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