Fundamentals of structural analysis 5th edition leet solutions manual 1 by letha.whitfill570

FUNDAMENTALS OF STRUCTURAL ANALYSIS

5th Edition

SOLUTIONS MANUAL

CHAPTER 5: BEAMS AND FRAMES

5-1

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Kenneth M. Leet, Chia-Ming Uang, Joel T. Lanning, and Anne M. Gilbert

P5.1. Write the equations for shear and moment between points B and C as a function of distance x along the longitudinal axis of the beam in Figure P5.1 for (a) origin of x at point A, and (b)

origin of x at D

ΣMA =-10(8)-25(0.5 10)-25+20C y = 0

C y = 34 kips

ΣF y = A y -8-0 5 10-15+34 = 0

A y =-6 kips

a) Cut beam with origin at A

ΣF y =-6-8-V = 0

V =-14 kips

ΣMA =-M +(10)⋅8+ x(-14)= 0

M =80-14x kip⋅ft

b) Cut beam with origin at D

ΣF y =-0.5(10)-15+V +34= 0

V =-14 kips

ΣMD = M +(x2 )⋅(-14)+10(34)-0.5(10)(5)+25 = 0

M =14x -340 kip⋅ft

5-2

w = 0.5

•

kips

kip/ft

kip

A B C D 10ʹ 10ʹ 10ʹ P5 1

Write the equations for shear and moment

No reproduction or distribution

A x ç ÷ x P5.2.

6ʹ B 8 kips w = 3 kips/ft C A E D 5ʹ 3ʹ x 10ʹ P5.2 ΣMA

E y = 0 E y

19 kips ΣF y

A y

E y = 0 A y =11 kips ΣF x = E x -8= 0 8 kips B C D 3 kips/ft E x =8 kips Ay Ey ΣF y = V x -3(10- x

V x

3x +11 kips ΣM = M + æ xö ÷ 3x -19(x) = 0 è2ø 3 kips/ft Vx Mx M =19x3 x 2 kip⋅ft 2 D x 10-x Ey

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=-6(8)+13(3 10)-18

-3 10+

)+19

No reproduction or

5-3

distribution without the prior written consent of McGraw-Hill Education.

P5.3. Write the equations for shear and moment between points A and B Select the origin at A. Plot the graph of each force under a sketch of the beam The rocker at A is equivalent to a roller

Check@ x = 6: M =12+6-36(6) 12 M =18-18= 0

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MA = 12 kN • m A 6 m P5.3 w 3 kN/m B + V and M Diagrams ΣM = 0; 12kN m + R 6 m-(9kN)2 m = 0 B A R = 6 -1 kN A 6 1 æ xö ÷ x 2 R = ´ = 2 ç è2÷ ø 4 W ¢ = 3 \ W ¢ = x x 6 2 kN 1 + ΣF y = 0; 1´W ¢-V = 0 2 V =11 æ x ö ÷ ´ ç ÷ 2 è2ø x 2 V =1- Ans. 4 + x 2 æ x ö ÷ ΣMO = 0; 1(x)+12ç ÷ - M = 0 4 è3ø M =12+ x - R æ xö ÷ ç è3÷ ø x 3 M =12+ x - Ans. 12

5-4

P5.4. Write the equations for shear V and moment M between points B and C Take the origin at point A. Evaluate V and M at point C using the equations.

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5-5

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z ç ÷

= 5 kips/ft C D A B 4ʹ 9ʹ 4ʹ P = 10 kips P5.4 For 4 ft < x < 13 ft A 4 ft 5 kips/ft B 10 kips x z V(x) M(x) ΣF y = 0 =-V(x)+10-5x V(x) =-5

V(13) = VC =

ΣM = 0 =-M

æ x

è2ø

x +10 kips

-55 kips

(x)-5x

ö ÷+10(x -4)

M(x) =5 x 2 +10(x -4)-40 kip ft

M(13) = MC =-332.5 kip ft

P5.5. Write the equations for moment between points B and C as a function of distance x along the longitudinal axis of the beam in Figure P5.5 for (a) origin of x at A and (b) origin of x at B

