Metal Structures Design Project I Steel truss

Metal Structures

Design Project I

Steel truss – examples of calculation

Today will be presented 9 examples of calculations various sub-parts of steel truss:

Ist example: roofing selection;

IInd example: purlin and wall girt in bi-axial bending;

IIIrd example: roofing column;

IVth example: truss member;

Vth example: welds;

VIth example: resistance of node;

VIIth example: field splice;

VIIIth example: support on masonry structure;

IXth example: roof bracing.

Theoretical explanations of above problems will be presented during lectures, here are

way of calculation only.

• tmin = 11 cm

• Loads I (dead-weight + snow + wind

pressure + imposed load) = 5,40 kN / m2

• Loads II (wind suction) = 0,50 kN / m2

• Distance between purlins = 2,10 m

• Fire resistance EI30

Ist example of calculations – roofing selection

Photo: steelprofil.eu

Tables propose 10 and 12 cm thick sandwich panels. With the minimum required thickness

of 11 cm, the second should be chosen.

Photo: pruszyński.com.pl

But such type of panels not satisfied condition for loads:

Epressure / Rpressure = 5,40 / 5,27 = 1,025 > 1,0

Esuction / Rsuction = 0,50 / 4,22 = 0,118 < 1,0

Next thickness, 16 / 20 cm: Photo: pruszyński.com.pl

Epressure / Rpressure = 5,40 / 5,55 = 0,973 < 1,0

Esuction / Rsuction = 0,50 / 4,23 = 0,118 < 1,0

Photo: pruszyński.com.pl

According to the manufacturer's declaration,

fire resistance is sufficient (min EI 30):

Sandwich panel PWD PIR 160 (14,20 kg / m2)

is applied to structure. Dead weight:

gk = 0,139 kN / m2

g = 0,188 kN / m2

s = 4,00 [kN / m2];

g = 0,40 [kN / m2];

w = 0,50 [kN / m2];

IPE 300

S 235

a = 5o

d = 2,50 m → d1 = d / cos a = 2,510 m

Photo: M. Łubiński, W. Żółtowski, Konstrukcje

Metalowe t. II, Arkady, Warszawa 2004

IInd example of calculations – purlin in bi-axial bending

d

d1

L

d1

y

z

qz

qy d1 = 2,510 m

L = 6,000 m

Photo: Author

IPE 300:

Jy = 8356,0 cm4

Jz = 603,8 cm4

Wpl, y = 628,4 cm3

Wpl, z = 125,2 cm3

h = 300 mm

d = 150 mm

tw = 7,1 mm

tf = 10,7 mm

AV, y = 32,1 cm2

AV, z = 21,3 cm2

Photo: Author

qH

Photo: Author

qV

s → qH, qV ; l

g → qH, qV ; l1

w → qy, qz ; l1

d1

d

qH(s) = 0,000 kN / m

qV(s) = s d = 10,000 kN / m

qH(g) = 0,000 kN / m

qV(g) = g d1 = 1,004 kN / m

qy(w) = 0,000 kN / m

qz(w) = w d1 = 1,255 kN / m

d1

d

l1 = 2 d1 / 2 = d1

l = 2 d / 2 = d

qH

Photo: Author

qV

qH → qy , qz

qV → qy , qz

qy(qH) = qH cos a

qz(qH) = - qH sin a

qy(qV) = qV sin a

qz(qV) = qV cos a

qy[qH(s)] = qH(s) cos a = 0,000 kN / m

qz[qH(s)] = - qH(s) sin a = 0,000 kN / m

qy[qV(s)] = qV(s) sin a = 0,872 kN / m

qz[qV(s)] = qV(s) cos a = 9,962 kN / m

qy[qH(g)] = qH(g) cos a = 0,000 kN / m

qz[qH(g)] = - qH(g) sin a = 0,000 kN / m

qy[qV(g)] = qV(g) sin a = 0,088 kN / m

qz[qV(g)] = qV(g) cos a = 1,000 kN / m

y Photo: Author

z

qz = 12,217 kN/m

qy = 0,960 kN/m

qy = qy[qH(s)] + qy[qV(s)] + qy[qH(g)] + qy[qV(g)] + qy(w) =

= 0,000 + 0,872 + 0,000 + 0,088 + 0,000 = 0,960 kN / m

qz = qz[qH(s)] + qz[qV(s)] + qz[qH(g)] + qz[qV(g)] + qz(w) =

= 0,000 + 9,962 + 0,000 + 1,000 + 1,255 = 12,217 kN / m

MEd, y = qz L2 / 8 = 54,977 kNm

MEd, z = qy L2 / 8 = 4,320 kNm

VEd, y = qy L / 2 = 2,880 kN

VEd, z = qz L / 2 = 36,651 kN

Resistances of cross-section:

MRd, y = fy Wpl, y / gM0 = 235 [MPa] ∙ 628,4 [cm3] / 1,00 = 147,674 kNm

MRd, z = fy Wpl, z / gM0 = 235 [MPa] ∙ 125,2 [cm3] / 1,00 = 29,422 kNm

VRd, y = fy AV, y / (gM0 √3) = 235 [MPa] ∙ 32,1 [cm2] / (1,732 ∙ 1,000) = 435,524 kN

