# Mechanics of Materials 7th Edition Beer Johnson Chapter 5

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## C

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B w

A

L

### PROBLEM 5.1

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION Reactions: B 0: 2 0 2 L wL M AL wL A        A 0: 2 0 2 L wL M BL wL B      

Free body diagram for determining reactions: Over whole beam, 0   x L

Place section at x.

Replace distributed load by equivalent concentrated load.

0: 0 2 y wL F wx V      2 L Vwx    0: 0 2 2 J wL x M x wx M      2 ( ) 2 w MLxx ( ) 2 w M x L  x Maximum bending moment occurs at .

2 L x 2 max 8 wL M 

(4)

B P C A L b a

### PROBLEM 5.2

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION   Reactions: 0: 0 C Pb M LA bP A L      0: 0 A Pa M LC aP C L      From A to B, 0   x a 0: 0 y Pb F V L     Pb V L   0: 0 J Pb M M x L    Pbx M L  From B to C, a   x L 0: 0 y Pa F V L     Pa V L    0: ( ) 0 K Pa M M L x L      ( ) Pa L x M L   At section B, M Pab2 L  

(5)

B w0

A

L

### PROBLEM 5.3

For the beam and loading shown, (a) draw the shear and moment diagrams, (b) determine the equations of the shear and bending-moment curves.

SOLUTION

Free body diagram for determining reactions.

Reactions: 0 0 0: 0 2 2  y A   Aw L w L F R R 0 2 0: 0 2 3             A A w L L M M 2 2 0 0 3 3    A w L w L M

Use portion to left of the section as the free body.

Replace distributed load with equivalent concentrated load.

0 1 0 0: 0 2 2  y     w L w x F x V L 2 0 0 2 2  w Lw x V L  2 0 0 0 0: 1 ( ) 0 3 2 2 3                 J M w L w L w x x x x M L 2 3 0 0 0 3 2 6  w Lw Lxw x M L 

(6)

B

w

L A

### PROBLEM 5.4

For the beam and loading shown, (a) draw the shear and moment diagrams, (b) determine the equations of the shear and bending-moment curves.

SOLUTION

Free body diagram for determining reactions.

Reactions: 0: 0  Fy RAwLA RwL 0: ( ) 0 2            A A L M M wL 2 0 2  A w L M

Use portion to the right of the section as the free body.

Replace distributed load by equivalent concentrated load.

0: ( ) 0  Fy Vw Lx  ( ) Vw Lx  0: ( ) 0 2            J L x M M w L x 2 ( ) 2   wM L x 

(7)

B P P C A a a

### PROBLEM 5.5

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION   From A to B: 0  x a 0 : 0  Fy  P VV   P 0 : 0 J M Px M     M  Px  From B to C: 2a x a 0 : 0  Fy   P P V  2 V   P  0 : ( ) 0 MJPxP xaM  2 M   PxPa 

(8)

D A B a a C L w w

### PROBLEM 5.6

For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION Reactions: ADwa From A to B, 0  x a 0: y F   wawx V  0 ( ) Vw a   x 0: J M ( ) 0 2 x wax wx M     2 2 x Mw ax      From B to C, a    x L a 0: y F wawa V  0 0 V   0: J M 0 2 a wax wa x  M       2 1 2 Mwa  From C to D, L   a x L 0: ( ) 0 y F V w L x wa       ( ) Vw L x a  0: ( ) ( ) 0 2 J L x M M w L x    wa L x           2 1 ( ) ( ) 2         M w a L x L x

(9)

B

A C D E

3 kN 2 kN 5 kN 2 kN

0.3 m 0.3 m 0.3 m 0.4 m

### PROBLEM 5.7

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION Origin at A: Reaction at A: 0: 3 2 5 2 0 2 kN y A A F R R         0: (3 kN)(0.3 m) (2 kN)(0.6 m) (5 kN)(0.9 m) (2 kN)(1.3 m) 0 MAMA       0.2 kN mMA From A to C: 0: 2 kN y F V    1 0: 0.2 kN m (2 kN) 0 M x M       0.2 2    M x From C to D: 0: 2 3 0 y F V      1 kN V   2 0: 0.2 kN m (2 kN) (3 kN)( 0.3) 0 M x x M          0.7 M   x From D to E: 0: 5 2 0 3 kN y F V V        3 0: 5(0.9 ) (2)(1.3 ) 0 M M x x         1.9 3 M   x

(10)