5-6

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Freebody Segment Origin Range V Equation M Equation Diagram BC A 4¢ £ x1 £10¢ + + ΣF y = 0; ΣMO = 0; k k /1 -4x1 - M1 -3(x1 -4 ) V =8-3x (x1 -4¢) 2 M =3(x -4¢)2 -4x1 BC B 0 £ x2 £6¢ + = 0;k + ¢ -æ x2 ö ÷ = ç è 2 ÷ ø 0 M =-16-4x3 x 2 2 2 2 2 + ΣFY = 0; k V =-4-3x 2 2 1 1 2 2 M

4 kips A 4ʹ w = 3 kips/ft B C D 6ʹ 5ʹ P5 5 -4 -V -3 (x -4¢) = 0 1 1 = 0 1 2 1 -4 -V -3x = 0 ΣMO 4 (x2 4 ) 2 3x2

P5.6. Write the equations required to express the Figure P5 6 Use an origin at point A, and then repeat computations using an origin at point D Verify that both procedures give the same value of

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5-7

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4 4 2 2 6 6 6 2 6 3 3 4 4 2 1 x

or distribution

moment

C 18 kips A 6ʹ C D 10ʹ 8ʹ P5.6 Origin at “A” é x ù Origin at “D” A-B M = 39.1x -2.4x ê 1 ú C-D M =17.3x 1 1 1 ê ú ë û (x -8) 2 B-C M =17.3x -(x -8)2.4 5 = 39.1x1 -1.2x1 5 5 5 2 é x ù =17.3x -1.2(x -8)2 B-C M = 39.1x -2.4x ê 2 ú-18(x -6) 5 5 2 2 2 ê ú 2 ë û B-A M =17.3k x -18k ( x -18) = 21.1+ x2 -1.2x2 +108 C-D M3 =39.1x3 -18(x3 -6)-2.4´16(8+ x3 -16) -2.4(x -8) (x6 -8) 2 M = 415.2-17.3x M = -1.2(x -8)2 -0.7x +324 6 6 6 Check Moment at “C” M =17.3x =17.3´8¢ =138.4 kip ft 2 18kips B 2.4kips/ft M = 21.1´16-1.2(16) +108=138.4 K G A C M1 M2 M3 x Ay = 39 1kips 2 x3 By = 17.3kips 18kips B 2 4 kips/ft A C

at point

5-8

No reproduction or distribution without the prior written consent of

Education. x x5 x M6 M5 M4 4 By = 17 3 kips A y = 39 1 kips 6

McGraw-Hill

P5.7. Write the equations for shear and moment the shear and moment at C, using the equations

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5-9

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2 3

10 kips C

. A B 32 kip ft D x x x 4ʹ 3ʹ 6ʹ P5.7 + ΣMD = 0 =-10´13-35´9.5+32+9RB RB = 47.83kips + ΣF = 0 = 47.83-10-35- R ; R = 2.83k y D D Segment AB: + 0 £ x1 £4 ΣE y = 0 =-10-5x1 -V \ V =-10-5x1 + ΣM x = 0 =10x +5x x1 + M 1 1 2 M =-10x5 x 2 1 2 1 Segment BC: + 0 £ x2 £3 ΣF y = 0 =-10-(4+ x2 )5+47.83-V V =-30-5x2 + 47.83 V =17 83-5x2 + æ + ö2 ΣM = 0 =-M -10(4+ x )-5ç 4 x2 ÷ +47.83x Z 2 ÷ ø 2 M =-40+37.83x5(4+ x )2 2 2 2 Segment DC: + 0 £ x3 £6¢ ΣF = 0; V -2.83k = 0; V = 2.83k y C C + ft k Σ MZ = 0; M +32 +2 83x3 = 0 M =-32ft k -2.83x MC (x3 -6¢)=-49 ft k

based on the origin at point D

P5.8. Write the equations for shear V and moment M in terms of distance x along the length of the beam in Figure P5.8. Take the origin at point A.