VRd, z = fy AV, z / (gM0 √3) = 235 [MPa] ∙ 21,3 [cm2] / (1,732 ∙ 1,000) = 288,993 kN

Checking resistances

E / R ≤ 1,0

MEd, y / MRd, y ≤ 1,0 → 54,977 / 147,674 = 0,372

MEd, z / MRd, z ≤ 1,0 → 4,320 / 29,422 = 0,147

VEd, y / VRd, y ≤ 1,0 → 2,880 / 435,524 = 0,007

VEd, z / VRd, z ≤ 1,0 → 36,651 / 288,993 = 0,127

Bi-axial bending of I-beam (EN 1993-1-1 (6.41)):

(MEd, y / MRd, y)2 + MEd, z / MRd, z ≤ 1,0 → 0,3722 + 0,147 = 0,285

Checking stability:

Lateral buckling can occurs as effect of bending. For Your range of design project you

can asume (not completely correct), that sanwich panels protect purlins from instability.

Checking deflections:

Dy = 5 qy L4 / (384 E Jz) =

= 5 ∙ 0,960 [kN/m] ∙ 6,04 [m4] / (384 ∙ 210 [GPa] ∙ 603,8 [cm4]) = 12,8 mm

Dz = 5 qz L4 / (584 E Jy) =

= 5 ∙ 12,217 [kN/m] ∙ 6,04 [m4] / (384 ∙ 210 [GPa] ∙ 8356 [cm4]) = 11,7 mm

D = √(Dz2 + Dz

2) = 17,3 mm

Dacc = EN 1993-1-1 N.A. 22 = L / 200 = 30 mm

D / Dacc = 0,578

Photo: Author

Dz

Dy

D

Conclusions:

E MEd, y MEd, z VEd, y VEd, z MEd, y + MEd, z LT D

E / R 0,372 0,147 0,007 0,127 0,285 0,578

1. For one-span beams longer than ~ 5,0 m the most important requirement concerns

deformations, not cross-sectional forces, nearly always.

2. Each results satisfy fundamental requirement E / R ≤ 1,0. But the highest value for analysed

case (0,578) is far from upper limit (1,000) – this means, cross-section is too big (too expensive)

for situation. It’s somethink like situation: total cost of structure = 10 M€, but only 5,780 M€ is

justified, the rest (4,220 M€) is the result of the designer's inability.

You should check smaller one I-beam (IPE 270? IPE 240?). You no need recalculated values of

loads: influence of change dead-weight of purlin for total loads (snow, wind…) is neglegible.

3. This purlin will be recalculated after analysis of roof bracings. Axial force in purlin occurs as

the effect of coopreation bracings-purlins. Max effort could be bigger that now. Because of this,

efforts for smaller one purlin (IPE 270? IPE 240?) should be:

• E / R ≤ 1,0 for D;

• E / R ≤ 0,8 all the rest conditions.

Wind action

on fron wall

Photo: Author

Roofing Front wall girts

Roofing

column

Purlin Top chord

Horizontal bracings

IIIrd example of calculations – roofing column

1,0 m

Photo: Author

1,0 m

1,0 m

1,0 m

0,80 kN

0,80 kN

0,80 kN

4,80 kN

4,80 kN

4,80 kN

2,40 kN

7,20 kN

7,20 kN

7,20 kNm

7,20 kNm

9,60 kNm MEd

VEd

NEd 7,20 kN

7,20 kN

2,40 kN

2,40 kN

0,80 kN

1,60 kN

2,40 kN

Hot-rolled RHS 80x8 (80 mm side, 8 mm thickness):

A = 22,4 cm2

Jy = Jz = 189,0 cm4

Wel, y = Wel, z = 47,3 cm3

Wpl, y = Wpl, z = 59,5 cm3

AV, y = AV, z = 12,8 cm2

iy = iz = 2,91 cm

S 235 → fy = 235 MPa

e = √ (235 / fy) = 1,0

l cr =

l =

4,0

m

S 235 + RHS hot rolled → buckling curve a → a = 0,21

(EN 1993-1-1 tab. 6.1, 6.2)

4,80 kN – from wind action

0,80 kN – from deadweigth


NRd = fy A / gM0 = 235 [MPa] ∙ 22,4 [cm2] / 1,00 = 526,400 kN

MRd, y = fy Wpl, y / gM0 = 235 [MPa] ∙ 59,5 [cm3] / 1,00 = 13,983 kNm

VRd, y = fy AV, y / (gM0 √3) = 235 [MPa] ∙ 12,8 [cm2] / (1,732 ∙ 1,000) = 173,672 kN


E / R ≤ 1,0

NEd / NRd ≤ 1,0 → 2,400 / 526,400 = 0,005

MEd, y / MRd, y ≤ 1,0 → 9,600 / 13,983 = 0,687

VEd, y / VRd, y ≤ 1,0 → 7,200 / 173,672 = 0,041

Checking interaction NEd ↔ MEd, y for RHS (EN 1993-1-1 (6.39)):

n = NEd / NRd = 0,005

aw = min [0,5 ; (A – 2 b t) / A] = min [0,5 ; (22,4 – 2 ∙ 8,0 ∙ 0,8) / 22,4] = 0,429