PROBLEM 5.7 (Continued) From E to B: 0: 2 kN y F V    4 0: 2(1.3 ) 0 M M x       2.6 2 M    x   (a) Vmax 3.00 kN  (b) M max  0.800 kN m 

(11)

100 lb 250 lb 100 lb 10 in. 25 in. 20 in. 15 in. A B C D E

### PROBLEM 5.8

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Reactions:

0: (45 in.) 100 lb(15 in.) 250 lb(20 in.) 100 lb(55 in.) 0

C E M R       200 lbRE  0: 200 lb 100 lb 250 lb 100 lb 0 y C F R        250 lbRC

At any point, V is the sum of the loads and reactions to the left (assuming + ) and M the sum of their moments about that point (assuming ).

(a) Vmax 150.0 lb 

(b) Mmax 1500 lb in.  

(12)

PROBLEM 5.8 (Continued)

Detailed computations of moments: 0 A M  (100 lb)(15 in.) 1500 lb in. C M     

(100 lb)(35 in.) (250 lb)(20 in.) 1500 lb in. D

M      

(100 lb)(60 in.) (250 lb)(45 in.) (250 lb)(25 in.) 1000 lb in.

      

E M

(100 lb)(70 in.) (250 lb)(55 in.) (250 lb)(35 in.) (200 lb)(10 in.) 0 B

M      

(13)

B A C D 25 kN/m 40 kN 40 kN 0.6 m 1.8 m 0.6 m

### PROBLEM 5.9

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

The distributed load is replaced with an equivalent concentrated load of 45 kN to compute the reactions. (25 kN/m)(1.8 m)  45 kN 0: (40 kN)(0.6 m) 45 kN(1.5 m) 40 kN(2.4 m) (3.0 m) 0 A B M R        RB 62.5 kN 0: 62.5 kN 40 kN 45 kN 40 kN 0 y A F R        RA  62.5 kN At C: 0: 62.5 kN y F V    1 0: (62.5kN)(0.6 m) 37.5kN m M M     

At centerline of the beam:

0: 62.5 kN 40 kN (25 kN/m)(0.9 m) 0 y F V       0 V  2 0: M   (62.5 kN)(1.5 m) (40 kN)(0.9 m) (25 kN/m)(0.9 m)(0.45 m) 0     M 47.625 kN mM  

(14)

PROBLEM 5.9 (Continued)

Shear and bending-moment diagrams:

(a) Vmax62.5 kN 

(b) M max47.6 kN m  From A to C and D to B, V is uniform; therefore M is linear.

(15)

B A C D 2.5 kips/ft 15 kips 6 ft 3 ft 6 ft

### PROBLEM 5.10

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION 0: 15 (12)(6)(2.5) (6)(15) 0 B A M R       18 kipsRA 0: 15 (3)(6)(2.5) (9)(15) 0 A B M R      12 kipsRB Shear: 18 kipsVA 18VC  (6)(2.5) 3 kips to : 3 kips C D V  to : 3 15 12 kips D B V    

Areas under shear diagram: 1 to : (6)(18 3) 63 kip ft 2 A C V dx     

to :C D

### 

V dx (3)(3) 9 kip ft to :D B

### 

V dx  (6)( 12)  72 kip ft Bending moments: MA  0 0 63 63 kip ft     C M 63 9MD  72 kip ft 72MB 72 0 max18.00 kips V    M max72.0 kip ft 

(16)

B A C D E 3 kN 3 kN 300 mm 300 mm 200 mm 450 N ? m

### PROBLEM 5.11

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION 0: (700)(3) 450 (300)(3) 1000 0 B M A       2.55 kN A 0: (300)(3) 450 (700)(3) 1000 0 A M B        3.45 kN B At A: V  2.55 kN M  0 A to C: V  2.55 kN At C: MC 0: (300)(2.55) M 0   765 N m M C to E: V  0.45 N m At D: MD 0: (500)(2.55) (200)(3) M 0    675 N m M At D: MD 0: (500)(2.55) (200)(3) 450 M 0     1125 N m M E to B: 3.45 kNV   At E: ME 0: (300)(3.45) 0 M   1035 N m M At B: 3.45 kN,VM  0 (a) Vmax3.45 kN  (b) Mmax1125 N m 

(17)

400 lb 1600 lb 400 lb

12 in. 12 in. 12 in. 12 in.