ΣF y =-V +8-4x = 0

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5-10

B C

distribution

ΣM =-M -ç ÷ 4x + x(8) = 0

w = 4 kips/ft A 15 kips 8 kips 10ʹ 6ʹ P5.8

ç è2÷

=8-4x kip

x ö ÷

ø M =8x -2x 2 kip ft

P5.9. Write the equation for moment between points B and C for the rigid frame in Figure

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5-11

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distribution

ΣMZ = 0 = 48x -6´10-6xç ÷ - M

P5.9. w = 6 kips/ft B C 10ʹ 6 kips A D 6 kips 48 kips 16ʹ 48 kips P5.9 Member BC 0 £ x £16¢ Section (1) + æ x ö ÷ ç è2ø ÷ M =-60+48x -3x 2

P5.10. Write the equations for moment as a function of distance along the longitudinal axes for members AB and BC of the frame in Figure P5 10. Origins for each member are

ΣM = æ7ö÷(4)7+7(60)-8-12C = 0

è2ø

C y = 42.5 kips

ΣF y = A y -4(7)-60+42.5 = 0

A y = 45.5 kips

ΣF x = A x +12= 0

A x =12 kips

Cut of segment AB

ΣF y = 45.5-4x -V = 0

V = 45.5-4x kips æ x ö ÷

ΣM = M + ç è2÷ ø ⋅4x - x(45 5) = 0

M x = 45 5x -2x kip⋅ft

Cut segment CB æ 169ö ÷

C =ç ÷ C = 46.04 kips

y¢ ç ÷ y 12

ΣF y = V +46.04 = 0

V =-46.04 kips

ΣMC = M x - x(46.04) = 0

M x = 46.04x kip ft

5-12

A ç ÷ y

w 4 kips/ft A x 60 kips B 12 kips

6ʹ 1 8 kip ft x2 6ʹ C 7ʹ 5ʹ P5.10

shown.

P5.11. Write the equations for shear and moment between points B and C for the rigid frame in Figure P5 11. Select the origin at point C

Reactions

ΣF x = 0; +

RAX = 4 kN ¬

ΣM = 0; 4kN ´6+18kN ´3- R 9 = 0 RC =8.67 kN

ΣF y = 0;

W x = 4 ;

RAY -18+8.67 = 0 RAY = 9.33 kN

W = 4 x x 9 9

5-13

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A C 2

or distribution

4 kN w = 4 kN/m B C x 6 m A 9 m P5.11

Shear Σ

2 V = 4 x ⋅ x -8.67 9 2 V = 2 x 2 -8.67 9 Moment + ΣMZ = 0; M +W x æ xö ÷-8.67x = 0 ç ÷ x 2ç è3÷ ø M 8.67x æ4 x ÷ ö x = -ç ÷ ç è9 6 3 M =8.67x2x 27

F y = 0;

x x +8.67 = 0

P5.12. Consider the beam shown in Figure P5.12.

(

a) Write the equations for shear and moment using an origin at end A.

(

b) Using the equations, evaluate the moment at section 1.

(

c) Locate the point of zero shear between B and C.

(

d) Evaluate the maximum moment between points B and C.

(

e) Write the equations for shear and moment using an origin at C.

(

f ) Evaluate the moment at section 1

(

g) Locate the section of maximum moment and evaluate M max .

