Question

1.- A Tee section is made of two plates welded together as shown below. Assume that no residual stresses are present. a) Compute the moment at first yield and the plastic moment for strong (x-x) axis bending of the section if both plates are made of A36 steel. PL 8" x 1" b) Repeat a) for weak (y-y) axis bending. c) Repeat a) if the plate used for the stem of the Tee is made of A36 steel, while the plate used for the flange is made of A514 Gr. 100. PL 10" x 12" --------- d) Repeat c) for bending about the weak (y-y) axis. --> e) Compute the shape factor (i.e., the ratio between the plastic moment and the moment at first yield Mp/My) for cases a) through d) above. Summarize your calculated values in a table. Briefly discuss your results.

Answer 1

ANSWER ::

Residual stresses in steel. ... DEFINITION:  Locked stresses in steel which are present prior to the application of stress are called “Residual stresses.”  These stresses are unintended and undesirable in structural steel which cause it to fail prematurely.

  
Given that: A36 Stee section 1 b=8" d=1" fy = 3600D psi Section 2 b=10" . d=0.5" Solution , for strong x-x axis: the plastic moment a) compute as shown in below; relation By using the For section I = 72000 lb.in For section a tip = (0.5 61012) 35000

9000 = 0.5 x 100 x36eme = 450,000lb.in calculate the total section : MP 72000 + 450000 - 592,000 lb.in b) yay axis : By using the in as shown delation below; Fot section a Mp = (both ty - 4x9000 = 576,000 lb-in Fot section 2 10(0.5) 9000 MP = 1 36000 * = 10(0.95) 9000 = 22,500 lb.in

Calculate the total section: HP = 576000 + 22500 = 598,5000 lbin c) Fob Steel Properties of A51466-100 yields sttength fy = 100,000 psi Fox section 1 - 200,000 lbs in for section 2 My = ( 05 (702) 10,000 0-25 X 100 X 100,000 = 1250,000 lb-in Calwlate the total section: My = 200000 + 1250000 - 1450,000 lbiin

d) Bending about the week yay axis using the belation as shown in below; For section 1 = (1072) 100,000 - 1600000 16. in For section a My = / 10(0.5) ²) 100,00 10x0.25 -X100,000 = 62,500 lb.in Glwlate the total section: My = 62500 +1600,000 = 662500 bin e) Determine the fol shape factor:

SF = 542, доо 1450 ; боб – 0-36 SF - 599 5000 1662,500 - 3-6 vawes: table Glculated Plot the АР (-4 axi) 529,000 jp F10 (-8 axis) S48 , 5000 1b.m Ну (х- і зія) 1950, 000 р. у гч (Ч-У цуі%) 1662 500 1.30 shape factor x-X axis o.36 у-у ад 19 3.3