MN, Rd, y = min [MRd, y ; MRd, y (1 – n) / (1 – 0,5 aw)] = min[13,983 ; 17,712] = 13,983 kNm

MEd, y / MN, Rd, y ≤ 1,0 → 9,600 / 13,983 = 0,687

Checking stability (EN 1993-1-1 6.3.1.2):

l = (lcr / i) / (93,9 e) = (4,0 [m] / 2,91 [cm]) / 93,3 = 1,463

F = [1 + a (l - 0,2) + l2] / 2 = [1 + 0,21 (1,463 - 0,2) + 1,4632] / 2 = 1,703

c = min {1 / [F + √ (F2 - l2)] ; 1,0} = min {1 / [1,703 + √ (1,7032 – 1,4632)] ; 1,0} =

= min {0,388 ; 1,0} = 0,388

NEd / (c NRd) = 0,013

Checking stability with interaction NEd ↔ MEd, y for RHS (EN 1993-1-1 NA.20):

NEd / (c NRd) + Cmy MEd, y / (cLT MRd, y) ≤ 1,0 - D0

Latreral buckling is not danger for SHS → cLT = 1,0

Cmy could be taken into consideration as 1,0 (rought approximation)

RHS 80x8 → Ist class of cross-section (Lec. #4) → D0 = 0,1 + 0,2 [ (Wpl, y / Wel, y) - 1) ] =

= 0,1 + 0,2 [(59,5 [cm3] / 47,3 [cm3]) – 1,0 ] = 0,152

NEd / (c NRd) + Cmy MEd, y / (cLT MRd, y) = 0,013 + 0,687

1,0 - D0 = 1,0 – 0,152 = 0,848

0,700 < 0,848

(0,943 < 1,000)

Checking deflections:

Rought approximation: deflection under 3 forces in point ≈ deflection under equivalent

continous load

qeq = 3 F / l = 3 ∙ 4,80 [kN] / 4 [m] = 3,60 [kN / m]

D = 5 qeq L4 / (384 E Jy) =

= 5 ∙ 3,6000 [kN/m] ∙ 4,04 [m4] / (384 ∙ 210 [GPa] ∙ 189 [cm4]) = 31 mm

Dacc = EN 1993-1-1 N.A. 22 = L / 150 = 27 mm

D / Dacc = 1,148

Conclusion:

E NEd MEd, y VEd, y MEd, y + Ned NEd stab. MEd, y + Ned stab. D

E / R 0,005 0,687 0,041 0,687 0,013 0,943 1,148

1. Deflection are bigger than accepted limit. You should analised bigger one cross-section

(RHS 80x10?), or check more accurate value of deflection from FEM calculations.

IVth example of calculations – truss members

Photo: Author

Chord Max tension [kN] Max compression [kN] Cross-section

Top -45,987 228,322 Hot-rolled

CHS 57x10 Bottom -235,621 43,181

Connections with roof bracings S 235 → fy = 235 MPa

e = √ (235 / fy) = 1,0

Hot-rolled CHS 57x10 (57 mm diameter, 10 mm thickness):

A = 14,8 cm2

Jy = Jz = 42,6 cm4

iy = iz = 1,70 cm

S 235 + CHS tor tolled→

buckling curve a → a = 0,21

(EN 1993-1-1 tab. 6.1, 6.2)

L = 24,000 m


NRd = fy A / gM0 = 235 [MPa] ∙ 14,8 [cm2] / 1,00 = 347,800 kN


E / R ≤ 1,0

NEd, top, comp / NRd ≤ 1,0 → 228,322 / 347,800 = 0,656

NEd, top, tens / NRd ≤ 1,0 → 45,957 / 347,800 = 0,132

NEd, bott, comp / NRd ≤ 1,0 → 43,181 / 347,800 = 0,124

NEd, bott, tens / NRd ≤ 1,0 → 235,621 / 347,800 = 0,677

Flexural buckling of chords:

Top chords in compression; buckling in plane: critical length = distance between nodes

Photo: Author

Critical length for compressed chord of truss is depend on plane of analysis and part of

structure. For web members, critical length = distance between nodes.

→ Des #1 / 48


Top chords in compression; buckling out of plane: critical length = distance between

horizontal bracings

Photo: Author

→ Des #1 / 49


Bottom chords in compression; buckling in plane: critical length = distance between

nodes

Photo: Author

→ Des #1 / 50


Bottom chords in compression; buckling out of plane: critical length = distance between

vertical bracings

Photo: Author

→ Des #1 / 51

Checking stability for CHS (EN 1993-1-1 6.3.1.2):

l = (lcr / i) / (93,9 e)

F = [1 + a (l - 0,2) + l2] / 2

Chord lcr in plane [m] lcr out of plane [m]

