Question

USE THREE-MOMENT EQUATION

Problem 1. Determine the reactions at the supports using Three-moment equation. Assume A is a pin and B and C are rollers. El is constant. 10k 2.5 k/it B - -10 ft--++-10 ft + 25 ft

Answer 1

Draw the free body diagram of the beam: 10K 2.5K.ft С A B 10 ft 10 ft 25 ft * C Apply equations of equilibrium: MA=0 C,(45)–10(10) - (x2.5425)*(* + 20 = 985.416 Cy 45 = 21.9k F.=0 -10 -x 2.5 x 25 2 Ay = 31.25 +10-Cy = 41.25 -21.9 =19.35k

Fr=0 A=0 Determine the internal moments Mof the primary structure using the figure below: x 1.X x 10K 2.5K.ft A С B Kx2 X1 X3 21.9K 19.35K 10 ft 10 ft X 25 ft K * > Bending moment at x-x which is between AB and to the left of 10k is M=19.35(x1) Bending moment at x-x which is between AB and to the right of 10k is M2 = 19.35(10+x2)-10 (x2) = 193.5+9.35x2 From similar triangles y 2.5 хз 25

2.5X3 y = 25 = 0.1 x3 Bending moment at x-x which is between BC and at a distance xz from end C xz Mz = 21.9 xz 0.1.xz XX3 2 3 = 21.9.xz -0.01667x3 Now consider the free body diagram of the redundant beam with 1k load at B X B A С 10ft Yoft ik 25ft =0.4444 K X1 X3 X Apply equations of equilibrium Taking moments about A and equate to zero MA=0 -C, (45)+1(20)=0 Cy=0.4444k

Sum of all vertical force is equal to zero F,=0 -A, -C, +1=0 Ay = 1-CY = 1-0.4444 = 0.5556k Bending moment at x-x which is at a distance x from and A m =-0.5556x1 Bending moment at x-x which is at a distance x2 m2 =-0.5556(10+x2) Bending moment at x-x which is at a distance xz from C m3 =-0.4444xz.

Choose the vertical reaction at B as the redundant. Let the reaction B, be applied in the upward direction. Apply the principle of superposition to the beam. The deflection of the beam is shown below: 2.5K.ft 110K WWW С A B C = 2.5K.ft 10K B AT С Дв Primary structure

+ B. y fi BB C A B B y redundant The compatibility equation for the beam is 48-B, SBB=0 Here, 43 is the displacement at joint B due to real loads, B, is the vertical reaction at joint B, and fbb is the flexibility coefficient at joint B. Use virtual work method to find the values of 43 and fBB. Consider real loads and find Ag using the equation pl Mm 13 = $ dx Here, Mis the moment in the beam caused by real loads, m is the virtual moment in the beam caused by the external virtual unit load, EI is the flexural rigidity of the beam. The internal consents (M) of the primary structure have to be determined. so Jo EI

Find fbb using the equation. 583 = S. EI L mm dx Jo EI •4 mm L2 m,m dx + EI JO ΕΙ 1(-0.5556.xz)? dxz + ΕΙ = - dx + do [? L3 m3 m3 - dxz ΕΙ 10 25 (-5.556-0.556x,)? - dx₂ + (-0.4444xz)? FBB = • dxz ΕΙ ΕΙ 0 10 10 10 10 25 0.309x ЗЕІ 0.197x + 30.87x2 EI 0.309x3 ЗЕІ + + 6.18.x3 2 EI + ЗЕІ 0 1 = (103+308.7 +103 +309+1026) ΕΙ 1849.7 ΕΙ

Find the value of B, using the compatibility equation 43 - B, SBB=0 -60255.3 1849.7 Substitute for fbb in the above equation ΕΙ ΕΙ for AB, -60255.3 1849.7 -B, =0 ΕΙ ΕΙ 60255.3 By 1849.7 = 32.58 k Therefore, the vertical reaction at joint B is 32.58 k Free body diagram of the beam is shown below: 10K 2.5 K.ft A Ą С 1B C 10 ft 10 ft 25 ft B

1 *2.5x25) { 3 Find the reaction C, by taking the moments about A MA=0 25 B,(20)+C, (45)-10(10)-1 = x2.5x 25 x - + 20 = 0 2 45C, = 985.417 – B, (20) 45C, =985.417 – 32.58(20) 335.817 cy 45 = 7.42 k Therefore, the vertical reaction at joint C is 7.42 k. Find the horizontal reaction at joint A using equation of equilibrium Fx=0 Ą=0 Therefore, the horizontal reaction at joint A is 0

Find the vertical reaction at joint A using equation of equilibrium F, =0 4,+By+C,-(<2.5x25 )-1 ין Ay = 41.25–32.58–7.42 4, = 1.25k Therefore, the vertical reaction at joint A is 1.25 k. The beam with the vertical reactions is shown below: 110K 2.5K/ft D AT С B 7.42K 10 ft 10 ft 25 ft 1.3K 32.5K

Compute the bending moment: Bending moment at A=0 Bending moment at D is Mp=1.3(10) =13 kft Consider a section x-x at a distance x from end C. Shear force at x-x is 1 =-7.42 + -x 2.5xx 2 =-7.42 +1.25 x The maximum bending moment occurs where the shear force is zero V=0 -7.42 +1.25x = 0 x=5.94 ft 2.5 у X K 25

= X Find the ordinate y from similar triangles y 2.5 25 2.5x y 25 = 0.1.x Find the moment at x-x M = 7.42x-1 (2x0.xx. x 0. 1x xx x 3 = 7.42 x -0.0167 x3 At x = 5.94 ft The maximum bending moment is M = 7.42(5.94)-0.0167(5.94) = 44.075 -3.5 = 40.575 kft Therefore, the maximum bending moment is 40.575 k ft. At r= 25 ft The bending moment is M = 7.42(25) – 0.0167(25) = 185.5-260.94 = -75.44 kft

Plot the bending moment diagram using the above values. (M)(K.ft) 40.5 13K.ft 0 45(ft) 10 20 ---75.4K.ft