Question

I= 940 in^4; E= 29050 ksi. Values for a= 9 and c= 7; please help!!

Problem #1 The Moment of Inertia I = (900 + 5 b) in^ and Modulus of Elasticity E = (29,000 + 50 d) ksi. Determine the vertical deflection at point B on the beam. (Note: Provide your answer in units of inches using 3 significant digits of accuracy.) (1 + a) kips/ft C! A B |--(4+ c) ft- (10 + C) ft

Answer 1

Given data, ta = 10 kips/bt I= 940 int. B E 29050 Ksi A 4+0=11ft 9, c= 10+ 0 = 1767 t To find: AB: 9t is recommended to use Moment Area Method to find slope and deflection at different locations of a cantilever beam. However use others methods too. we can Let's first solve this problem using Moment Area Method. Steps Computations for Computations for M diagram. EI Sign Convention: A +

Mx = -10 x x x -502 2 It is an equation of 2° curve (ie Parabolic sprandle) whose d My which is five and increasing. slope -10x dre O Moreover Mx=oft Mx = lift = -5x112 = -605k-ft. Mx = 17 ft -5x172 - 1445 k-ft. Based on EI above informations; plot the M diagram along with the deflected profile of beam. -605 Er B A - 1445 EI B А 8 Initial Position of beam Able Deflected position. LAAK AB/C we can → Deflection at B with respect to tangent drawn from c (ie the tangent from c is horizontal) Therefore directly say that Able is equal to deflection at B ie 4B and similarily at location A. This why Moment Area Method is suggested for cantilever beams; as fangent from fixed support only becomes the reference Axis

Step 2: Deflection Calculations: Moment of area of M diagram between Band - ABIC EI C about B. we we Note 96 would have required to find slope at B je OR then would have done the same thing as Ob/c there dok = Slope at B with respect to tangent drawn from c. However Oc = 0. (due to Fixity at support c). Hence, 08-Oc diagram between B and c. OBICE Oble Area of a A B EI Area of ACDA A =-17 X 1445 x1 = 1445 Ег 8188:33 ET EI 3 D CG of A, from a z ft = 4.25ft. x = 17 Note For Parabolic sprandler Area X rea of A BEA =-118605x EI b A2 - 2218.33 EI xt x = b , A = bh. 4. CG of Az from C X = 6+ 4 = 8.75ff. For Parabola. b A = 2 bh, x = 36

X = Ax - A₂X2 A - A₂ -8188.334.25-(-2218.33x8.75 (-8188.33) (-22 18.33) Ť = 2.58ft from c. of Anet A, -A2 = _5970 EI from ß = 6-2.58 = 3.42ft. 5970 x 3.42 ff EI 20417.4X120 Kain} 290504 940K-in2 Note - see the unit of Anet AB - 1.292 inches (59704 Er Here fine sign indicates deflection is down ward. 'This has unit of (K-ft) xft -This has unit of Kain2. Method 2: Using Double Integration Methode lo kips/ft Mae = - 10 x x x = -5x² A B x Line Diagram dy -502 dre EI EI Integrating twice; we get Erdy - 55 x + fG dre

245 12. 11 + 8188.3? at C₂ > GF-104401.193 C is in Kafts, EI y = - + C, ata Apply Boundary conditions, At n = 17ff, y=0 and dy = 0 C = 8188.33 0 = -5x17 12. X17 So, EI Y = t 8188.332 104401. 193 72 At a= lift ie at Bi Ey = I x 11" + 8188.33x11 - 1044 01193 -20429,980 EFY Note- Here Unit of Enly = (-20429.98) sa4 y = -1.293 inches "It has unit of Krin?, So be carefull with units and But the proper units int the final step to obtain correct results.