Bending and Shear in Beams

Transcription

Bending and Shear in Beams
Lecture 3
5th October 2016
Contents – Lecture 3
• Bending/ Flexure
–
–
–
–
Section analysis, singly and doubly reinforced
Tension reinforcement, As
neutral axis depth limit & K’
Compression reinforcement, As2
• Flexure Worked Example – Doubly reinforced
• Shear in Beams - Variable strut method
• Beam Examples – Bending, Shear & High shear
• Exercise - Design a beam for flexure and shear
EC2 Webinar – Autumn 2016
Lecture 3/1
Bending/ Flexure
Section Design: Bending
• In principal flexural design is generally the same as
BS8110
• EC2 presents the principles only
• Design manuals will provide the standard solutions
for basic design cases.
• There are modifications for high strength concrete
( fck > 50 MPa )
Note: TCC How to guide equations and equations used on
this course are based on a concrete fck ≤ 50 MPa
Lecture 3/2
Section Analysis to determine
Tension & Compression Reinforcement
EC2 contains information on:
•
•
•
•
Concrete stress blocks
Reinforcement stress/strain curves
The maximum depth of the neutral axis, x. This depends on
the moment redistribution ratio used, δ.
The design stress for concrete, fcd and reinforcement, fyd
In EC2 there are no equations to determine As, tension steel, and As2,
compression steel, for a given ultimate moment, M, on a section.
Equations, similar to those in BS 8110, are derived in the following
slides. As in BS8110 the terms K and K’ are used:
M
bd 2 f ck
K =
Value of K for maximum value of M
with no compression steel and
when x is at its maximum value.
If K > K’ Compression steel required
Rectangular Concrete Stress Block
EC2: Cl 3.1.7, Fig 3.5
εcu3
η fcd
Fc
Ac
Concise:
Fig 6.1
λx
x
Remember this
from last week?
d
As
Fs
εs
fck ≤ 50 MPa
50 < fck ≤ 90 MPa
λ
0.8
= 0.8 – (fck – 50)/400
η
1.0
= 1,0 – (fck – 50)/200
fck
50
55
60
70
80
90
λ
0.8
0.79
0.78
0.75
0.73
0.7
η
1
0.98
0.95
0.9
0.85
0.8
fcd = αcc fck /γc = 0.85 fck /1.5
For fck ≤ 50 MPa failure concrete strain, εcu, = 0.0035
Lecture 3/3
Reinforcement
Design Stress/Strain Curve
EC2: Cl 3.2.7, Fig 3.8
Idealised
σ
kfyk
In UK fyk = 500 MPa
kfyk/γs
fyk
fyd = fyk/γs
fyd = fyk/γs = 500/1.15 = 435 MPa
Design
Es may be taken to be 200 GPa
Steel yield strain
(εs at yield point)
= fyd/Es
= 435/200000
= 0.0022
fyd/ Es
ε ud εuk
ε
At failure concrete strain is 0.0035 for fck ≤ 50 MPa.
If x/d is 0.6 steel strain is 0.0023 and this is past the yield point.
Design steel stress is 435 MPa if neutral axis, x, is less than 0.6d.
Analysis of a singly reinforced beam
EC2: Cl 3.1.7
Design equations can be derived as follows:
b
M
For grades of concrete up to C50/60, εcu= 0.0035, η = 1 and λ = 0.8.