(

h) Write the equations for shear and moment between B and C using an origin at B

(

) Evaluate the moment at section 1

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5-14

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w = 3 kips/ft ips 3 kips/ft 4’ 5’ max 1 max

P = 8 kips 1 A B 5ʹ 4ʹ P5 12 C 16ʹ 1 + ΣMB = 0; 8´4-48´8+14RC = 0 8 k ΣF y = 0; RC = 22 kips -8-48+ RB +22 = 0 A C 16’ RB =34 kips (a) A–B Origin at “A;” 0 £ x £ 4 By = 34 kips 8 kips 8 kips Cy = 22 kips 3(x – 4) M(x) + ΣF y = 0; ΣM z = 0; -8-V = 0 V =-8k M +8x = 0 V(x) x z M(x) A 4’ x – 4 B y = 34kips x z V(x) + M =-8x N k + (c) Locate Point V = 0 B -C ΣF y = 0; -8+34-3(x -4)-V = 0 V =-3x +38 (EQ. 1) UseEQ 1 0 =-3x +38 x =12.67 + ΣM 0; 8x 34(x 4) 3(x 4) (x -4) M 0 = - - + - + = 2 M =-1.5x 2 +38x -160 (EQ 2) (d) M max : Set x =12.67in (EQ 2) 2 M =-1.5(12.67) +38´12.67-160 (b) Moment at (1) set x = 9 2 =-240.79+481.46-160 M =-1.5(9) +38 9-160 M = 80.67 kip⋅ft

=-121.5+340-160

= 60.5 kip ft

5-15

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distribution

5-16

M M

(

+ ΣF

+ ΣM

0; V

0 V

M +3

0 2 M = 22

1 x 2 2 M(x) z V(x) 3x x Cy = 22 kips A–

+ ΣF

0; + Σ

0; -8

0 V =-8k -8(20-

0 M

8 kips A 20– x V(x) z M(x)

ft 2

M max

0; V

7.332 3 -22+3x = 0 max max +

2 2

0; + -8+34-3

0 V

8 kips 3x ΣM 0; 8(4 K) 34x 3x x M 0 = -

= 2

3 x 2 A 2 4’ M(x) z x

÷ 2

5 By = 34 kips x V(x)

reproduction or distribution without the prior written consent of McGraw-Hill Education.

z P5.12. Continued

e) B–C

y = 0;

z =

-3x +22=

=-22+

x x -22

y =

MZ =

k -V =

x)-

=8x -160

(

)

at section (1) R

+ x =11

M = 22(11)3 (11)

= 60.5 kip

(

)

, Set

= 22 =

= 22´7.33-

(7.33)

=161.26-80.59 = 80.67 kip⋅ft (

) ΣF y =

x -V =

= 26-3x

+ + - -

M =-32+26x -

(i) Moment at section (1) Let x = 5¢ æ3ö

M =-32+26(5)-çè2÷ ø

M = 60.5 kip ft

P5.13. For each beam, with given moment A B C diagram: (a) sketch the deflected shape; and (b) determine, qualitatively, the applied loading.

Constant negative shear results in a linear moment diagram with negative slope. Likewise, constant positive shear results in a linear moment diagram with positive slope. Constant shear occurs when no distributed load is present. A jump in moment diagram arises from an applied concentrated moment.

5-17

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moment P5 13

P5.14. For each beam, with given moment diagram: (a) sketch the deflected shape; and (b) determine, qualitatively, the applied loading. moment

5-18

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or distribution

A B C D

P5.14

P5.15. For each beam, draw the shear and moment curves label the maximum values of

shear and moment, locate points of inflection,

and sketch the deflected shape.

without the prior written consent of McGraw-Hill Education.

5-19

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or distribution

40 kips w = 4 kips/ft

A D

C 6ʹ 4ʹ 10ʹ 4ʹ P5.15

P5.16. For each beam, draw the shear and moment diagrams, label the maximum values of shear and moment, locate points of inﬂection,

and sketch the deﬂected shape

ΣMA =-200+170-100-44B y = 0

B y =-2.95 kips

ΣF y = A y -2.95= 0 A y = 2.95 kips

Deflect

kip • ft 170 kip • ft 100 kip • ft

5-20

or distribution without the prior written consent of McGraw-Hill Education.

reproduction

200

A B C

10ʹ 12ʹ 12ʹ 10ʹ P5.16

D E

y By Shear Moment

P5.17. For each beam, draw the shear and moment curves label the maximum values of shear and moment, locate points of inflection, and sketch the deflected shape.

without the prior written consent of McGraw-Hill Education.

