TUTORIAL 4 FLEXURAL MEMBERS • Bending •Shear 06 March 2012
Summary of design process The design process for a beam can be summarised as follows (a) Determination of all forces and moments on critical section (b) Selection of UB or UC (c) Classification of section (d) Check shear strength; if unsatisfactory return to (b) (e) Check bending capacity; if unsatisfactory return to (b) (f) Check deflection; if unsatisfactory return to (b) (g) Check web bearing and buckling at supports or concentrated load; if unsatisfactory provide web stiffener (h) Check lateral torsional buckling; if unsatisfactory return to (b) or provide lateral restraints (i) Summarise results
Problem 1: A simply supported beam 406 x 178 x 74 UB S 275 is required to span 4.5 m and to carry an ultimate design load of 40 kN/m. Check the suitability of the section with respect to shear. ANSWER: •
STEP 1: Section Properties: A= 95 cm2 ; b=179.7mm; tw=9.5 mm; tf=16mm; tr=10.2mm; D=412.8mm; d=360.4 mm; d/t=37.9; b/T=5.62
•
STEP 2: Design shear force at the end of the beam,
wl 40 × 4.5 Fv = = = 90kN 2 2 • STEP 3: Calculation of Shear Area:
For rolled UB section: Av= 9500 -2x179.7x16 +(9.5+2x10.2)16 = 9500-5750.4+478.4 = 4228mm2 Which is more than Dt = 412.8x9.5 = 3922 mm2
• STEP 4: Section Classification Since web thickness tw=9.5mm and t ≤40mm, from table 3.1 page 26: fy = 275 N/mm2 , fu = 430 N/mm2
235 ε = fy
0 .5
235 = 275
0 .5
= 0 . 92
Since the beam is subject to pure bending the neutral axis will be at mid depth Referring to table 5.2: c/t=d/t=37.9 < 72ε :
The web is plastic… note 72ε=72 x 0.92=66.24
Since c/t < 72ε , No need to check shear buckling
• STEP 5: Calculation of Shear Capacity:
V
pl , Rd
=
4228
(275 1 . 00
/
3
)=
V pl , Rd = 671285
(
Av f y / 3
γM0
)
N = 671 . 3 kN
• Vpl,Rd = 671.3 kN ≥ 90kN, shear due to load; The value shows that excessive reserve of shear strength in the web
Problem 2: A simple supported beam of 5 m span, carries a reinforced concrete floor capable of providing lateral restraint to the top compression flange. The uniformly distributed load is made up of 20 kN/m imposed load plus 20 kN/m dead load. Choose a suitable I section beam of S275 steel. It may be assumed that the section is held on web cleats and that web bearing and buckling are not critical
Answer: • STEP 1:Factored Loads Imposed load : 20 x 1.5 Dead Load: 20 x 1.35 Total load:
= 30.0 kN/m = 27.0 kN/m = 57.0 kN/m
Feebody diagram 57 kN/m
5m
• STEP 2: Maximum bending moment = wL2/8 MEd= 57 x 5 x 5/8 = 178.125 kNm • STEP 3 : Plastic modulus required = M / fy = 178.125 x 1000000/275 = 647727.3 mm3 Wpl,y required = 647.8 cm3 < 681.82 cm3 • STEP 4: Looking at the tables for UB, choose a section that has Wpl,x value higher than 647.8 cm3 – 305 x 127 x 48 UB Wpl,y = 711 cm3 weight/m is 48.1 kg/m – 305 x 165 x 46 UB Wpl,y = 720 cm3 weight/m is 46.1 kg/m – 356 x 171 x 45 UB Wpl,y = 775 cm3 weight/m is 45.0kg/m • The deeper beam is lighter and more efficient in carrying the bending moment and would normally be the first choice. • Engineer’s choice may be affected by the available space – in many structures there is a tight limit on head room and the storey height, which may dictate the adoption of a shallower beam
• STEP 5: Considering also the self weight of the beam, the loading on the beam = 57 + 0.45 x 1.35 = 57.6075 kN/m Maximum bending moment : MEd= 57.6075 x 5x5/8 = 180.02 kNm Wpl,y required = 180.02x1000000/275 =6654618 mm3 = 665.46cm3
Wpl,y required < Wpl,y provided. Hence UB is satisfactory • STEP 6: Properties of 356 x 171 x 45 UB h=351.4 mm; b=171.1 mm; tw=7.0; tf=9.7; r=10.2; d=311.6; d/t=44.5; Iy=12070; Iz=811; ry=14.5; rz=3.76; Wel,y=687; Wel,z=94.8; Wpl,y=775 ; Wpl,y=147; u=0.874; x=36.8; A = 57.3cm2
• STEP 7: Maximum shear force: VEd= ( 57.61 x 5 )/ 2 = 144.03 kN • STEP 8: Design value of Shear Capacity = Vc,Rd •
η can be taken as 1.2 (except for steel grades higher than S460 where η=1.00 is recommended.
