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STEEL STRUCTURE I / STRUKTUR BAJA I
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INTRODUCTIONStructure Material : wood , concrete and
steelStrength : wood ( < 50 MPa), Concrete ( 50
MPa) Steel ( 500 MPa)Tension members are define as structural
elements that are subjected to axial tensile forces.
Shape use as tension members :
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TENSION MEMBER IN STRUCTURE
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BEHAVIOUR AND STRENGH OF TENSION MEMBERSLOAD – DEFORMATION RELATIONSHIP
to get the diagram Tensile test (ASTM)According to SNI & ASTM
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Lo
Diameter of specimen
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The determination of stress and elongation is based on the following Formulas:
Ultimate strain : ε = ∆l / lYield stress : σ yield = F yield / AUltimate stress : σ ult = F ult / AWhere :
Lo : The specimen length (mm)
L : Ultimate elongation (mm)Fyield : Yield force (N)
Fult : The ultimate force (N)
A : Section area of the specimen (mm2)
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STEEL TYPE (SNI) Fy (MPa) Fu (MPa)
BI 37 240 370
BJ 41 250 410
BJ 50 290 500
BJ 55 410 550
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DESIGN STRENGTHA tension member can fail by
reaching:excessive deformation or fracture.Stress P/A must be less than a limiting stress
F :P/A < FPn = Fy Ag (nominal stress in yielding)Pn = Fu Ae (nominal strength in fracture)Resistance factor (φ) : for yielding = 0,90
for fracture = 0,75Fy Ag = Fu Ae
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EFFECTIVE NET AREA
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DESIGN OF TENSION MEMBERSPu < φ Pn Pu : Sum of factored load
To prevent yielding :0,90 Fy Ag > Pu or Ag > Pu / 0,90 Fy
To avoid fracture : 0,75 Fu Ae > Pu or Ae > Pu / 0,75 Fu
The slenderness ratio limitation : r > L / 240 (utama)r > L / 300(sekunder)r : minimum radius of giration
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FLOWCHART : TENSION MEMBER DESIGN TENSION MEMBER DESIGN
FRIGHTTENED LOAD
ADDITIONAL LIFE LOAD
DETERMINE Fy, Fu
CALC. A needed = P/Fy
SELECT SHAPE TYPE r < L/300
CALC. Ag and Ae
CALC. ACTUAL STRESS F = P/Ae
F < Fy
FINISH
NO
YES
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2. Given : A double line of standard holes 7/8 in. bolts are placed in a 10 x ¾ plate.Determine the gross and net areasGross area Ag = 10 (3/4) = 6.50 in2 Effective hole size for a 7/8 in diameter bolt :dc = 7/8 + 1/16 + 1/16 = 1 inNet area An = ( 10 – 2 (1)) (3,4) = 6 in2
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EXAMPLE1. Given : A single line of standard holes for ¾
in. bolts are placed in a 6 x ½ plate. Determine the gross and net area.Solution :Gross area Ag = 6 (1/2) = 3 in2
Net are An = (b - dc) tdc = ¾ + 1/16 + 1/16 = 7/8 inAn = (6 – 7/8) (1/2) = 2,56 in2
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Sambungan geser baut : Rn = m x r x fu x A r = 0.4 (pada ulir ) ; 0.5 (tidak pd ulir)
fu : kuat tarik baut, A luas penampang baut pd tempat tidak berulirm : jumlah bidang geser
Sambungan tumpu : Rn = 2.4 x d x t x fu d : diameter lubang ; t : tebal profil ; fu : kuat tarik profil
Sambungan tarik : Rn = 0.75 x fu x A
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