FE ( Fundamentals of Engineering ) Supplied-Reference by Ncees

By Ncees

This can be the authentic reference fabric utilized in the FE examination room. evaluation it sooner than examination day and get yourself up to speed with the charts, formulation, tables, and different reference info supplied. word that non-public copies aren't allowed within the examination room. New copies should be provided on the examination website.

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0h 25. # dx = 1 tan- 1 x a a a2 + x2 26. # dx = 1 tan- 1 b x ac l, ac ax + c 27a. # dx = ax 2 + bx + c 2 tan- 1 2ax + b 2 4ac - b 2 4ac - b dx = ax + bx + c 2ax + b - b 2 - 4ac 1 1 n b 2 - 4ac 2ax + b + b 2 - 4ac -1 _0 < sec- 1u < r/2i _- r # sec- 1u < - r/2i d _csc- 1ui du 1 =- 1. 2. 3. 4. u u - 1 dx 2 _ 4ac - b 2 > 0i 2 _0 < csc u # r/2i _- r < csc u # - r/2i -1 -1 27b. # 27c. # 2 ^a > 0, c > 0h dx 2 , =2ax + b ax 2 + bx + c _b 2 - 4ac > 0i _b 2 - 4ac = 0i MATHEMATICS 27 MENSURATION OF AREAS AND VOLUMES Circular Sector ♦ Nomenclature A = total surface area P = perimeter V = volume Parabola A = zr 2/2 = sr/2 z = s/r Sphere ♦ Ellipse ♦ V = 4rr3/3 = rd3/6 A = 4rr 2 = rd 2 Parallelogram Papprox = 2r _a 2 + b 2i /2 R 2 2 S1 + _ 1 2i m 2 + _ 1 2 # 1 4i m 4 S 2 P = r ^a + bh S+ _ 1 2 # 1 4 # 3 6 i m6 + _ 1 2 # 1 4 # 3 6 # S SS+ _ 1 # 1 # 3 # 5 # 7 i2 m10 + f 2 4 6 8 10 T where m = ^a - bh / ^a + bh V W W 2 5 i m8 W, 8 W WW X P = 2 ^a + bh d1 = a 2 + b 2 - 2ab ^cos zh d2 = a 2 + b 2 + 2ab ^cos zh d12 + d22 = 2 _a 2 + b 2i Circular Segment ♦ A = ah = ab ^sin zh A If a = b, the parallelogram is a rhombus.

3. All of the external loads lie in the plane of the truss and are applied at the joints only. 4. The truss reactions and member forces can be determined using the equations of equilibrium. Σ F = 0; Σ M = 0 5. A truss is statically indeterminate if the reactions and member forces cannot be solved with the equations of equilibrium. Plane Truss: Method of Joints The method consists of solving for the forces in the members by writing the two equilibrium equations for each joint of the truss. Σ FV = 0 and Σ FH = 0, where FH = horizontal forces and member components and FV = vertical forces and member components.

Donald E. Hudson, Applied Mechanics Dynamics, D. , Princeton, NJ, 1959. W. E. Hudson. y y y y y y Figure 52 STATICS y y a a a b a C 2a C C C C x x x x x 2 [ 2a sin 3θ 3 θ − sinθ cosθ yc = 0 xc = A = a2 θ − sin 2θ 2 A = a 2θ 2a sinθ xc = 3 θ yc = 0 yc = 4a/(3π) xc = a A = πa /2 yc = a xc = a A = π (a2 – b2) ( ) Iy = Ix = 4 2sin 3 θ cosθ Aa 2 1+ 4 θ − sin θ cosθ ] 2sin 3θ cosθ Aa 2 1− 4 3θ − 3sinθ cosθ [ [ Iy = a4(θ + sinθ cosθ)/4 Ix = a (θ – sinθ cosθ)/4 4 I x = πa 4 8 I y = 5π a 4 8 I yc ( ) a 4 9π 2 − 64 72π = πa 4 8 I xc = ( 5πa πb Ix = Iy = − πa 2b 2 − 4 4 J = π a4 − b4 2 4 I xc = I y c = π a 4 − b 4 4 ] ) +b ) 4 = ) 2 ) 2 ry2 = rx2 = 2sin 3θ cosθ a2 1+ 4 θ − sinθ cosθ [ [ ] 2sin 3θ cosθ a2 1− 4 3θ − 3sinθ cosθ a 2 (θ − sinθ cosθ ) θ 4 a 2 (θ + sinθ cosθ ) 2 ry = θ 4 rx2 = ry2 = 5a 2 4 rx2 = a 2 4 ( a 2 9π 2 − 64 36π 2 2 2 ryc = a 4 rx2c = ( ry2 rp2 = a 2 + b 2 2 rx2 ( = (5a rx2c = ry2c = a 2 + b 2 4 =a 2 rp2 J = πa 4 2 yc = a 2 rx2 = ry2 = 5a 2 4 I x = I y = 5π a 4 4 xc = a rx2c = ry2c = a 2 4 ] (Radius of Gyration)2 I xc = I y c = π a 4 4 ) Area Moment of Inertia A = πa2 ] Area & Centroid Housner, George W.

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