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STEEL STRUCTURE I / STRUKTUR BAJA I
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 :
TENSION MEMBER IN STRUCTURE
BEHAVIOUR AND STRENGH OF TENSION MEMBERSLOAD – DEFORMATION RELATIONSHIP
to get the diagram Tensile test (ASTM)According to SNI & ASTM
Lo
Diameter of specimen
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)
STEEL TYPE (SNI) Fy (MPa) Fu (MPa)
BI 37 240 370
BJ 41 250 410
BJ 50 290 500
BJ 55 410 550
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
EFFECTIVE NET AREA
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
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
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
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
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