V pl , Rd =
(
Av f y /
3
γM0
A v = A − 2 bt
f
)
In which Av is the shear area, taken as follows
+ (t w + 2 r )t f ≥ η h w t w
Av = 5730 − 2 × 171 . 1 × 9 . 7 + (7 . 0 + 2 × 10 . 2 )9 . 7 ≥ 1 . 0 × 311 . 6 × 7
Av = 2676.44 ≥ 2181.2
V pl , Rd
(
)
2676 .44 275 / 3 = = 424941 .9 N = 424 .9 kN 1.00
The design value of shear force VEd=144.03 kN is well within the plastic shear resistance, Vpl,Rd = 424.9kN. Note: In practice, shear stress is seldom critical. It may, however become so when a very short span carries a heavy load, or when carrying a heavy concentrated load near support.
• STEP 9:DEFLECTION • The assessment of the structure’s behaviour for deflection is a matter of engineering judgement
• • • •
Unfactored Loads Imposed load = 20.0 kN/m Dead Load = (20+0.45) = 20.45 kN/m For variable action
5WL3 5 × 20 × 5 ×1000 × 50003 = Deflection, δ = 384 EI 384 × 205000 ×12070 × 10000 δ=6.58 mm < L/350 =14.3 mm Hence OK • For variable +permanent action
5WL3 5 × 40.45 × 5 × 1000 × 50003 Deflection, δ = = 384 EI 384 × 205000 ×12070 × 10000 δ=13.30 mm < L/250 = 20 mm Hence OK
• Deflection is well within the allowable deflection • STEP 12: Restraint is provided by the floor and no further check is required. • STEP 13: The beam is assumed to be supported on web cleats; web bearing and buckling are not therefore design criteria. • A check should be carried out on the maximum deflection of the beam due to the most adverse realistic combination of unfactored imposed serviceability loading. • In BS EN 1993-1-1:2005 this is covered by Section 7.0, and the following clauses are addressed. • Clause 7.1: General • Clause 7.2: Serviceability limit state for buildings • Very little coverage in EC3 • Reference to EN 1990 on the basis that many serviceability criteria are independent of the structural material • Clause 3.4, Clause 6.5 and A1.4 of En 1990
• Calculation of deflections from first principles has to be done using the Area-Moment Method, Macaulay's Method, or some other similar approach, a subject of structural mechanics/ analysis. • However, many calculations of deflection are carried out using formulae for standard cases, which can be combined as necessary to give the answer for more complicated situations.
Problem 3: A simply supported beam of 7 m span carries a reinforced concrete floor capable of providing lateral restraint to the top compression flange. The total UDL is made up of 100 kN dead load (including self weight) plus 150 kN imposed load. In addition the beam carries a point load (PL) at midspan made up of 50kN dead load and 50 kN imposed load. Choose a suitable Universal beam using S275 steel. Also carry out the web buckling and web bearing checks, assuming a stiff bearing length of 75 mm.