8 in. 8 in. C A D E F G B

### PROBLEM 5.12

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION 0: 16 (36)(400) (12)(1600) G M C      (12)(400)  1800 0 C lb 0: 0 x x F C G      1800 Gx  lb 0: 400 1600 400 0 y y F G        2400 Gy lb A to E: 400 V   lb E to F: 2000 V   lb F to B: 400 V  lb At A and B, 0M  At ,D MD 0: (12)(400)M  4800 0 M   lb in. AtD +, 0: D M   (12)(400) (8)(1800)M  9600 0 M lb in. At E, ME 0: (24)(400)(8)(1800)M  4800 0 M  lb in. At ,FMF 0:  M (8)(1800) (12)(400) 0 M  19, 200 lb in. AtF ,+ MF  0:  M (12)(400)  4800 0 M   lb in. (a) Maximum | |V 2000 lb  (b) Maximum | |M 19, 200 lb in. 

(18)

B A C D 1.5 kN 1.5 kN 0.9 m 0.3 m 0.3 m

### PROBLEM 5.13

Assuming that the reaction of the ground is uniformly distributed, draw the shear and bending-moment diagrams for the beam AB and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Over the whole beam,

0: 1.5 1.5 1.5 0 y F w      2 w kN/m A to C: 0 x 0.3 m 0: 2 0 y F x V     (2 ) Vx kN 0: (2 ) 0 2 J x M x   M         2 ( ) kN m Mx  At ,C 0.3 x  m 0.6 kN, 0.090 kN m 90 N m VM     C to D: 0.3 m  x 1.2 m 0: 2 1.5 0 y F x V      (2Vx1.5) kN 0: (2 ) (1.5)( 0.3) 0 2 J x M x   x M          2 ( 1.5 0.45) kN m Mxx 

At the center of the beam, x  0.75 m

0 0.1125 kN m 112.5 N m VM       AtC +, x 0.3 m, V  0.9 kN (a) Maximum | |V 0.9 kN 900 N  (b) Maximum | |M 112.5 N m 

(19)

B C D E 2 kips/ft 24 kips A 3 ft 3 ft 3 ft 3 ft 2 kips/ft

### PROBLEM 5.14

Assuming that the reaction of the ground is uniformly distributed, draw the shear and bending-moment diagrams for the beam AB and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Over the whole beam,

0: 12 (3)(2) 24 (3)(2) 0 y F w       3 w kips/ft A to C: (0 x 3 ft) 0: 3 2 0 y F x x V      ( ) Vx kips 0: (3 ) (2 ) 0 2 2 J x x M x x M       M (0.5 ) kip ftx2 At C, x  3 ft 3 kips, 4.5 kip ft VM   C to D: (3 ft  x 6 ft) 0: 3 (2)(3) 0 y F x V      (3Vx6) kips 3 0: (3 ) (2)(3) 0 2 2 x MK x   xM           2 (1.5 6 9) kip ft M x x At ,D 6 x ft 12 kips, 27 kip ft VM  

D to B: Use symmetry to evaluate.

(a) | |Vmax 12.00 kips  (b) | |M  27.0 kip ft 

(20)

B A C 3 kN/m 1.5 m 1.5 m 2.2 m 100 mm 200 mm 10 kN

### PROBLEM 5.15

For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION

Using CB as a free body,

3 0: (2.2)(3 10 )(1.1) 0 C M M       3 7.26 10 N m M   

Section modulus for rectangle:

2 1 6 Sbh 2 3 3 6 3 1 (100)(200) 666.7 10 mm 6 666.7 10 m      Normal stress: 3 6 6 7.26 10 10.8895 10 Pa 666.7 10 M S         10.89 MPa 

(21)

750 lb B A C D 150 lb/ft 750 lb 3 in. 12 in. 4 ft 4 ft 4 ft

### PROBLEM 5.16

For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION Reactions: CA by symmetry. 0: (2)(750) (12)(150) 0 y F A C       1650 lb AC

Use left half of beam as free body. 0: E M   (1650)(6) (750)(2) (150)(6)(3) M 0      3 5700 lb ft 68.4 10 lb in.      M Section modulus: 1 2 1 (3)(12)2 72 in3 6 6 Sbh       Normal stress: 3 68.4 10 950 psi 72 M S      950 psi 

(22)

B A C D E 150 kN 150 kN 2.4 m 0.8 m 0.8 m 0.8 m W460 ⫻ 113 90 kN/m

### PROBLEM 5.17

For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION

Use entire beam as free body. 0: B M   4.8A (3.6)(216) (1.6)(150) (0.8)(150) 0      237 kN  A

Use portion AC as free body.