fcd = 0.85fck/1.5
fyd = fyk/1.15 = 0.87 fyk
Fc = (0.85 fck / 1.5) b (0.8 x) = 0.453 fck b x
For no compression
Fst = 0.87As fyk
reinforcement Fsc = 0
Methods to find As:
• Iterative, trial and error method – simple but not practical
• Direct method of calculating z, the lever arm, and then As
Lecture 3/4
Determine As – Iterative method
b
M
For horizontal equilibrium
Fc= Fst
0.453 fck b x = 0.87As fyk
Guess As
Solve for x
z = d - 0.4 x
M = Fc z
Stop when design applied BM, MEd ≃ M
Determine As – Direct method
Take moments about the centre of the tension force, Fst:
M
Now z
= Fc z = 0.453 fck b x z
(1)
= d - 0.4 x
∴
x
= 2.5(d - z)
∴
M
= 0.453 fck b 2.5(d - z) z
M
= 1.1333 (fck b z d - fck b z2)
Let
K = M / (fck b d 2)
K=
 fckbdz
M
fckbz 2 

= 1.1333 
2
2
fckbd
fckbd 2 
 fckbd
(K may be considered as the normalised bending resistance)
∴
0
= 1.1333 [(z/d)2 – (z/d)] + K
0
= (z/d)2 – (z/d) + 0.88235K
Lecture 3/5
0
= (z/d)2 – (z/d) + 0.88235K
Solving the quadratic equation:
= [1 + (1 -
z/d
z
M
3.53K)0.5]/2
= d [ 1 + (1 - 3.53K)0.5]/2
The lever arm for an applied moment is now known
Quadratic formula
Higher Concrete Strengths
fck ≤ 50MPa
z = d[1 + (1− 3.529K )]/2
fck = 60MPa
z = d[1 + (1− 3.715K )]/2
fck = 70MPa
z = d[1+ (1− 3.922K )]/2
fck = 80MPa
z = d[1+ (1− 4.152K )]/2
fck = 90MPa
z = d[1+ (1− 4.412K )]/2
Normal strength
Lecture 3/6
Tension steel, As
Concise: 6.2.1
Take moments about the centre of the compression force, Fc:
= Fst z = 0.87As fyk z
M
Rearranging
As
= M /(0.87 fyk z)
The required area of reinforcement can now be found using three
methods:
a) calculated using these expressions
b) obtained from Tables of z/d (eg Table 5 of How to beams or
Concise Table 15.5, see next slide)
c) obtained from graphs (eg from the ‘Green Book’ or Fig B.3 in
Concrete Buildings Scheme Design Manual, next slide but one)
Design aids for flexure K = M / (fck b d
2)
method (b)
Concise: Table 15.5
.
Traditionally z/d was
limited to 0.95 max to
avoid issues with the
quality of ‘covercrete’.
‘Normal’ tables and
charts are only valid
up to C50/60
Lecture 3/7
Design aids for flexure-
method (c)
K = M / (fck b d 2)
TCC Concrete Buildings Scheme Design Manual, Fig B.3
Design chart for singly reinforced beam
Maximum neutral axis depth
EC2: Cl 5.5 Linear elastic analysis with limited redistribution
Concise:
Table 6.1
According to Cl 5.5(4) the depth of the neutral axis is limited, viz:
δ
≥ k1 + k2 xu/d
where
k1
= 0.4
k2
= 0.6 + 0.0014/ εcu2 = 0.6 + 0.0014/0.0035 = 1
xu = depth to NA after redistribution
δ =
Redistributed Bending Moment
Elastic Bending Moment
= Redistribution ratio
∴ xu ≤ d (δ - 0.4)
Therefore there are limits on K and
this limit is denoted K’
For K > K’ Compression steel needed
Lecture 3/8
K’ and Beams with Compression
Reinforcement, As2
Concise: 6.2.1
The limiting value for K (denoted K’) can be calculated as follows:
As before
and
… (1)
M = 0.453 fck b x z
K = M / (fck b d 2)
&
z = d – 0.4 x
& xu = d (δ – 0.4)
Substituting xu for x in eqn (1) and rearranging:
M’ = b d2 fck (0.6 δ – 0.18 δ 2 - 0.21)
∴
K’ = M’ /(b d2 fck)
= (0.6 δ – 0.18 δ 2 - 0.21)
Min δ = 0.7 (30% redistribution). Steel to be either Class B or C for 20% to
30% redistribution.
Some engineers advocate taking x/d < 0.45, and ∴K’ < 0.168. It is often
considered good practice to limit the depth of the neutral axis to avoid
‘over-reinforcement’ to ensure a ductile failure. This is not an EC2
requirement and is not accepted by all engineers.
Note: For plastic analysis xu/d must be ≤ 0.25 for normal strength concrete,
EC2 cl 5.6.2 (2).
Compression steel, As2
EC2: Fig 3.5
Concise: 6.2.1
For K > K’ compression reinforcement As2 is required.
As2 can be calculated by taking moments about the centre of the
tension force:
M = K’ fck b d 2 + 0.87 fyk As2 (d - d2)
Rearranging
As2 = (K - K’) fck b d 2 / (0.87 fyk (d - d2))
Lecture 3/9
Tension steel, As
for beams with Compression Reinforcement,
The concrete in compression is at its design
capacity and is reinforced with compression
reinforcement. So now there is an extra force:
Fsc = 0.87As2 fyk
The area of tension reinforcement can now be considered in two
parts.