Top 3,047 2 x 3,047 = 6,094

Bottom 3,047 4 x 3,047 = 12,188

Chord l in plane l out of plane

Top 1,909 3,818

Bottom 1,909 7,635

Photo: Author

Chord F in plane F out of plane

Top 2,502 16,337

Bottom 2,502 30,427

Checking stability for CHS (EN 1993-1-1 6.3.1.2):

c = min {1 / [F + √ (F2 - l2)] ; 1,0}

NEd / (c NRd)

Chord c in plane c out of plane

Top 0,243 0,031

Bottom 0,243 0,017

Chord E / R in plane E / R out of plane

Top 2,700 21,161

Bottom 0,510 7,291

Checking deflection:

Max vertical deformation under loads (FEM calculation): 52 mm

Dacc = EN 1993-1-1 N.A. 22 = L / 250 = 96 mm

D / Dacc = 0,542

Conclusions:

1. Stiffness of bars is too small and make instability. Cross-section should be much more

massive (CHS 219,1 x 6,3 ?).

2. Smaller distances between bracings should be considered (→ smaller lcr → bigger c).

Chord E / R tension E / R compression

Top 0,132 0,656

Bottom 0,677 0,124

Resistance:

Stability:

Chord E / R in plane E / R out of plane

Top 2,700 21,161

Bottom 0,510 7,291

Deflection: 0,542

Vth example of calculations – welds

Photo: Author

S235 → fu = 360 MPa, βw = 0,80

(EN 1993-1-8 tab 4.1)

CHS diameter 51 mm

CHS thickness of flange 3,2 mm

Thickness of weld 3 mm

A = p · [(51 / 2 + 3)2 - (51 / 2)2] = 509 mm2

NEd = 64,73 kN

s = NEd / A = 127,171 MPa

t = 0 MPa

σ┴ = t┴ = σ / √2 = 89,923 MPa

t║ = t = 0,000 MPa

γM2 = 1,25

EN 1993-1-8 (4.1):

Condition 1: √[(σ┴)2 + 3(t║2 + t┴

2)] = 254,342 MPa < fu / (βw γM2) = 360,000 MPa

Condition 2: σ┴ = 89,923 MPa < 0,9 fu / γM2 = 259,200 MPa

Photo: Author

Two types of joints: T and KT will be presented in calculations.

Photo: EN 1993-1-8 fig. 7.1

VIth example of calculations – resistance of node

Photo: Author

Truss

Model 3

Model 2

Model 1a 1b 1c

e = 0

e inside

limits

e outside

limits

satisfy

satisfy

Generally, five various models of truss

structure must be taken into

consideration. The choice of model is

determined by the satisfy or dissatisfy of

many conditions. You don’t check these

conditions for your range of project. You

take into consideration model 1a (ideal

truss).

Six various models of joint’s destruction

must be analised. Not all are equally

important for various shapes of truss

joints. Satisfaction of conditions

presented in EN 1993-1-8 tab. 7.1 allows

taken into consideration 1-2 mechanism

only of destruction for each type of joint.

You don’t check these conditions for

your range of project. You make initial

assumptions that these conditions are

satisfied.

Photo: scielo.br

Photo: offshoremechanics.asmedigitalcollection.asme.org

Photo: EN 1993-1-8 fig 7.3, 7.4

General formula of resistance for joints in truss chord – web members as CHS-CHS

(EN 1993-1-8 (7.3)):

Ni, Ed / Ni, Rd + (Mip, i, Ed / Mip, i, Rd)2 + Mop, i, Ed / Mop, i, Rd ≤ 1,0

Ni, Ed – force in web member;

Ni, Rd – resistance of joint under force;

Mip, i, Ed , Mop, i, Rd - local bending moments in joints, in plane (ip) and out of plane (op);

effects of eccenticities e 0 (→ #t / 34). In your range of project, Mip, i, Ed = Mop, i, Rd = 0.

Formulas for Ni, Rd for truss CHS-CHS are presented in EN 1993-1-8 tab. 7.2, 7.3, 7.4, 7.5,

7.6, 7,7. Joints T and KT are listed in:

• 7.2;

• 7.5 – but this table is important for joints under bending, not important for your range of

project;

• 7.6.

Photo: EN 1993-1-8 tab. 7.2

Photo: EN 1993-1-8 tab. 7.2

Photo: EN 1993-1-8 tab. 7.6

T joint, condition 1

T joint, condition 2

KT joint, condition 1

KT joint, condition 2

T joint, condition 1:

Photo: Author

76,1 x 3,2

NP, Ed = more important (Nleft ; Nright)

A0 = 7,33 cm2

t0 = 3,2 mm

d0 = 76,1 mm

51 x 3,2

N1, Ed

d1 = 51 mm

EN 1993-1-8 tab. 7.2:

N1, Rd = g0,2 kp fy0 t02 (2,8 + 14,2 b) / (gM5 sin q)

kp =

for compressive NP, Ed :

min[ 1,0 ; 1,0 – 0,3 np (1 + np)]

for tensile NP, Ed : 1,0

EN 1993-1-8 1.5:

g = d0 / 2 t0

b = d1 / d0

np = NP, Ed / (A0 fy0 gM5)