5-21

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distribution

P 20 kips A B hinge C w 12 kips/ft D 10ʹ 5ʹ 15ʹ P5 17

P5.18. For each beam, draw the shear and

kips 12 kips moment curves label the maximum values of shear and moment, locate points of inflection,

and sketch the deflected shape.

FBD of segment CDE

ΣMC = 0 = 3(5)-195+12(15)- D y

(10)

D y = 0 kips

ΣF y = 0 =-VC -3-12

VC =-15 kips

FBD of segment ABC

ΣMA = 0 =-B y (25)+15(30)

B y =18 kips

ΣF y = 0 =-15+18+ A y

VC =-3 kips

5-22

No reproduction or distribution without

prior

consent

Education. VC = 15 kips C 25ft 5 ft -3 kips B = 18kips 15 15 -3 75 7-5 POI

the

written

of McGraw-Hill

C 195 kip ft B D E hinge 25ʹ 5ʹ 5ʹ 5ʹ 5ʹ P5.18

VC 3 kips C 195kips 12kips D E Dy

Ay = y 5 ft 5 ft 5 ft

Shear (kips)

POI

Moment (kip-ft) Deflected Shape

moment curves label the maximum values of shear and moment locate points of inflection, and sketch the deflected shape.

w1 = 2 kips/ft

w2 = 5 kips/ft

B = 0; -26(6.5¢)+65(6.5¢)+104(19.5¢)- D y (26¢)= 0

D =87.5k + ΣF = 0; -26k + B -65k -104k - D = 0 y y y

B =107.25k

y y

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5-23

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distribution

A B C D 13ʹ

13ʹ

+ Σ

13ʹ

P5 19

P5.20. For each beam, draw the shear and moment curves label the maximum values of shear and moment locate points of inflection, and sketch the deflected shape.

5-24

No reproduction or distribution

A k /1 2 2 2 = A = @A y y

without the prior written consent of McGraw-Hill Education.

3ʹ w = 2 kips/ft B C D A P = 12 kips rocker support hinge 4ʹ 20ʹ 20ʹ P5 20 Shear @ A: Entire Structure: 12k 5¢ VA 4¢ V = 9.6k @B x = 5¢ 2k /1(5¢) 5¢ = VBW 4¢ VBW =8k \ VB = 9.6k +8k =17.6k Axial Force @ A: 12k 5¢ AA 3¢ = 7.2k @B, x 5¢: 2 (5¢) k (3¢) 6 = = 5 \ A = 7.2k +6k =13.2k FBD “BC” + ΣM = 0; -68ft k -22k (20¢)-2k /1 (20) + B (20¢)= 0 C y B = 45 4k + ΣF = 0; -22k -2k /1(20¢)+45.4k +C = 0 y y C =16.6k FBD “CD” + ΣF = 0; -16.6k -2(20¢)+ D = 0 D = 56.6k y y y + ΣM = 0; -2 k /1(20) 2 -16 6 k (20¢)+ M = 0 M = 732ft k D D D

P5.21. For each beam, draw the shear and moment curves label the maximum values of shear and moment, locate points of inflection, and sketch the deflected shape.

without the prior written consent of McGraw-Hill Education.

5-25

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B 2 y y y

P = 12 kips w = 5 kips/ft A hinge B C D 20ʹ 6ʹ 24ʹ P5.21 FBD “AB” + ΣM = 0; 5k /1(20¢)2 +12k (20¢)-V (20¢) = 0 A B 2 2 V = 62k 2 + ΣF = 0; A -F k /1(20¢)+62k -12k = 0 y y A = 50k FBD “BCD” + ΣM = 0; 5k /1(30¢)2 -62k (30¢)- + C (24¢)= 0 D y C =171.25k + ΣF = 0; -62k -5k (30¢)+C +171.25k = 0 y y D = 40.75k

for the beam in Figure P5 22, and sketch the deflected shape. Find the vertical displacement

