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eccentric moments in trusses 1
- Thread starter kags
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Hi Kags,
The moment is usually distributed into the chord each side of the joint as the bracings are considered pinned and chord continuous.
The moment is the combined horizontal component of the braces (along the axis of the chord) is multiplyed by the eccentricity ( distance between the chord centreline and point at which the bracings centrelines cross).
This is then split into the chord each side of the joint with regard to the chord stiffness (I/LE)
where I = chord second moment of area
and LE = chord effective length
Example
Chord SHS 150x150x10 S355 EN10210 I=1773 cm^4
Two Braces at 45deg. to chord forming K-joint with 6mm eccentricity.
Brace 1 = 267 kN Tension
Brace 2 = 300 kN Comp
Effective length of chord side 1, LE1 = 270cm
Effective length of chord side 2, LE2 = 135cm
Horiz component of braces, F = 267 cos 45° + 300 cos 45°
F = 401 kN (both acting in same direction so added)
Moment due to eccentricity, M = 401 x 0.006 = 2.406 kNm
Chord side 1 stiffness = I/LE1 = 1773/270 = 6.57
Chord side 2 stiffness = I/LE2 = 1773/135 = 13.13
Moment chord side 1 = 2.406 kNm 6.57/(6.57+13.13)= 0.80 kNm
Moment chord side 2 = 2.406 kNm 13.13/(6.57+13.13)= 1.60 kNm
Hope this helps
The moment is usually distributed into the chord each side of the joint as the bracings are considered pinned and chord continuous.
The moment is the combined horizontal component of the braces (along the axis of the chord) is multiplyed by the eccentricity ( distance between the chord centreline and point at which the bracings centrelines cross).
This is then split into the chord each side of the joint with regard to the chord stiffness (I/LE)
where I = chord second moment of area
and LE = chord effective length
Example
Chord SHS 150x150x10 S355 EN10210 I=1773 cm^4
Two Braces at 45deg. to chord forming K-joint with 6mm eccentricity.
Brace 1 = 267 kN Tension
Brace 2 = 300 kN Comp
Effective length of chord side 1, LE1 = 270cm
Effective length of chord side 2, LE2 = 135cm
Horiz component of braces, F = 267 cos 45° + 300 cos 45°
F = 401 kN (both acting in same direction so added)
Moment due to eccentricity, M = 401 x 0.006 = 2.406 kNm
Chord side 1 stiffness = I/LE1 = 1773/270 = 6.57
Chord side 2 stiffness = I/LE2 = 1773/135 = 13.13
Moment chord side 1 = 2.406 kNm 6.57/(6.57+13.13)= 0.80 kNm
Moment chord side 2 = 2.406 kNm 13.13/(6.57+13.13)= 1.60 kNm
Hope this helps
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