Case NP, Ed [kN] N1, Ed [kN] np kp g b N1, Rd [kN] N1, Ed / N1, Rd

Comp. 37,584 -10,752 0,218 0,920 11,891 1,492 87,129 0,123

Tens. -192,754 58,235 -1,119 1,000 94,706 0,615

S235

q = 90o

T joint, condition 2:

Condition 2 is important for initial assumption only:

d1 ≤ d0 – 2 t0

d1 = 51 mm

t0 = 3,2 mm

d0 = 76,1 mm

d0 – 2 t0 = 69,7 mm

Condition important for checking

N1, Rd = fy0 t0 p d1 (1 + sin q) / [ 2 (√3) gM5 (sin q)2] = 69,563 kN

Case NP, Ed [kN] N1, Ed [kN] N1, Rd [kN] N1, Ed / N1, Rd

Comp. 37,584 -10,752 69,563 0,155

Tens. -192,754 58,235 0,837

KT joint, condition 1:

Condition 1 is important for initial assumption only:

di ≤ d0 – 2 t0

d1 = d2 = d3 = 51 mm

t0 = 3,2 mm

d0 = 76,1 mm

d0 – 2 t0 = 69,7 mm

Condition important for checking

76,1 x 3,2

NP, Ed = more important (Nleft ; Nright)

A0 = 7,33 cm2

t0 = 3,2 mm

d0 = 76,1 mm

51 x 3,2

N1, Ed

d1 = 51 mm

51 x 3,2

N3, Ed

d1 = 51 mm

51 x 3,2

N2, Ed

d1 = 51 mm

Photo: Author q1 = 45o

q2 = 90o

q3 = 45o

Case NP, Ed [kN] N1, Ed [kN] N2, Ed [kN] N3, Ed [kN]

Comp. 42,322 14,008 -9,221 -17,955

Tens. -207,155 -78,252 71,072 81,321




Member Ni, Ed Ni, Ed / Ni, Rd

1 14,008 0,039

-78,252 0,220

2 -9,221 0,133

71,072 1,022

3 -17,955 0,050

81,321 0,228

KT joint, condition 2: Condition important for situation:

N1, Ed compressive (> 0);

N2, Ed compressive (> 0);

N3, Ed tensile (< 0);

Photo: EN 1993-1-8 tab. 7.6

Case NP, Ed [kN] N1, Ed [kN] N2, Ed [kN] N3, Ed [kN]

Comp. 42,322 14,008 -9,221 -17,955

Tens. -207,155 -78,252 71,072 81,321

Such one situation doesn’t exist in analised case

Conclusion:

1. Joint KT, member 2, condition 1: exceeding resistance. Two parameters are important

for resistance it this case: t0 and di . Resistance and stability for member 2 and chord is

satisfied, but, because of problem in joint, one of them / both must have more massive

cross-section.

Algorithm

Photo: Author

→ Lec #3 / 96

Photo: Author

Max compressive force: not important for such one joint.

Max tensile force: NEd = -297,345 kN

Bolts: M24, class 8.8

M24 → dbolt = 24 mm

Class: X.Y

X = fub / 100 fub = 100 X = 800 MPa

Y = 10 fyb / fub fyb = 10 X Y = 640 MPa

VIIth example of calculations – field splice

tp tp

Photo: Author

76,1x3,2 → D = 76,1 mm, t0 = 3,2 mm

a = 5 mm

d = 24 mm

d0 = 24 + 2 = 26 mm

x = 40 mm

y = 35 mm

tp = 30 mm

p2

e1 e2

r0

ri

re

rp e1 = y = 35 mm

e2 = x + a √2 = 47 mm

r0 = D / 2 + e2 = 85 mm

p2 = L

ri = D / 2 – t0 = 35 mm

re = D / 2 = 38

rp = r0 + e1 = 120 mm

S235

A new Eurocode (working name EN 1993-1-801) is being developed; will contain

information about the calculation of joints for CHS / RHS. But at now calculation

procedure is presented based on literature, not on Eurocode.

k1 = ln (r0 / ri) = 0,887

k2 = k1 + 2 = 2,887

k = [k1 + √(k22 - k1

2)] / (2 k1) = 2,049

eeff = min (e2 ; 1,25 e1) = 44 mm

reff = re + e2 + eeff = 129 mm

k5 = ln (reff / r0) = 0,482

k3 = 1 - 1 / k + 1 / (k k5) = 0,417

Circumference along a line of bolts: c = 2 p r0 = 534 mm

Max distance between bolts p2 = L = 200 mm

Number of bolts: n = c / L ≈ 4 p2 = L = 134 mm

d0 = d + 2 mm (d ≤ 24 mm)

d0 = d + 3 mm (d > 24 mm)

Initial assumptions: d0 = d ; recommended value d0 = d + 2 or 3 mm is consistent with

EN 1090-2, technical requirements for steel structures. Value d0 = 26 mm is accepted.

2,2 d0 ≤ p2 ≤ min (14 tp ; 200 mm)

2,2 d0 = 57 mm < 134 mm < min (420 mm ; 200 mm) ; OK

1,2 d0 ≤ e2 ≤ (1,5 - 2,0) d

1,2 d0 = 31 mm < 47 mm < 48 mm ; OK.