0: C M   (2.4)(237) (1.2)(216) 0 309.6 kN m M M      For W460 113, S 2390 10 mm 6 3 Normal stress: 3 6 3 6 309.6 10 N m 2390 10 m 129.5 10 Pa M S          129.5   MPa 

(23)

B A a a 30 kN 50 kN 50 kN 30 kN 2 m 5 @ 0.8 m 5 4 m W310 3 52

### PROBLEM 5.18

For the beam and loading shown, determine the maximum normal stress due to bending on section a-a.

SOLUTION

Reactions: By symmetry, AB

0 : 80 kN

FyAB

Using left half of beam as free body,

0: J M   (80)(2) (30)(1.2) (50)(0.4) M 0      3 104 kN m 104 10 N m M     For 3 3 6 3 W310 52, 747 10 mm 747 10 m S       Normal stress: 3 6 6 104 10 139.2 10 Pa 747 10 M S        139.2 MPa   

(24)

B A C 8 kN 1.5 m 2.1 m W310 ⫻ 60 3 kN/m

### PROBLEM 5.19

For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION

Use portion CB as free body.

3 0: (3)(2.1)(1.05) (8)(2.1) 0 23.415 kN m 23.415 10 N m C M M M            For 3 3 6 3 W310 60, 844 10 mm 844 10 m S       Normal stress: 3 6 6 23.415 10 27.7 10 Pa 844 10 M S        27.7 MPa   

(25)

B A C D E F G 5 kips 5 kips 2 kips 2 kips 2 kips 6 @ 15 in. 5 90 in. S8 3 18.4

### PROBLEM 5.20

For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION

Use entire beam as free body. 0: B M   90A (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0        9.5 kipsA

Use portion AC as free body.

0: (15)(9.5) 0 142.5 kip in. C M M M       For S8 18.4, S 14.4 in3 Normal stress: 142.5 14.4 M S    9.90 ksi   

(26)

B A

C D E

25 kips 25 kips 25 kips

2 ft

1 ft 6 ft 2 ft

S12 ⫻ 35

### PROBLEM 5.21

Draw the shear and bending-moment diagrams for the beam and loading shown and determine the maximum normal stress due to bending. SOLUTION 0: B M   (11)(25) 10 C(8)(25)(2)(25) 0 C  52.5 kips 0: C M   (1)(25)(2)(25)(8)(25) 10 B  0 B  22.5 kips Shear: A to C: 25 V   kips C to D: 27.5 V kips D to E: 2.5 V  kips E to B: 22.5 V   kips Bending moments: At C, MC 0: (1)(25)M  0 25 kip ft M    At D, MD 0: (3)(25)(2)(52.5)M  0 30 kip ft M At E, ME 0: M (2)(22.5)  0 M  45 kip ft

maxM  45 kip ft  540 kip in.

For S12 35 rolled steel section, S 38.1 in3

Normal stress: 540 14.17 ksi

38.1 M

S

(27)

Hinge 2.4 m 0.6 m 1.5 m 1.5 m C B A E D 80 kN/m 160 kN W310 ⫻ 60

### PROBLEM 5.22

Draw the shear and bending-moment diagrams for the beam and loading shown and determine the maximum normal stress due to bending.

SOLUTION

Statics: Consider portion AB and BE separately.

Portion BE: 0: E M   (96)(3.6)(48)(3.3)C(3)(160)(1.5)  0 248kNC   56kNE  0 A B E MMM At midpoint of AB: 0: 0 0: (96)(1.2) (96)(0.6) 57.6 kN m y F V M M         

Just to the left of C:

0: 96 48 144 kN

y

F V

      

0:MCM  (96)(0.6)(48)(0.3)  72 kN

Just to the left of D:

0: 160 56 104 kN 0: (56)(1.5) 84 kN m y D F V M M             

(28)

PROBLEM 5.22 (Continued)

From the diagram,

M max 84 kN m 84 10 N m 3  

For W310 60 rolled-steel shape,

3 3 6 3 844 10 mm 844 10 m x S      Stress: max m M S   3 6 6 84 10 99.5 10 Pa 844 10 m       99.5 MPa m   

(29)

H A 7 @ 200 mm ⫽ 1400 mm Hinge 30 mm 20 mm C B D E F G 300 N 300 N 40 N 300 N

### PROBLEM 5.23

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum normal stress due to bending.