The first part balances the compressive force in the concrete
(with the neutral axis at xu).
The second part balances the force in the compression steel.
The area of reinforcement required is therefore:
As = K’ fck b d 2 /(0.87 fyk z) + As2
where z is calculated using K’ instead of K
Design Flowchart
The following flowchart outlines the design procedure for rectangular
beams with concrete classes up to C50/60 and grade 500 reinforcement
Carry out analysis to determine design moments (M)
Determine K and K’ from:
M
K=
& K ' = 0.6δ − 0.18δ 2 − 0.21
b d 2 fck
Note: δ =1.0 means no redistribution and δ = 0.8 means 20% moment redistribution.
Yes
Is K ≤ K’ ?
No compression steel
needed – singly reinforced
No
δ
K’
1.00
0.208
0.95
0.195
0.90
0.182
0.85
0.168
0.80
0.153
0.75
0.137
0.70
0.120
Compression steel needed doubly reinforced
It is often recommended in the UK that K’ is limited to 0.168 to ensure ductile failure
Lecture 3/10
Flow Chart for Singly-reinforced
Beam/Slab
K ≤ K’
Calculate lever arm z from:
z=
d
[1 + 1 − 3.53K ] ≤ 0.95d *
2
(Or look up z/d from table or from chart.)
* A limit of 0.95d is considered good practice, it is not a requirement of Eurocode 2.
Calculate tension steel required from:
As =
M
fyd z
Check minimum reinforcement requirements:
As,min ≥
0.26 fctm bt d
≥ 0.0013 bt d
fyk
(Cl.9.2.1.1)
Exp. (9.1N)
Check max reinforcement provided As,max ≤ 0.04Ac (Cl. 9.2.1.1)
Check min spacing between bars > Øbar > 20 > Agg + 5
Check max spacing between bars
Minimum Reinforcement Area
EC2: Cl. 9.2.1.1, Exp. 9.1N
The minimum area of reinforcement for beams and slabs is given by:
As,min ≥
0.26 fctm bt d
≥ 0.0013 bt d
fyk
Lecture 3/11
Flow Chart for DoublyReinforced Beam
K > K’
Calculate lever arm z from:
Calculate excess moment from:
z=
d
1 + 1 − 3.53K '
2
[
]
M ' = bd 2fck (K − K ' )
Calculate compression steel required from:
M'
As 2 =
fyd (d − d 2 )
Calculate tension steel required from:
As =
K ' fck bd 2
+ As 2
fyd z
Check max reinforcement provided As,max ≤ 0.04Ac (Cl. 9.2.1.1)
Check min spacing between bars > Øbar > 20 > Agg + 5
Flexure Worked Example
(Doubly reinforced)
Lecture 3/12
Worked Example 1
Design the section below to resist a sagging moment of 370 kNm
assuming 15% moment redistribution (i.e. δ = 0.85).
Take fck = 30 MPa and fyk = 500 MPa.
d
Worked Example 1
Initially assume 32 mm φ for tension reinforcement with 30 mm
nominal cover to the link all round (allow 10 mm for link) and assume
20mm φ for compression reinforcement.
= 50
d = h – cnom - Ølink - 0.5Ø
= 500 – 30 - 10 – 16
= 444 mm
= 444
d2 = cnom + Ølink + 0.5Ø
= 30 + 10 + 10
= 50 mm
Lecture 3/13
Worked Example 1
K ' = 0.168
M
K= 2
bd f ck
370 × 10 6
=
300 × 444 2 × 30
= 0.209 > K '
∴ provide compression steel
[
K’
0.208
0.95
0.195
0.90
0.182
0.85
0.168
0.80
0.153
0.75
0.137
0.70
0.120
]
d
1 + 1 − 3.53K '
2
444
=
1 + 1 − 3.53 × 0.168
2
= 363 mm
z=
δ
1.00
[
]
Worked Example 1
M ' = bd 2fck (K − K ' )
= 300 × 444 2 × 30 × (0.209 − 0.168 ) × 10 −6
= 72.7 kNm
As 2 =
=
M'
f yd (d − d 2 )
72.7 x 10 6
435 × (444 – 50)
= 424 mm2
As =
=
M −M'
+ As 2
f yd z
(370 − 72.7) × 10 6
+ 424
435 × 363
= 2307 mm2
Lecture 3/14
Worked Example 1
Provide
2 H20 for compression steel = 628mm2 (424 mm2 req’d)
and
3 H32 tension steel = 2412mm2 (2307 mm2 req’d)
By inspection does not exceed maximum area (0.04 Ac) or maximum
spacing of reinforcement rules (cracking see week 6 notes)
Check minimum spacing, assuming H10 links
Space between bars = (300 – 30 x 2 - 10 x 2 - 32 x 3)/2
= 62 mm > 32 mm*
…OK
* EC2 Cl 8.2 (2) Spacing of bars for bond:
Clear distance between bars > Ф bar > 20 mm > Agg + 5 mm
Poll Q1:
Design reinforcement strength, fyd
For H type bar reinforcement what is fyd?