1,2 d0 ≤ e1

1,2 d0 = 31 mm < 35 mm ; OK.

Verification of geometrical parameters, recommended in literature:

Resistance of joint is the effect of plate resistance (NRd1) and bolts resistance (NRd2):

NRd = min (NRd1 ; NRd2)

NRd1 = tp2 fyp p k / (2 gM0) = 394,349 kN

NRd2 = n Ft, Rd / k3

EN 1993-1-8 tab. 3.4

Ft,Rd = k2 fub As / gM2

k2 = 0,90

As - area of threaded portion of bolt; M24 → As = 3,53 cm2

Ft,Rd = 162,662 kN

NRd2 = 1560,307 kN

NRd = 394,349 kN

NEd / NRd = 0,754

Photo: Author

Top plate

Rocker

Bottom plate

VIIIth example of calculations – support on masonry structure

Top and bottom plate: tp = 30 mm, dp x dp = 400x400 mm

Rocker: tr = 30 mm, br x dr = 200x400 mm, rr = 500 mm

Masonry structure: fd = 4,0 MPa

Masonry wall thickness: t = 400 mm

Vertical distance floor – support: hc = 4,70 m

S235

Photo: Author

Vertical action from the first truss (vertical reaction

from red area) REdc, 1 = 295,328 kN

Vertical action from the second truss (vertical reaction

from blue area) REdc, 2 = 590,696 kN

Photo: EN 1996-1-1 fig 6.2

Resistance depends on

distane between support and

end of wall:

a1 = 0,00 for first truss,

a1 = 5,80 for the second one.

first truss second truss not important case

not important

case

NEdc / NRdc ≤ 1,0

NRdc = b Ab fd

b = min { 1,25 + a1 / (2 hc) ; 1,5 ; max [(1 + 0,3 a1 / hc) ∙ (1,5 - 1,1 Ab / Aef) ; 0 ] }

Aef = lefm t

lefm, 1 = dp + (hc / 2) / tg 60o = 1,757 m

lefm, 2 = dp + 2 (hc / 2) / tg 60o = 3,114 m

Truss Aef [m2] Ab [m

2] b NRdc [kN] NEdc / NRdc

1 0,703 0,160 min(1,250 ; 1,500 ; 1,250) = 1,250 800,000 0,396

2 1,246 min(2,484 ; 1,500 ; 1,862) = 1,500 960,000 0,584

Photo: Author

tp tr

br

rr br

dr

dp

dp dp

dp

Rocker-beam compression:

EN 1337-6, Structural bearings; rocker

bearings

Plate in contact with round rocker:

(NEd / bp) / [23 rr fu2 / (E γ

M)] ≤ 1,0

γM

= 1,0

1476,740 [kN/m] / 7097,143 [kN/m] =

= 0,208

Rocker in contact with plate:

(NEd / bp) / [fy (2 tr + bp) / γM ] ≤ 1,0

γM

= 1,1

1476,740 [kN/m] / 98272,727 [kN/m] =

= 0,015

Three various types of actions must be analised.

Photo: Author

1. Wind action on front wall. Load depends on area of front wall and wind pressure.

Additionally friction of wind on roof could be taken into consideration.

IXth example of calculations – roof bracing

Photo: Author

2. Instability of roof girders.

Bracing prevent roof girder

from instability. Witout them,

buckling of girder occurs on

total length. Bracings act as

complex of forces, make girder

more straight. These forces – as

reaction – must be applied to

bracings. Walue of these forces

is equivalent load, in proportion

to compressive force in girder.

Girders lose stability randolmy,

in left or right. Total effect for

entire roof is not multiplied

[(single effect) by (number of

girders m)]

but by (reduction factor am).

Photo: Author

3. Imperfections of roof girders.

Roof girders are not perfectly

straight. Bracings act as complex

of forces, make girder more

straight. These forces – as

reaction – must be applied to

bracings. Walue of these forces

depends on initial bow

imperfection of girder and

compresive force in girder.

Girders are imperfected

randolmly, in left or right. Total

effect for entire roof is not

multiplied

[(single effect) by (number of

girders m)]

but by (reduction factor am).

Way of calculation - 2D vs. 3D - is very important for algorithm.

Photo: mesilo.pl

3D model in FEM calculations: we have full information about cooperation between trusses,

purlins and bracing bars right away. Calculation is made in two steps:

• Initial calculation: dead weight, climatic actions, live loads etc.

• Equivalent forces: for bracing bars additionally important are equivalent forces from

imperfections of trusses and from prevention of instability of trusses. Values of these

forces are calculated based on forces in trusses after initial calculations. Cumulative

effect: loads from initial calculations and equivalent forces is final result of static

calculations.

Rys: Autor

Purlin: bi-axial bending

Flat frame

Roof bracings

Loads as in initial calculations

Equivalent forces

Cooperation frame-bracings:

additional forces in purlins,

additional forces in frame.

