FUNDAMENTALS OF STRUCTURAL ANALYSIS

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1 FUNMENTLS OF STRUTURL NLYSIS 5th Edition Kenneth M. Leet, hia-ming Uang, Joel T. Lanning, and nne M. Gilbert SOLUTIONS MNUL HPTER 5: EMS N FRMES 5-1 opright 018 McGraw-Hill Education. ll rights reserved.

2 P5.. Write the equations for shear and moment between points and E. Select the origin at. 6ʹ 8 ips w = 3 ips/ft E 5ʹ 3ʹ 10ʹ P5. =- 6(8) 13(3 10) -18E = 0 ΣF ΣF = E = 0 E = 19 ips = 11 ips = E - 8 = 0 E = 8 ips 8 ips 3 ips/ft E ΣF = V -3(10 - ) 19 V = 3 11 ips æö = M 3 19( ) 0 ç è - = ø 3 M = - 19 ip ft V M 3 ips/ft 10- E 5-3 opright 018 McGraw-Hill Education. ll rights reserved.

3 P5.1. onsider the beam shown in Figure P5.1. (a) Write the equations for shear and moment using an origin at end. (b) Using the equations, evaluate the moment at section 1. (c) Locate the point of ero shear between and. (d) Evaluate the maimum moment between points and. (e) Write the equations for shear and moment using an origin at. (f ) Evaluate the moment at section 1. (g) Locate the section of maimum moment and evaluate M ma. (h) Write the equations for shear and moment between and using an origin at. (i) Evaluate the moment at section 1. P = 8 ips 4ʹ 5ʹ 1 w = 3 ips/ft 16ʹ P5.1 = 0; R = 0 = ips ΣF = 0; R = 0 R R = 34 ips 8 ips 4 5 = 34 ips 1 3ips/ft 16 = ips (a) Origin at ; 0 4 ΣF = 0; -8- V = 0 V =-8 = 0; M 8= 0 M =-8 N 8 ips V() M() 8 ips (c) Locate Point V = 0-4 = 34 ips 3( 4) 4 M() V() ΣF = 0; ( -4) - V = 0 V = (EQ. 1) ( -4) = 0; 8-34( - 4) 3( - 4) M = 0 M = (EQ. ) (b) Moment at (1) set = 9 M =- - = (9) Use EQ. 1 0 = = 1.67 ( d) M : Set = 1.67 in (EQ. ) ma M =- - ma = M = ip ft ma 1.5(1.67) = 60.5 ip ft 5-13 opright 018 McGraw-Hill Education. ll rights reserved.

4 P5.1. ontinued (e) ΣF = 0; V- 3 = 0 V =- 3 = 0; M 3 - = 0 1 M = - ΣF = 0; -8 - V = 0 V =-8 = 0; -8(0-)- M = 0 Z M = M() V() 8 ips 3 V() 0 = ips M() (f ) M at section (1) R = 11 3 M = - = (11) (11) 60.5 ip ft ( g) M, Set V = 0; - 3= 0 M ma ma V = = = (7.33) = M ma = ip ft ( h) ΣF = 0; V = 0 V = = 0; - 8(4 K) M = 0 3 M = (i) Moment at section (1) Let = 5 æ3ö M =- 3 6(5) - 5 ç è ø M = 60.5 ip ft 8 ips 4 = 34 ips 3 M() V() 5-14 opright 018 McGraw-Hill Education. ll rights reserved.

5 P5.17. For each beam, draw the shear and moment curves label the maimum values of shear and moment, locate points of inflection, and setch the deflected shape. 10ʹ P = 0 ips hinge 5ʹ w = 1 ips/ft 15ʹ P opright 018 McGraw-Hill Education. ll rights reserved.

6 P5.1. For each beam, draw the shear and moment curves label the maimum values of shear and moment, locate points of inflection, and setch the deflected shape. P = 1 ips w = 5 ips/ft hinge 0ʹ 6ʹ 4ʹ P5.1 F /1 5 (0 ) = 0; 1 (0 )- V (0 ) = 0 V = 6 /1 ΣF = 0; - F (0 ) 6-1 = 0 F = 50 /1 5 (30 ) = 0; -6 (30 )- (4 ) = 0 = ΣF = 0; -6-5 (30 ) = 0 = opright 018 McGraw-Hill Education. ll rights reserved.

7 P5.5. raw the shear and moment curves for each member of the frame in Figure P5.5. Setch the deflected shape hinges at and. 4 m 30 N m w = 5 N/m 3 m 3 m 6 m P5.5 Lin arries ial Load Onl. Member Member eflected Shaps 5-7 opright 018 McGraw-Hill Education. ll rights reserved.

P5.40. (a) raw the shear and moment curves for the frame in Figure P5.40. Setch the deflected shape. (b) Write the equations for shear and moment in column. Tae the origin at.

8 P5.40. (a) raw the shear and moment curves for the frame in Figure P5.40. Setch the deflected shape. (b) Write the equations for shear and moment in column. Tae the origin at. (c) Write the shear and moment equations for girder. Tae the origin at joint. 4 ips 15ʹ 1 w =.4 ips/ft 1 ips 5ʹ F 10ʹ E 4ʹ 3ʹ P5.40 Segment : Origin ; Range 0 =.4 ΣF = 0; V = 0 V = = 0; O æ ö (15 )-.4 M 0 ç - = çè ø M = Segment, Range 0, 15 Segment = 0; 4 - M = 0 M = 4 O Segment E 5-4 opright 018 McGraw-Hill Education. ll rights reserved.

9 P5.47. For the frame in Figure P5.47, draw the shear and moment curves for all members. Net setch the deflected shape of the frame. Show all forces acting on a free-bod diagram of joint. 0 ips 4ʹ 6ʹ w = 5 ips/ft E hinge 6ʹ 4ʹ 4ʹ P5.47 M iagrams eflected Share Joint 5-49 opright 018 McGraw-Hill Education. ll rights reserved.

PROBLEM 8.3. S F = 0: N -(250 lb)cos 30 -(50 lb)sin 30 = SOLUTION

PROBLEM 8.3. S F = 0: N -(250 lb)cos 30 -(50 lb)sin 30 = SOLUTION PROLEM 8. Determine whether the block shown is in equilibrium and find the magnitude and direction of the friction force when θ = and P = 5 lb. ssume equilibrium: S F = : F-(5 lb)sin + (5 lb)cos = S F