a. 435 MPa
b. 460 MPa
c. 476 MPa
d. 500 MPa
Lecture 3/15
Poll Q2:
Neutral axis depth, x
A beam section has an effective depth of 500mm and the
ultimate elastic bending moment has been reduced by 30%.
What is the maximum depth of the neutral axis, xu?
a. 150 mm
b. 250 mm
c. 300 mm
d. 450 mm
Shear in Beams
Variable strut method
Lecture 3/16
Shear
EC2: Cl 6.2.2, 6.2.3, 6.4
Concise: 7.2, 7.3, 8.0
There are three approaches to designing for shear:
• When shear reinforcement is not required e.g. slabs, week 5
Shear check uses VRd,c
• When shear reinforcement is required e.g. Beams
Variable strut method is used to check shear in beams
Strut strength check using VRd,max Links strength using VRd,s
• Punching shear requirements e.g. flat slabs, week 5
The maximum shear strength in the UK should not exceed that of
class C50/60 concrete. Cl 3.1.2 (2) P and NA.
Shear in Beams
EC2: Cl 6.2.3
Concise: 7.3
Shear design is different from BS8110. EC2 uses the variable strut
method to check a member with shear reinforcement.
Definitions:
VEd
- Applied shear force. For predominately UDL, shear may be checked
at d from face of support
VRd,c – Resistance of member without shear reinforcement
VRd,s - Resistance of member governed by the yielding of shear
reinforcement
VRd,max - Resistance of member governed by the crushing of compression
struts
Whilst Eurocode 2 deals in Resistances (capacities), VRd,c ,VRd,s ,VRd,max and
Effect of actions, VEd in kN, in practice, it is often easier to consider shear
strengths vRd, vRd,max and shear stresses, vEd, in MPa.
Lecture 3/17
Members Requiring Shear
Reinforcement
Concise: Fig 7.3
EC2: 6.2.3(1)
compression chord
compression
strut
compression
chord
V(cot θ - cotα )
Fcd
α
d
½z
θ
z = 0.9d
V
s
shear reinforcement
N
½z
M
V
Ftd
tension chord
angle between shear reinforcement and the beam axis
α
Normally links are vertical. α = 90o and cot α is zero
θ
angle between the concrete compression strut and the beam axis
z
inner lever arm. In the shear analysis of reinforced concrete
without axial force, the approximate value z = 0,9d may
normally be used.
Members Requiring Shear
Reinforcement
Concise: Fig 7.3
EC2: 6.2.3(1)
bw
is the minimum width
Asw
Area of the shear reinforcement
fywd
design yield strength = fyk/1.15
fcd
design compressive strength = αccfck/1.5
= fck/1.5
(αcc = 1.0 for shear)
αcw
= 1.0 Coefficient for stress in compression chord
ν1
strength reduction factor concrete cracked in shear
ν1 = ν = 0.6(1-fck/250) Exp (6.6N)
Lecture 3/18
Strut Inclination Method
EC2: Equ. 6.8 & 6.9 for Vertical links
Strut angle limits
Equ 6.9
VRd,max =
α cw bw z ν 1 fcd
cot θ + tan θ
VEd
21.8°° < θ < 45°°
Cot θ = 2.5
Cot θ = 1
Equ 6.8
VRd, s
A
= sw z f ywd cot θ
s
Eurocode 2 vs BS8110:
Shear
Safer
Shear
reinforcement
density
Asfyd/s
Fewer links
Eurocode 2:
BS8110: VR = VC + VS
VRmax
Test results VR
(but more critical)
Minimum links
Shear Strength, VR
Lecture 3/19
Shear Design: Links
Variable strut method allows a shallower strut angle –
hence activating more links.