DL=50kN LL=50kN DL=100 kN LL=150kN
7m
• • • •
STEP 1: Load factors,γf: DL=1.35 and LL=1.5 Factored loads: UDL =100x1.35 +150 x 1.5=135+225=360kN Point Load = 50 x 1.35 + 50 x 1.5 = 142.5kN
• STEP 2: Load diagram PL=142.5kN UDL=360kN
7m
STEP 3: Maximum BM=My=WL/8+WL/4=360 x7/8+142.5x 7/4=315+249.375 = 564.375kNm STEP 4: Sy required = My / fy=564.375x1000000/275=2052272.7mm3 = 2053cm3
From tables choose a suitable section. • Choosing 533 x 210 x 92 UB (Wpl=2360 cm3) • Properties h=533.1 mm, b=209.3 mm, d=476.5 mm, r=12.7 mm, tf=15.6mm, Iz=55230cm4, tw=10.1mm, Wel,y=2072cm3 ; A = 118cm2 • STEP 5: Check the design strength,fy • Since T=15.6mm<40 mm Therefore fy=275N/mm2 ε=0.924 • STEP 6 section classification • Flange: c = (209.3 -10.1-2x12.7)/2=86.9 mm ‘c/tf = 86.9/15.6=5.57 <9 ε <9x 0.924=8.32……………….plastic • Web: part subject to bending • ‘d/t=476.5/10.1=47.2 < 72 ε ………….neutral axis at mid-depth 72x 0.924=72…………………………….plastic • The section is plastic. Note that a compact section would be sufficient as there is no plastic re-distribution of moments.
Considering the Self weight =92kg/m = 92x10/1000 = 0.92kN/m • Self weight =0.92 x 7= 6.44kN • UDL = (100+6.44)x1.35 +150 x 1.5=135+225=368.7kN • Point Load = 50 x 1.35 + 50 x 1.5 = 142.5kN • Maximum BM=Mx=WL/8+WL/4=368.7 x7/8+142.5x 7/4=322.62+249.375= = 571.99kNm • Maximum SF = (368.7+142.5)/2 = 255.6kN = 255600 N • Sx required = Mx/fy=571.99x1000000/275=2079954.545mm3=2080mm3
• STEP 7: check whether Sx will be satisfactory taking self weight into consideration • Sx required< Sx provided • 2060 < 2360
• STEP 8: Factored shear force Fv=(368.7+142.5)/2 = 255.6 kN
A − 2 bt
• STEP 9: Av =
f
+ (tw + 2 r )t
f
• Shear area = 11800 – 2 x 209.3x15.6 + (10.1+2x 12.7)15.6 • = 11800 – 6530.16 + 553.8 =5823.64 mm2 • STEP 10: Design Shear plastic resistance
V pl , Rd =
(
Av f y /
γ
M 0
3
)=
(
5823 . 64 275 / 1 .0
• 255.600 kN < 924.65 kN Hence OK
3
) = 924 . 65 kN
STEP 11:
Check for Design Resistance for bending Mc,Rd, allowing for effects of coexisting shear • Factored Fv at maximum moment = 71.25 kN • Maximum moment = 571.99kN • Clause 6.2.8 (3) : When Shear force is less than half the plastic shear resistance, its effect on the moment resistance maybe neglected except where shear buckling reduces the section resistance. •
Vpl,Rd = 924.65 kN
•
Note that 0.5Pv=0.5x924.65=462.32kN
•
Since Fv < 0.5Pv, the Shear Force is LOW
•
Mc,Rd=py x Sx= 275 x 2360 x1000/1000000=649 kNm
To avoid irreversible deformation under serviceability loads, the value of Mc should be limited to 1.5pyZ generally and to 1.2pyZ in the case of simply supported beam or a cantilever (Clause 4.2.5.1)
1.2 × 275 × 2072 ×1000 1 .2 p y Z x = = 684kNm 1000000
PL=142.5kN UDL=368.7kN
7m 255.6kN
+
71.25kN 71.25kN
-
571.99kNm
255.6kN
• The lesser of the two values is 649kNm • Since Mc,Rd=649kNm> My=571.99kNm, the section is Satisfactory. • • • • • •
STEP 12: Deflection Serviceability loads Case 1: unfactored imposed loads only P=50kN (imposed point load) W=150kN (imposed UDL) Actual deflection=δ=PL3/48EI+5WL3/384EI
50 ×1000 × 70003 5 ×150 × 70003 + δ= = 9.10mm 4 4 48 × 205000 × 55230 ×10 384 × 205000 × 55230 ×10 •
STEP 13:
Allowable deflection =L/350=7000/350=20 mm The deflection is satisfactory.