5-26

reserved. No reproduction or distribution without the prior written consent of McGraw-Hill Education. B 5 C 3 75 2.5 ft 2 5 ft 46 875 -37 5 0 0 -375 y y y y

All rights

of joint D A hinge 5ʹ 10ʹ 150 kip ft B C P5.22 D spring, E k = 30 kips/in. 10ʹ Left Side: 5 Displacement: F =-kx ΣM A = + 2 (15 5)-(5)B y = 0 26.25 =-30 x Right Side: ΣM B = A = 37.5 kips =150-5(37.5)+10(30)-10D = 0 x = 0.875 in x = 7 ¢¢ C y 8 D = 26.25 kips ΣF y =-37.5+C y -30+26.25 = 0 C = 41.25 kips 15 kips/ft A Dy 37 Shear (kips) 0 Moment (kip-ft) A B C Deflected Shape Dy

P5.23. Draw the shear and moment curves for each member of the frame in Figure P5.23. Sketch the deflected shape.

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5-27

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ʹ E y y y

6 kips A 3 B C 3 kips/ft D 12ʹ E 6ʹ 18ʹ P5 23 Compute Reactions: + ΣM = 0; 6k (15¢)+3KLF (18)(9¢)- D (18) = 0 D = 32k + ΣF = 0; -3k / FR (18¢)+32k + E = 0 y y ΣF = 0; E = 22k E = 6k ¬ x x

P5.24. Draw the shear and moment curves for each member of the frame in Figure P5.24. Sketch the deflected shape

5-28

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6 m C 10 kN B w = 2 kN/m D E A 3 m 3 m 6 m 6 m P5.24 Reactions + ΣF x = 0 : RAy = 0 + ΣMA = 0 : 10´3+24(12)-R D12 = 0 RD = 26.5 kN + ΣF y = 0 : RAY -34+26.5= 0 RAY = 7 5kN

P5.25. Draw the shear and moment curves for

each member of the frame in Figure P5.25.

Sketch the deflected shape hinges at B and C.

Member CD

Member AB

Deflected Shaps

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5-29

w = 5 kN/m

No reproduction

B 30 kN m

4 m C D 3 m 3 m 6 m P5.25

Link BC Carries Axial Load Only.

P5.26. Draw the shear and moment curves for the beam in Figure P5 26 Sketch the deflected shape.

without the prior written consent of McGraw-Hill Education.

5-30

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ç y y

w2 = 6 kips/ft w1 3 kips/ft w3 = 2 kips/ft A C 30 kip ft 9ʹ B 9ʹ 9ʹ P5.26 + ΣM = 0; -30 ft k +27(4.5)+13 5(6¢)+36k (15¢+18¢)- B A y B = 75 58k + ΣF y = 0; -27-13.5-2(36)+75.58+ A y = 0 FBD: Sim-Triangles A = 36.92k 3k/ ft 9¢ = y x Y = x 3 ΣF = 0; 36.92k -3x1 x æ xö ÷ ÷= 0 y V = 0@x = 8.394 2 ç è3÷ ø x 2 Shear @x = 9¢ V =-36.92+3x + = -3.58k 6 @x =8 4¢ DM = ò Vdx x 2 ΔM = ò 36.92 dx +ò-3xdx +ò- dx 6 = 36.92xò 8 40 3x 2 ò 2 8 4¢0 x 3 ò 15 8 4¢ 0 =171.5 M@8.4¢ =171.4 ft k-30 ft k = 141.5 ft k Max + M Moment Deflected Shape

P5.27. Draw the shear and moment curves for each member of the frame in Figure P5 27. Sketch the deflected shape.

