Question

A force and a temperature change were applied to a member. What is the elongation of the following member in inches (to the nearest 0.0001)? a=15 x 1061/°F P = 452 kip L = 33 in To = 62°F Tp = 56 °F E = 10,877 ksi A = 1 in 2 Р b

Answer 1

Solution:- Whole solution is shown in below images.

Part (a) :- First of all find the elongation of the member due to tensile load (P) and due to change in temperature (∆T) separately. The final total elongation of the member is equal to the sum of the elongation due to load and the elongation due to change in temperature.

The elongation of the member is obtained as 1.3742 inches.

Part (a):- Given: a = 15x100 per of P = 452 kip L = 33 in To = 62°F T = 567 E = 10,877 Ksi A I in² SP a To find - Elongation in given bar in inches (to the nearest 0.0001) Solution - Elongation in the memberi due to tensile load (P) is given PL AE (452 kip) x ( 33 in.) (1 in) x(10877 ksi) Top = 1.3713 in. . ture To - Tt Elongation due to change in Temperature :- AT = x.&T: L (2) Here, AT = change in Temperature = 62°F – 564 OT = 6°F Substituting in equation @, we get, Klongation, At = (15 X166 )*(6°F)*(33 in) = 0.0029 in LOT Final in the member, 0 = elongation Opt of = 1-3713 + 0.0029 10 = 1.3742 inches] It is the required elongation in the member.

Part (b) :- First of all find the moment of inertia of the symmetrical I- section about the horizontal axis at the centroid. Substitute the respective values in the given formula to find the maximum bending tensile stress of the cross-section.

The maximum tensile stress is obtained as 33.1 Ksi.

Part (b): Given: a = 0.5 in. b = 6 in. c = 0.4 in. d=3in. M= 59 K.ft {mo Moment is applied around the local x-axis i.e. horizontal axis at the centroid. Ia=0.5in 3.5 in b=bin Х a=0.5 in c! K d 3 in d 3 in. 0.4in. nearest O. 1 ksi To find - Maximum tensile stress in ksi to the Solution:- Maxinum tensile stress of the cross section is given by :- Я M Ixx Ixx I [(a+c+d)(20 +5°] - [@+a) (69] Los [ {{3+04+3)(2905 «c} - {(8-456] Tas [?(64700"-{10}] 1 [(2195-2) - (1230)] Ixx 11 12 4 > Ixx 74.933 in 2

Part (c) :- First of all find the reaction at support in the simply supported beam and then from the shear force diagram, find the corresponding maximum shear force (Vmax) value. Now, the maximum shear stress of the rectangular cross-section is 1.5 times the average shear stress. Substituting all the respective values, find the maximum shear stress.

The maximum shear stress obtained is 18.7 Ksi.