As strut angle reduces concrete stress increases
Min curtailment
for 45o strut
s
Vhigh
z
Vlow
d
x
Angle = 45
°
z
θ
d
x
V carried on 3 links
Max angle - max shear resistance
Angle = 21.8
°
V carried on 6 links
Min strut angle - Minimum links
Shear Resistance of Sections with
Vertical Shear Reinforcement Concise: 7.3.3
V
z
x
s
V
d
1 ≤ cot θ ≤ 2,5
θ
z
d
x
Basic equations
shear reinforcement control – Exp (6.8)
VRd,s = Asw z fywd cot θ /s
concrete strut control – Exp (6.9)
where:
fywd = fyk/1.15
ν1 = ν = 0.6(1-fck/250)
αcw = 1.0
VRd,max = αcwz bw ν1 fcd /(cotθ + tanθ) = 0.5 z bw ν1 fcd sin 2θ
Lecture 3/20
Shear links
EC2: Cl 6.2.3
Equation 6.9 is first used to determine the strut
angle θ and then equation 6.8 is used to find the
shear link area, Asw, and spacing s.
fck
Equation 6.9 gives VRd,max values for a given strut
angle θ
°
e.g. when cot θ = 2.5 (θ = 21.8 ) Equ 6.9 becomes
VRd,max = 0.138 bw z fck (1 - fck/250)
or in terms of stress:
vRd ,max= 0.138 fck (1 - fck/250)
Values are in the middle column of the table.
Re-arranging equation 6.9 to find θ:
θ = 0.5 sin-1[vRd /(0.20 fck(1 - fck/250))]
Suitable shear links are found from equation 6.8:
Asw /s = vEdbw/( fywd cot θ)
vRd, cot θ vRd, cot θ
= 2.5
= 1.0
20
2.54
3.68
25
3.10
4.50
28
3.43
4.97
30
3.64
5.28
32
3.84
5.58
35
4.15
6.02
40
4.63
6.72
45
5.08
7.38
50
5.51
8.00
vRd ,max values, MPa, for
cot θ = 1.0 and 2.5
EC2 – Shear Flow Chart
for vertical links
Determine vEd where:
vEd = design shear stress [vEd = VEd/(bwz) = VEd/(bw 0.9d)]
Determine the concrete strut capacity vRd when cot θ = 2.5
vRdcot θ = 2.5 = 0.138fck(1-fck/250)
(or look up from table)
Is vRd,cot θ = 2.5 > vEd?
Yes
No
(cot θ = 2.5)
Calculate area of shear
reinforcement:
Asw/s = vEd bw/(fywd cot θ)
Check minimum area, cl 9.2.2:
Asw/s ≥ bw ρw,min
ρw,min = (0.08 √fck)/fyk ≈ 0.001
Is vRd,cot θ = 1.0 > vEd?
Yes
No
Re-size
(cot θ > 1.0)
Determine θ from:
θ = 0.5 sin-1[(vEd/(0.20fck(1-fck/250))]
Check maximum spacing of shear
reinforcement, cl 9.2.2:
sl,max = st,max = 0.75 d
Lecture 3/21
Design aids for shear
Concise Fig 15.1 a)
Short Shear Spans with Direct
Strut Action
EC2: 6.2.3
d
av
d
av
• Where av ≤ 2d the applied shear force, VEd, for a point load
(eg, corbel, pile cap etc) may be reduced by a factor av/2d
where 0.5 ≤ av ≤ 2d provided:
− The longitudinal reinforcement is fully anchored at the support.
− Only that shear reinforcement provided within the central 0.75av is
included in the resistance.
Note: see PD6687-1:2010
Cl 2.14 for more information
Lecture 3/22
Beam examples
Bending, Shear and High Shear
Beam Example 1
Gk = 75 kN/m, Qk = 50 kN/m, assume no redistribution and use EC0
equation 6.10 to calculate ULS loads.