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5-31

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A y x

4 kips B 10 kips w 2 kips/ft C D 15ʹ A 8ʹ 8ʹ P5.27 + ΣM = 0; -4 k (15¢)-10k (8¢)+ M = 0

A A M =1401 k + ΣF = 0; -10k -32k + A = 0 y y + A = 42k ΣF = 0; -4k + A = 0 x x A = 4k

Column “AC”:

Beam: “BCD”

P5.28. Draw the shear and moment curves for each member of the frame in Figure P5 28. Sketch the deflected shape

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5-32

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2 kN/m A 10 kN/m D C 20 kN 5 m 3 m B 5 m 5 m P5.28 + æ20 ΣM 0; 20(5m) 40 5R 20(2m) 0 = + ç ÷- + = A ç è 3 ÷ ø B RB =81.33 kN + ΣF y = 0; -60 kN

RAY = 0 RAY =

+81.33-

21.33

P5.29. Draw the shear and moment curves for each member of the frame in Figure P5 29. Sketch the deflected shape

Reactions:

+ ΣMA = 0 =-C x 8+8´2 C x = 2 kips + ΣF x = 0 = A x -2 \ A x = 2 kips

+ ΣF = 0; A =8 kips y y

Member AB

Member BC

Deflected Shape

5-33

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3ʹ 5ʹ B w 2 kips/ft A 4ʹ P5.29 C

No reproduction

P5.30. Draw the shear and moment curves for each member of the beam in Figure P5.30 Sketch the

cted shape The shear connection at B acts as a hinge.

without the prior written consent of McGraw-Hill Education.

5-34

reproduction or distribution

w = 12 kips/ft 6 kips/ft 6 kips A B C D hinge 6ʹ 3ʹ 3ʹ 2ʹ P5.30 FBD of segment BCD ΣMB = 3(6 6)-3C y +8(6)= 0 MA -156 kip-ft 12 kips/ft 6 kips/ft 6 kips C y = 52 kips ΣF y = B y -6⋅6+52-6 = 0 B y =-10 kips FBD of segment ABC ΣMA = MA +3(12⋅6)-6(10) = 0 MA =-156 kip ft ΣF y = A y -12 6+10

A y

kips Shear (kips) Moment (kip-ft) Ay = 62 kips 62 -152 4 2 0 84’ B -10 C Cy = 52 kips 23 8 -28 -57 D 6 -12 Deflected Shape

deﬂe

= 0

=62

P5.31. Draw the shear and moment curves for each member of the beam in Figure P5.31 Sketch the deflected shape.

Member DE

All Forces = Zero

Since No Loads Applied

Member BCD

Reactions + ΣM = 0: 25´ 20 +40(15)-10R = 0 3

RC = 76.67 kN + ΣF y = 0: -B y -25-40+76.67 = 0

B y =11.67 kN Entire Structure

ΣF x = 0: B x = 0

Member AB

5-35

B C

5 kN/m 5 kN/m 40 kN E A B C D 10 m 10 m 5 m 10 m P5.31

P5.32. Draw the shear and moment curves for the beam in Figure P5 32 Sketch the deflected

w = 9 kN/m B shape. A C

w = 9 kN/m

6 m 6 m

P5.32

Reactions

Resultant of distni loads each side of £

R = 1 6´9 = 27 kN

ΣMA = 0; 27´4-27(8)+12 RC = 0

RC = 9 kN , RA = 9 kN

5-36

P5.33. Draw the shear and moment curves for the indeterminate beam in Figure P5.33 Reactions at support A are given. Sketch the deflected shape.

ΔMAB =-1.69´8=-13.52 kip ft

4.5-13.52 = -9.02 kip⋅ft

Locate Point of Inflection

13.5 = 3.375

y x 4x = y

ΔM = AreaV -Diagram

13.76 = 1 yx = 1 (4x)x 2 2

x 2 = 13.76 = 6.88 2

x = 2.62¢

5-37

or distribution without the prior written consent of McGraw-Hill Education.

No reproduction

4.5 kip ft A 1.69 kips w = 4 kips/ft B C 8ʹ 6ʹ 15.19 kips 10.5 kips P5 33

P5.34. Draw the shear and moment curves for the beam in Figure P5 34 Sketch the deflected shape.

without the prior written consent of McGraw-Hill Education.