Purlin: bi-axial bending +

compressive axial force

(recalculation)

Frame: additional axial forces

from cooperation with

bracings (recalculation)

2D model (method of equivalent flat frames) is much more complicated:

Photo: Author

6 x 3,70 [m]

12 x

6,0

0 [

m]

1,50 [m]

2,15 [m]

Example is made for 2D model.

Wind pressure and winf suction od

both front walls are beared by roof

bracings.

Roof bracings prevent trusses

chords from buckling → buckling

equivalent forces from chords act

on bracings.

Trusses chords are not ideal

straight (imperfection) →

imperfection equivalent forces act

on roof bracings.

Photo: Author


Top -55,112 302,157 Hot-rolled

CHS 108x10 Bottom -304,853 52,789

Connections with roof bracings S 235 → fy = 235 MPa

e = √ (235 / fy) = 1,0

Hot-rolled CHS 108x10 (108 mm diameter, 10 mm thickness):

A = 30,8 cm2

Jy = Jz = 373,0 cm4

iy = iz = 3,48 cm

S 235 + CHS hot rolled→

buckling curve a → a = 0,21

(EN 1993-1-1 tab. 6.1, 6.2)

L = 22,200 m

Chord E / R - tension E / R - compression in plane E / R - compression out of plane

Top 0,076 0,503 0,851

Bottom 0,421 0,081 0,412

Purlin: IPE 240

A = 37,22 cm2

Jy = 3675,10 cm4

Jz = 282,68 cm4

Wel, y = 306,26 cm3

Wel, z = 47,11 cm3

Wpl, y = 346,39 cm3

Wpl, z = 72,69 cm3

iy = 9,94 cm

iz = 2,76 cm

Photo: Author

y

z

qy = 0,942 kN/m

qz = 4,842 kN/m

E MEd, y MEd, z MEd, y + MEd, z D

E / R 0,268 0,248 0,320 0,960

Photo: Author

Roofing columns collect wind action and transport them to roof bracings and to columns’s

supports on masonry wall. Only half of action from each column acts on roof bracings: wind

action from blue area of fron wall.

Photo: Author

Purlins and roof bracings are in different parallel planes. Cooperation occurs for olny such

purlins, which have common nodes with bracings.

Bracing bars ( 25) are very slender and flexible. They lose stability under very small

compressive force. After buckling they not cooperate with resyt of structure. Only tensed

bars are taken into calculations.

Photo: Author

Photo: Author

Because of shape of truss, areas for each column are different. But it could be approximated

as equvalent areas: awerage height of columns x distance between columns.

Acol = [(1,50 + 1,50 + 2,15) /2 ] x 3,70 = 11,932 m2

Atotal = 6 Acol = 71,595 m2

Wind pressure + wind suction at the both ends of roof: qw = 0,8 kN / m2

Additionally important is wind friction on roof. It depends on wind action on roof,

qw1 = 0,5 kN / m2

Photo: Author

L = 12 x 6,0 m h = 1,50 m + 2,15 m (truss only; masonry structure is not taken into consideration) A (for roof bracings) = 1533,390 m2

h

d

h / 2

h / 2

L1

L1 = L – min (2d ; 4h)

EN 1991-1-4 tab 7.10

Surface cfr

steel, smooth concrete 0,01

rough concrete, tar-boards 0,02

ripples, ribs, folds 0,04

Friction of wind (EN 1991-1-4 7.5)

Wind total: Ftotal = Atotal qw + A cfr qw1 = 88,235 kN qw-t = Ftotal / d = 3,975 kN / m

Photo: Author

12 bands between trusses → 13 trusses. 2 braced bands in roof → m = 6,5 trusses per braced

band.

12 x

6,0

0 [

m]

Reduction factor for effects from instability

and imperfections (EN 1993-1-1 (5.12)):

am = √[ 0,5 (1 + 1 / m)]

m = 6,5

am = 0,760

Max compressive force in top chord of

vertical truss: NEd, c = 302, 157 kN

Buckling equivalent force from truss chord to

bracings:

Fbuck = am NEd, c / 100 = 2,295 kN

Initial horizontal bow inperfection for truss

chord: e0 = am d / 500 = 33,744 mm.

d = 6 x 3,70 = 22,20 [m]

Photo: Author

Two alternative schemes of action are taken into consideration:

• wind and equivalent effects from buckling;

• wind and equivalent effects from imperfections.

For final analysis more important will be bigger value of both above.

qw-t

Fbuck Fbuck Fbuck Fbuck Fbuck Fbuck Fbuck

qimperf

qw-t

Calculation of equivalet loads from imperfection qimperf for roof bracings is complicated.

According to EN 1993-1-1 (5.13), total eqiuvalent load (from wind and imperfection) is

calculated as:

qimperf = qd = S [8 NEd, c (e0 + dq) / L2]

dq – deformation comes from load qd

Total load qd depends on dq ,but dq is the effect of qd . Such problem can be solved by iteration

procedure:

dq0 = d(qw-t) = dw-t

qd(0) = qd

(0)(e0 + dw-t)

dq(1) = dq

(1)(qd(0))

qd(1) = qd

(1)(e0 + dq(1) + dw-t)

dq(2) = dq

(2)(qd(1))

qd(2) = qd

(2)(e0 + dq(2) + dw-t)

…

Static calculations of bracings must be made few times for iteration procedure.

