𝐰𝐰𝐰𝐰𝐰𝐰𝐰 𝐧𝐧𝐧𝐧𝐧𝐧 𝐞𝐞𝐞𝐞𝐞𝐞𝐞𝐞𝐞𝐞𝐞𝐞 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟕𝟕𝟕𝟕 𝐤𝐤𝐤𝐤 − 𝐦𝐦 𝐢𝐢𝐢𝐢 𝟏𝟏𝟏𝟏𝟏𝟏. 𝟐𝟐𝟐𝟐 𝐤𝐤𝐤𝐤
Prepared by: Engr. John Mark G. Payawal, MSCE 165
Sample Problem 6: Analysis of SRRB Strength Capacity
A rectangular beam has b = 300 mm and d = 500 mm. It is reinforced in tension only at the bottom
with As = 6-32 mm bars. f ’c = 27.6 MPa and fy = 414 MPa. Calculate the ultimate moment capacity
of the beam.
Solution:
Assume “yielding” (fs ≥ fy ), state and equate C = T
to solve α.
0.85f ′ cαb = AS fy (yielding)
Check if fs ≥ fy . If yes, accept α (depth of
compression block).
600 d − c
fs =
c
If no, Use C = T (non-yielding) to solve α.
0.85f ′ cαb = As fs
Determine type of failure to know ∅.
Solve for ∅Mn
′
α
∅M = ∅0.85f cαb � d −
2
α
∅M = ∅As fy � d − (YIELDING)
2
α
∅M = ∅As fs � d − (NON-YIELDING)
2
Prepared by: Engr. John Mark G. Payawal, MSCE 166
Solution to Problem 6: Analysis of SRRB Strength Capacity
0.85f ′ c εc = 0.003 α
α
600 d −
2
α β1
x 𝐂𝐂
= 𝟎𝟎. 𝟖𝟖𝟖𝟖𝐟𝐟 ′ 𝐜𝐜𝛂𝛂𝐛𝐛
c fs = α
N.A
α β1
d−
d = 500 2 d−c 𝐟𝐟𝐬𝐬 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟑𝟑𝟑𝟑 𝐌𝐌𝐌𝐌𝐌𝐌
mm
6 − 32 𝑚𝑚𝑚𝑚 𝑏𝑏𝑏𝑏𝑏𝑏𝑏𝑏 𝐓𝐓 = 𝐀𝐀𝐬𝐬 𝐟𝐟𝐲𝐲
fs
Since fs < fy , assumption in step 1 is incorrect.
or T = As fs εt =
STRESS BLOCK (non-yielding) Es
b = 300
We need to recompute α using the equation:
STRAIN DIAGRAM
Step 1: Solve for 𝛂𝛂 from 𝐂𝐂 = 𝐓𝐓, Assume steel
yields first, 𝐟𝐟𝐬𝐬 ≥ 𝐟𝐟𝐲𝐲 0.85f ′ cαb = As fs
600 d − c
0.85f ′ cαb = AS fy ′
0.85f cαb = As
AS fy c
α= ′
600 d − c
0.85f ′ cb 0.85f ccβ1 b = As
𝛂𝛂 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟖𝟖𝟖𝟖𝟖𝟖 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬 c
𝜋𝜋 2
600 d − c
0.85 27.6 𝑐𝑐 0.85 300 = 6 � � 32
4 c
Step 2: Check if 𝐟𝐟𝐬𝐬 ≥ 𝐟𝐟𝐲𝐲 𝐜𝐜 = 𝟑𝟑𝟑𝟑𝟔𝟔. 𝟐𝟐𝟐𝟐𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
600 d − c
fs = 𝛂𝛂 = 𝟐𝟐𝟐𝟐𝟐𝟐. 𝟐𝟐𝟐𝟐𝟐𝟐 𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬𝐬
c
α = cβ1
Since 17 ≤ f′c = 27.6 MPa ≤ 28,
𝛃𝛃𝟏𝟏 = 𝟎𝟎. 𝟖𝟖𝟖𝟖
Prepared by: Engr. John Mark G. Payawal, MSCE 167
Solution to Problem 6: Analysis of SRRB Strength Capacity
Step 3: Determine type of failure to know ∅.
0.003 d − c
εs =
c
𝛆𝛆𝐬𝐬 = 𝟎𝟎. 𝟎𝟎𝟎𝟎𝟎𝟎𝟎𝟎 (store)
fy
Sinceεs ≤ εty 𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦𝑦 𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠𝑠 =
Es
fy 414 𝑀𝑀𝑀𝑀𝑀𝑀
= = 0.00207
Es 200000 𝑀𝑀𝑀𝑀𝑀𝑀
𝐂𝐂. 𝐂𝐂. 𝐅𝐅, ∅ = 𝟎𝟎. 𝟔𝟔𝟔𝟔. (see page 145 of lecture)
Prepared by: Engr. John Mark G. Payawal, MSCE 168
Solution to Problem 6: Analysis of SRRB Strength Capacity
STEP 4: Solve for ∅𝐌𝐌𝐧𝐧
α ′
∅M = ∅0.85f cαb � d −
2
260.298
∅M = (0.65)(0.85)(27.6)(260.298)(300) � 500 −
2
∅𝐌𝐌 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟒𝟒𝟒𝟒 𝐤𝐤𝐤𝐤 − 𝐦𝐦
α
∅M = ∅As fy � d − (YIELDING)
2
α
∅M = ∅As fs � d − (NON-YIELDING)
2
Since fs < fy , we’ll use
α
∅M = ∅As fs � d − (NON-YIELDING)
2
α
π 600 d − 260.298
2 β1
∅M = 0.65 (6 � � 32 )( α ) � 500 −
4 2
β1
∅𝐌𝐌 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟒𝟒𝟒𝟒 𝐤𝐤𝐤𝐤 − 𝐦𝐦
∴, The ultimate moment capacity of the beam, ∅𝐌𝐌𝐧𝐧 = 𝟒𝟒𝟒𝟒𝟒𝟒. 𝟒𝟒𝟒𝟒 𝐤𝐤𝐤𝐤 − 𝐦𝐦
Prepared by: Engr. John Mark G. Payawal, MSCE 169