5-38

reproduction

distribution

P = 36 kips 3ʹ E F P = 24 kips w = 12 kips/ft 4ʹ 2ʹ B C A D 3ʹ 3ʹ 6ʹ P5 34

Structure

Entire

Isolate Cantilever EF Isolate AE

the beam in Figure P5 35 Reaction at support B is given. Locate all points of zero shear and moment. Sketch the deflected shape. A

without the prior written consent of McGraw-Hill Education.

5-39

No reproduction or distribution

= 3 kips/ft C B 6ʹ 6ʹ 6ʹ 24ʹ RB = 85.49 kips P5.35

P5.36. Draw the shear and moment curves for each indeterminate beam Reactions are given Label maximum values of shear and moment Locate all inflection points, and sketch the

without the prior written consent of McGraw-Hill Education.

5-40

No reproduction

or distribution

43.5 kN m A w = 6 kN • m C B deflected shape. RA = 28.31 kN 8 m RB 28.69 3 m P5.36

P5.37. Draw the shear and moment curves for

each indeterminate beam. Reactions are given. Label maximum values of shear and moment. Locate all inflection points, and sketch the

5-41

or distribution without the prior written consent of McGraw-Hill Education.

No reproduction

30 kips 13.93 kip ft w 4 kips/ft

A B C D deflected shape. 9ʹ 6ʹ 4.64 kips 17.97 kips 6ʹ 9ʹ 40 kips 12.67 kips P5 37

P5.38. Draw the shear and moment curves and sketch the deflected shape.

Since Structure Symmetrical Analyze Left Half

5-42

without the prior written consent of McGraw-Hill Education.

No reproduction

w 8 kN/m 30 kN 30 kN

or distribution

3 m 3 m A D B C RA 9 kN RB 105 kN RC 105 kN RD 9 kN 6 m 9 m 6 m P5 38

P5.39. Draw the shear and moment curves and sketch the deflected shape.

5-43

the prior written consent of McGraw-Hill Education.

No reproduction or distribution

y y

without

P = 10 kips w 2 kips/ft G A B C D E F 10ʹ 8ʹ 4ʹ 4ʹ 4ʹ 10ʹ 5ʹ P5.39 FBD “CD”: By Symmetry: V = V = 1 (10k +2k/1(8¢))= 13k CR DL 2 FBD “ABC”: + ΣM = 0; 13k (18¢)- B (10¢)= 0 B = 23.4k A y y + ΣF = 0; - A +23.4k -13k = 0 A =10.4k y y y

“DEFG”: ΣM = 0; -13k (4¢)-F (10¢)= 0 E y F = 5.2k + ΣF = 0; -13k + E -5.2k = 0 Entire STR: y y E =18.2k

FBD

Deflected Shape:

P5.40. (a) Draw the shear and moment curves for the frame in Figure P5.40 Sketch the deflected shape. (b) Write the equations for 4 kips

shear and moment in column AB Take the origin at A (c) Write the shear and moment equations for girder BC. Take the origin at

without the prior written consent of McGraw-Hill Education.

5-44

No reproduction

2 2 æ x ö 1 1 x

or distribution

B C 12 kips

x2 D 15ʹ 5ʹ F 10ʹ joint B 1 A E 24ʹ 3ʹ P5.40 Segment ABC: Origin CD; Range 0 ≤ x2 = 2 4 + ΣF y = 0; k 24.8 -2.4x2 -V2 = 0 V = 24.8-2.4x + ΣMO = 0; 24.8x +4k (15¢)-2.4x 2 ÷ - M = 0 2 2 ç è ÷ 2 ÷ ø 2 M = 60+24.8x -1.2x 2 2 2 2 Segment AB, Orgin @ A, Range 0 ≤ x, ‒ 15′ Segment BC + ΣMO = 0; +4x1 - M1 = 0 M = 4k Segment CE

= 2.4 kips/ft