Table of Contents Table of Contents
Previous Page  203 / 228 Next Page
Information
Show Menu
Previous Page 203 / 228 Next Page
Page Background

9. CALCULATION EXAMPLES OF LVL STRUCTURES

Geometry conditions:

Minimum distance to edge

a

2,CG

in stud ≥ 4

d

= 4∙6,0 mm = 24 mm. Stud thickness 51 mm/2 = 25,5 mm

→ Screw size 6,0x140 mm is OK for the stud

Min. screw spacing

a

1

in stud ≥ 10

d

= 60 mm

Min. screw spacing

a

2

in beam ≥ 5

d

= 30 mm

a

1

in the stud is more critical

Distance to edge of the beam

a

2,CG

≥ 4

d

= 24 mm. When the screwing angle is 45°, the beam thickness

t

1

/2 = 25,5 mm gives the minimum distance.

Maximum number of screws in the connection:

1+((h_beam – 2 ∙ min a_(2,CG) ))/((min a_(1,stud) / sin 45°) )=1+((200 mm – 2 ∙ 25,5 mm))/((60 mm

/ sin 45°) )=1+(149 mm)/(85 mm)=2,79

→ 2 screws are chosen for the connection, so that the heads of the screws are 20 mm and 110 mm from

the bottom edge of the beam.

1 + �ℎ

beam

– 2 ∙ min

2,CG

�min

1,stud

/ sin 45°� = 1 + (200 mm – 2 ∙ 25,5 mm)

(60 mm / sin 45°) = 1 + 149 mm 85 mm = 2,79

Minimum distance to the beam end

a

1,CG

≥ 10d = 60 mm. Therefore the end of the bean shall exceed the

stud edge.

Effective penetration length

l

ef,1

in ledger beam is

l_(ef,1)=l_(g,1)=t_1/sin⁡〖45°〗 -l_u=(51 mm)/(sin 45°)-17 mm=55 mm

Penetration length in wall stud l_(g,2)=l-t_1/sin⁡〖45°〗 =140 mm-(51 mm)/(sin45°)=68 mm

For the beam the angles in the connections are:

α

= 45°,

β

= 45° and

ε

= 90° and for the stud they are:

α

= 45°,

β

= 0° and

ε

= 45°.

ef,1

=

g,1

= t

1

sin 45° − l

u

= 51 mm sin 45° − 17 mm = 55 mm

enetration length in wall stud

g,2

= −

1

sin45°

= 140 mm −

51 mm sin45°

= 68 mm

k

=

0,9

T,k

(cos + sin )

(5.33)

T,k

= min ⎩⎨ ⎧max �

ax,90,1,k g,1

;

head,k h2

k a

0,8

ax,ε,2,k g,2

tens,k

(5.31)

ax,ε,k

=

ax

ax,90,k

1,5 cos

2

+ sin

2

k a

0,8

(5.32)

ef,1

=

g,1

= t

1

si 45° − l

u

= 1 m sin 45° − 17 m = 5 m

Penetration lengt in all st d

g,2

= −

1

sin45°

= 140 m −

51 m sin45°

= 68 mm

k

=

0,9

T,k

(cos + sin )

(5.3 )

T,k

= min ⎩⎨ ⎧max �

ax,90,1 k g,1

;

head,k h2

k a

0,8

ax,ε,2 k g,2

tens,k

(5.31)

ax,ε,k

=

ax

ax,90,k

1,5 cos

2

+ sin

2

k a

0,8

(5.32)

Connection capacity

The characteristic load-carrying capacity of the tension screw connection, see Figure 5.11(b), is cal-

culated by the equation:

R

k

=

n

0,9

R

T,k

(cos

α

+

μ

sin

α

)

(5.33)

LVL Handbook Europe

201