8m
Cover = 40mm to each face
fck = 30
1000
Determine the flexural and shear
reinforcement required
450
(try 10mm links and 32mm main steel)
Lecture 3/23
Beam Example 1 – Bending
ULS load per m = (75 x 1.35 + 50 x 1.5) = 176.25
Mult = 176.25 x 82/8
= 1410 kNm
d
= 1000 - 40 - 10 – 32/2
= 934
K=
M
1410 × 106
= 0.120
=
bd 2fck 450 × 934 2 × 30
K’
= 0.208
K
< K’ ⇒ No compression reinforcement required
z=
934
d
1 + 1 − 3.53K =
1 + 1 − 3.53 x 0.120 = 822 ≤ 0.95d
2
2
[
]
[
]
6
As =
M
1410 x 10
= 3943 mm2
=
fyd z 435 x 822
Provide 5 H32 (4021 mm2)
Beam Example 1 – Shear
Shear force, VEd = 176.25 x 8/2 = 705 kN
(We could take 505 kN @ d from face)
Shear stress:
vEd
= VEd/(bw 0.9d) = 705 x 103/(450 x 0.9 x 934)
= 1.68 MPa
vRdcot θ = 2.5 = 3.64 MPa
fck
vRdcot θ = 2.5 > vEd
vRd, cot θ
=
vRd, cot θ
2.5
1.0
20
2.54
3.68
∴ cot θ = 2.5
25
3.10
4.50
Asw/s = vEd bw/(fywd cot θ)
28
3.43
4.97
Asw/s = 1.68 x 450 /(435 x 2.5)
30
3.64
5.28
Asw/s = 0.70 mm
32
3.84
5.58
35
4.15
6.02
40
4.63
6.72
45
5.08
7.38
50
5.51
8.00
Try H10 links with 4 legs.
Asw
= 314
mm2
s < 314 /0.70 = 448 mm
⇒ provide H10 links at 450 mm spacing
=
Lecture 3/24
Beam Example 1
Provide 5 H32 (4021) mm2)
with
H10 links at 450 mm spacing
Max spacing is 0.75d = 934 x 0.75
= 700 mm
Beam Example 2 – High shear
UDL not dominant
Find the minimum area of
shear reinforcement
required to resist the
design shear force using
EC2.
Assume that:
fck = 30 MPa and
fyd = 500/1.15 = 435 MPa
Lecture 3/25
Find the minimum area of shear reinforcement required to resist
the design shear force using EC2.
fck = 30 MPa and
fck
fyd = 435 MPa
Shear stress:
vEd = VEd/(bw 0.9d)
vRd, cot θ
=
vRd, cot θ
2.5
1.0
20
2.54
3.68
25
3.10
4.50
28
3.43
4.97
= 312.5 x 103/(140 x 0.9 x 500)
30
3.64
5.28
= 4.96 MPa
32
3.84
5.58
35
4.15
6.02
40
4.63
6.72
vRdcot θ = 2.5 < vEd < vRdcot θ = 1.0
45
5.08
7.38
50
5.51
8.00
∴ 2.5 > cot θ > 1.0 ⇒ Calculate θ
=
Calculate θ


v Ed

0
.
20
f
(
1
−
f
/
250
)


ck
ck
θ = 0.5 sin − 1 

4.96


(
)
 0.20 x 30 1- 30 / 250 
θ = 0.5 sin − 1 
= 35.0°
∴ cot θ = 1.43
Asw/s = vEd bw/(fywd cot θ )
Asw/s = 4.96 x 140 /(435 x 1.43)
Asw/s = 1.12 mm
Try H10 links with 2 legs.
Asw = 157 mm2
s < 157 /1.12 = 140 mm
⇒ provide H10 links at 125 mm spacing
Lecture 3/26
Exercise
Lecture 3
Design a beam for flexure and shear
Beam Exercise – Flexure & Shear
Gk = 10 kN/m, Qk = 6.5 kN/m (Use EC0 eq. 6.10)
8m
Cover = 35 mm to each face
450
fck = 30MPa
Design the beam in flexure and shear
300
Lecture 3/27
Aide memoire
Exp (6.10)
Remember
this from the
first week?
Or
Concise
Table 15.5
Workings:- Load, Mult, d, K, K’, (z/d,) z, As, VEd, Asw/s
Φ
mm
Area,
mm2
8
50
10
78.5
12
113
16
201
20
314
25
491
32
804
Lecture 3/28
Working space
Working space
Lecture 3/29
Working space
End of Lecture 3
Lecture 3/30

Bending and Shear in Beams

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