First of all, deformation under wind action dq0 = d(qw-t) = dw-t must be calculated

Photo: Author

Fw-t = 3,975 [kN / m] ∙ 3,70 [m]

0,5 Fw-t 0,5 Fw-t

Steps of iteration:

dq0 = d(qw-t) = dw-t = 8,31 mm

qd(0) = S [8 NEd, c (e0 + dw-t) / L

2] =

= 6,5 ∙ 8 ∙ 302,157 [kN] ∙ (33,74 [mm] + 8,31 [mm]) / (22,2 [m])2 = 1,332 kN / m

Static callculation of deformation

dq(1) = dq

(1)(qd(0)) = 2,78 mm

qd(1) = S [8 NEd, c (e0 + dq

(1) + dw-t) / L2] =

= 6,5 ∙ 8 ∙ 302,157 [kN] ∙ (33,74 [mm] + 8,31 [mm] + 2,78 [mm]) / (22,2 [m])2 = 1,429 kN / m


dq(2) = dq

(2)(qd(1)) = 2,99 mm

qd(2) = S [8 NEd, c (e0 + dq

(2) + dw-t) / L2] = 1,435 kN / m


dq(3) = dq

(3)(qd(2)) = 3,00 mm

qd(3) = S [8 NEd, c (e0 + dq

(3) + dw-t) / L2] = 1,436 kN / m

Very small difference, end of iteration.

Photo: Author

Photo: Author

Photo: Author

Two situations must be taken into consideration: wind + equivalent effects of buckling…

Photo: Author

9,649 kN 9,649 kN

17,003 kN

17,003 kN

17,003 kN

17,003 kN

17,003 kN

…and wind + equivalent effects of imperfections.

Photo: Author

10,011 kN 10,011 kN

20,021 kN

20,021 kN

20,021 kN

20,021 kN

20,021 kN

Forces for the second one are bigger the second is important for calculations of

horizontal bracing..

Results (signs of forces in opposite way to Eurocode):

Photo: Author

Max tensile force in bracing bar: NEd, t, bb = 58,8 kN

Max compressive force in purlin: NEd, c, p = 60,1 kN

Max compressive force in chord: NEd, c, ch = 55,6 kN

Max tensile force in chord: NEd, t, ch = 49,4 kN

Checking of resistance: bracing bar

NRd, bb = fy A / gM0 = 115,355 kN

NEd, c, bb / NEd, bb = 0,510

Purlin: bi-axial bendint and axial force is calculated according to the same way as roofing

column, IInd example.

Checking resistance of cross-section:

NRd, p = fy A / gM0 = 874,670 kN

NEd, c, p / NRd, p = 0,069

Checking interaction NEd ↔ MEd, y for I-beam (EN 1993-1-1 (6.33) – (6.38)):

Influence of axial force for bending resistance of cross-section:

n = NEd / Npl, Rd

a = min [ 0,5 ; (A - 2 b tf) / A]

NEd ≤ min ( 0,25 Npl, Rd ;

0,5 hw tw fy / gM0)

NEd > min (0,25 Npl, Rd ;

0,5 hw tw fy / gM0)

MN, y, Rd Mpl, y, Rd min [ Mpl, y, Rd ; Mpl, y, Rd (1 - n) / (1 - 0,5 a) ]

NEd ≤ hw tw fy / gM0 NEd > hw tw fy / gM0

MN, z, Rd

Mpl, z, Rd

n ≤ a n > a

Mpl, z, Rd Mpl, z, Rd {1 - [(n - a) / (1 - a)]2}

min ( 0,25 Npl, Rd ; 0,5 hw tw fy / gM0) = 218,668 kN

hw tw fy / gM0 = 490,680 kN

Limits for both directions are bigger than axial force no influence of axial force for resistance of cross-section.

Checking stability for NEd and for interaction NEd ↔ MEd, y

For Your range of design project you can asume (not completely correct), that sanwich panels

protect purlins from instability.

Chords of roof bracings.

Both chords of roof bracings are simultaneously top chords of vertical trusses. New values of

forces in trusses must be taken into consideration as additional forces applied to trusses.

Max compressive force from bracings together with combination of loads which gives max

compression in top chord.

Max tensile force from bracings together with combination of loads which gives max tension

(or min compression) in top chord.

Photo: Author


Top -103,362 356,222 Hot-rolled

CHS 108x10 Bottom -315,973 57,729

New forces in chords

Chord E / R - tension E / R - compression in plane E / R - compression out of plane

Top 0,143 0,591 1,003

Bottom 0,436 0,089 0,451

More massive cross-section must be applied for top chord. For such a litte difference change

from CHS 108x10 to CHS 108x11 is enough. Bottom chord should be the same more

massive to avoid mistakes durnig manufacturing proces (two very similar cross-sections:

CHS 108x10 and 108x11 could be difficult to distinguish).

Of course, recalculation must be made for each truss members, nort only for chords.

Thank you for attention

© 2021 Tomasz Michałowski, PhD

[email protected]

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Metal Structures Design Project I Steel truss