By Barry S. Onouye, Kevin Kane

Be aware: Scanned B/W PDF with OCR. details lower than refers back to the textbook that accompanies this answer manual.

Statics and energy of fabrics for structure and development building, Fourth variation, bargains scholars an available, visually orientated creation to structural thought that doesn't depend on calculus. as an alternative, illustrations and examples of creating frameworks and elements allow scholars to raised visualize the relationship among theoretical ideas and the experiential nature of actual structures and fabrics. This re-creation contains totally labored examples in each one bankruptcy, a better half website with additional perform difficulties, and accelerated remedy of load tracing.

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Extra resources for Instructor's Solution Manual for Statics and Strength of Materials for Architecture and Building Construction (4th Edition)

Example text

4 From the solution for Prob. 3, it appears that the force in cable AB is maximum. Therefore, assume the following: reaction @ A = 20k and AB = 20k. Using the original FBD of Prob. 73k il£. 96' B [E F x = 0 ] C B X = 19k [S F y = 0 ] + C B y + 3 . 2 7 k hc - h B 2 ff CBy 3 ,2 7 k “ C B X ~~ 1 9 k (h‘= -hB) = 2{ l ^ i ) =3-44, h c = h B + 3 . 7k [ZFy = 0 ] + A - 2 k - 2 k - ( 0 . 11 Ri R2 Ra 8 . 12 1000# Step 1: FBD of the entire truss. Step 2: Solve for the external support reactions. ,, = 1-1433* [ZFX...

56 y ! 22CB 70? 8kN(6m) = 0 M RC = 8. 5#(f) Chapter 3 Problem Solutions : ■ E X _V 29’ [Z M a = 0] y “ V 29 + ^ - ^ - J ( 4 0 ,) - ( 3 0 0 # ) ( 3 0 ' ) - ( 3 0 0 ^) ( 2 0 ,) - ( 3 0 0 # )( 1 0 ,) = 0 E = +22SV 29 = +1212* E x = 1125#; Ey =450# [Z F x = 0 ] C B X = E X = 1125* [Z F y = 0 ] + C B y -3 0 0 * -3 0 0 * + 450* = 0 C B y = +150* .. (15QJX10') n3. 7 k ( 5 2 . 1m) .... 7k (52. -. 4 From the solution for Prob. 3, it appears that the force in cable AB is maximum. Therefore, assume the following: reaction @ A = 20k and AB = 20k.

9 dlx = d! 3 3 m ) M a = - 9 . 9 5 k N - m - 5 . 8# Ma = +T£x(30’) + T2y(6*) - T 1y(35') - T1y(4') =0 M. 51 R = 2 F y = 10 # + 7# + 6# - 18# = + 5 # M0 = +(7#)(4”) + (6#)(9”) - (18#)(17” = - 224#-in. 52 150N Weight of wood member: 100N co = 30N/m Assume the member weight is located at the center of the length. 4m to the left of the origin. 53 y 400# 200# 11 100# 20#/ft TT~T I I I l I I H r ~ r r origin 4' Total beam weight equals (20%)(16') = 320# at the center of the beam length. For the beam to remain stationary and horizontal, the moments taken about points A and B should be balanced by the opposing moments due to B and A respectively, resulting in no resultant moment.

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