• Case 2: unfactored dead load and imposed loads • Udl =256.44 kN and point load = 100kN
100 ×1000 × 70003 5 × 256.44 × 70003 δ= + 4 48 × 205000 × 55230 ×10 384 × 205000 × 55230 ×104
δ = 6.31 + 10.11 = 16.42 • Should be less than L/250 = 7000/250 = 28 mm Hence OK
STEP 14: Check for web bearing/crushing At locations of heavy concentrated loads such as support reactions or where columns are supported on a beam flange, additional stress concentrations occur in the web. This introduces the possibility of the web failing in a buckling mode similar to the vertical strut or by localised bearing failure at the top of the root fillet.
Critical section for bearing
• The code specifies that the bearing capacity Ry,Rd of the unstiffened web at the web to flange connection is given by
R y , Rd
( S =
s
In which Ss=length of stiff bearing Tw= web thickness Fyw= yield strength of web γM1=material partial safety factor(1.05)
bf s y = 2t f tw
0 .5
+ S y )t w f yw
γ M1
f yf σ f , Ed 1 − f f yw yf 0.5
2 0.5
2 0.5
209.3 275 0 s y = 2 ×15.6 1− 10.1 275 f yf 0 .5
s y = 142.03
0 .5
R y , Rd
( 75 + 142 .3 )10 .1 × 275 = 1 . 05 × 1000
= 579 .8 kN
This is greater than the support reaction =255.6kN
STEP 15: Check for Crippling resistance
0 .5 t R
a , Rd
2 w
(Ef
=
)
0 .5
yw
t t
γ
f w M
0 .5
t + 3 t
w f
s s d
1
Ss is limited to 0.2d (for 533 x 210 x 92 UB d=476.5 mm. So 0.2 x 476.5=95.3mm) Since the actual Ss=75 mm OK.
R a , Rd
Ra,Rd
0 .5 15 . 6 10 . 1 75 0 .5 2 0 . 5 × 10 . 1 (210000 × 275 ) + 3 10 . 1 15 . 6 476 . 5 = 1 . 05
51.005(7599.3421)[1.2428+ 0.305715] = = 571629.7 N 1.05
This is greater than the support reaction =255.6kN . Hence OK.
STEP 16 Check for web buckling : The buckling resistance is determined by taking a length of web as a strut The length of web is taken from Eurocode 3 which gives following equation
beff = 0 .5( h + s s ) 2
2 0 .5
ss +a+ 2
(
but ≤ h + ss 2
)
2 0.5
Where a=distance from end of bearing to end of beam, assumed to be zero Depth of beam, h=533.1mm
75 beff = 0.5(533.1 + 75 ) + 0 + = 306.7mm 2 2
2 0.5
• The top flange is held by a slab and the bottom by seating cleats. Assuming they provide restraint against rotation and displacement; the effective height of the web for buckling can be taken as 0.7x distance between fillets • ‘l = 476.5x0.7=333.55mm t • Radius of gyration for web = i = w = 0.29t w = 0.29 × 6.9 = 2.001 12
• Slenderness =
l 333.55 λ= = = 166.69 i 2.001
λ 0.5 λ = (β A ) λ1 • Where λ1=93.9ε=93.9x0.924=86.8 • Assume βA=1.0 λ
λ=
λ1
(β A )
0 .5
166.69 (1.0)0.5 = 1.92 = 86.8
• Since the web strut can be considered as a solid section; use strut curve “C”
From graph, corresponding to λ=1.92; Χ=0.23
• Buckling resistance Nb,Rd=
N b , Rd
0.23 × 275 × 306.7 ×10.1 = = 186.6kN 1.05 × 1000
• Support reaction = 255.6kN Therefore not safe from buckling
• Problem 4:
• Step 2: Design Values
• Step 3: Section Selection and Properties
• Step 4: Section Classification
• Step 5: Shear Resistance Check • Shear Area:
A v = A − 2 bt
f
+ (tw + 2 r )t
f
A v = 5860 − 2 × 142 . 2 × 11 . 2 + ( 6 . 9 + 2 × 10 . 2 ) × 11 . 2
Av = 2980 . 48 mm
V pl , Rd
(
2
2980 . 48 275 / = 1 .0
3
) = 473214
. 756 N = 473 . 2 kN
• Step 6: Web Crushing Resistance Check
• Step 7: Web Crippling Resistance Check
• Step 8: Web Buckling Resistance Check
• Step 9: Deflection Check
