Question

1) As I have have provided figure below, which is made of Steel (E=200 GPa) with a solid rectangular cross section somm wide and lcomm creep.. derive an att algebraic expression for the deflection as a functi of x = the distance from the left end. Hence also calculate the slope at the left support, and the deflection am from the left support 20 kalim 12 kN/m sm

Answer 1

12 kry 20 kNom Dk sm A od I GOKA 20Kons 2.5m TAY . Ay tby = 60 - EMATO -20 + 60 (25) = By (5) 130 = By . By = 26kN) . Esan 2 : Ay = 60-26= 34 kal. Akryy zoknow 26kn. 34kn

Sigularity functions - Win) = -20 <x-o> 20<x-03 734 24-017-12<x-oj +34 2 - + 26 <X-55 v(x) = f wix) Vex) = -202x}+\+34 2x3° – 12 2xy +262x-3° MAX) = -20.433° + 34 <x>' - 12 xxx' + 2624-99 2 5-20 <x3° + 34 <xy! – 6 <xy +26 2x-53' Mix) = Mos e os < > they á (<20483+34 23) –62x3°+262853 hebben (-ocx)+34483_4234242x-3) +4 · = (-2018' +172x)=24x3+134x=8374) slope en. 1 . Intigrate slope an 3 3 4 - 2X) +7LX 20LXS +) -defle din Equatilin, at x=0, y 8.8 x=2.5, dy (o=o.

96) == 0 +0 -0 to to tea) 0 = (2) 425) = 1 / 3 ( 10 (55+ 17 (833 _// 1557 +594) OF ( - 65 44 145.83459) O = 145763+54 G = -1453 = -29.16. 5 ET .y==(-1068* +43 (0.03 04 (my tu cas! . -29.1691) ► deflection Emm ho pe=(-20 () ' +17 (2)*—20°+13 (45 - 29.16 ) slope . .. slope at left support (250) op (-20(0)+1960) –0 +131-5j25.06) son a 6,66810 out 290784 200x109 x 6.66x10 6 = 2:224,07 m 21222450 ty

deflection am from left Support fr= 2m) 412)=4-1082 414 (3) 3-4 (29) #3 (2-5 -29.16 (33) z EI -- = -1.33x10 m -177.98 200 Holy 6.664106 deflection in downward y (2) == 1.33x10 tus