SOLUTION         

Free body EFGH. Note that ME  due to hinge. 0 0: 0.6 (0.2)(40) (0.40)(300) 0 213.33 N E M H H       0: 40 300 213.33 0 126.67 N y E E F V V        Shear: to : 126.67 N m to : 86.67 N m to : 213.33 N m E F V F G V G H V        Bending moment at F: 0: (0.2)(126.67) 0 25.33 N m F F F M M M       Bending moment at G: 0: (0.2)(213.33) 0 42.67 N m G G G M M M       

Free body ABCDE.

0: 0.6 (0.4)(300) (0.2)(300) (0.2)(126.63) 0 257.78 N B M A A        0: (0.2)(300) (0.4)(300) (0.8)(126.67) 0.6 0 468.89 N A M D D        

(30)

PROBLEM 5.23 (Continued) Bending moment at B. 0: (0.2)(257.78) 0 51.56 N m B B B M M M        Bending moment at C. 0: (0.4)(257.78) (0.2)(300) 0 43.11 N m C C C M M M         Bending moment at D. 0: (0.2)(213.33) 0 25.33 N m         D D D M M M max M 51.56 N m  2 2 3 3 6 3 1 1 (20)(30) 6 6 3 10 mm 3 10 m       S bh Normal stress: 6 6 51.56 17.19 10 Pa 3 10      17.19 MPa max 342 N V max 516 N m M

(31)

24 kN/m 64 kN ? m B A C D 2 m 2 m 2 m S250 ⫻ 52

### PROBLEM 5.24

Draw the shear and bending-moment diagrams for the beam and loading shown and determine the maximum normal stress due to bending. SOLUTION Reactions: 0: 4 64 (24)(2)(1) 0 28 kN D M A      A 0: 28 (24)(2) 0 76 kN y F D       DA to C: 0  x 2m 0: 28 0 28 kN y F V V        0: 28 0 ( 28 ) kN m J M M x M x        C to D: 2m  x 4m 0: 28 0 28 kN y F V V        0: 28 64 0 ( 28 64) kN m J M M x M x          D to B: 4m  x 6m 0: 24(6 ) 0 ( 24 144) kN y F V x V x         2 0: 6 24(6 ) 0 2 12(6 ) kN m J M x M x M x               3 maxM 56 kN m 56 10 N m For S250 52 section, S 482 10 mm 3 3 Normal stress: 3 6 6 3 56 10 N m 116.2 10 Pa 482 10 m M S         116.2 MPa 

(32)

B A C D 5 ft 8 ft 5 ft W14 ⫻ 22 10 kips 5 kips

### PROBLEM 5.25

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum normal stress due to bending. SOLUTION Reaction at C: 0: (18)(5) 13 +(5)(10) 0 10.769 kips B M C C      Reaction at B: 0: (5)(5) (8)(10) 13 0 4.231 kips C M B B      Shear diagram: to : 5 kips to : 5 10.769 5.769 kips to : 5.769 10 4.231 kips A C V C D V D B V               At A and B, 0MAt C, 0: (5)(5) 0 25 kip ft C C C M M M       At D, 0: (5)(4.231) 21.155 kip ft D D D M M M       max 5.77 kips V   max

| |M occurs at C. | |Mmax  25 kip ft 300 kip in. 

For W14 22 rolled-steel section, S  29.0 in3

Normal stress: 300

29.0 M

S

(33)

B C D E A 8 kN 8 kN W310 ⫻ 23.8 1 m 1 m 1 m 1 m W

### PROBLEM 5.26

Knowing that W 12 kN, draw the shear and bending-moment diagrams for beam AB and determine the maximum normal stress due to bending.

SOLUTION By symmetry, AB 0: 8 12 8 0 2 kN y F A B A B          Shear: to A C: V  2 kN   C to :DV  6 kN   D to :EV 6 kN   E to :B V  2 kN 

Bending moment: Vmax6.00 kN

At C, MC 0: MC (1)(2)  0 MC 2 kN m At D, 0:MDMD (2)(2)(8)(1) 0 4 kN mMD     By symmetry, M 2 kN m at . D 2 kN mME max| |M  4.00 kN m occurs at E.  For W310 23.8, Sx  280 10 mm 3 3  280 10 m 6 3 Normal stress: 3 max max 6 | | 4 10 280 10 x M S      6 14.29 10 Pa   max 14.29 MPa 

(34)

B C D E A 8 kN 8 kN W310 ⫻ 23.8 1 m 1 m 1 m 1 m W

### PROBLEM 5.27

Determine (a) the magnitude of the counterweight W for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding maximum normal stress due to bending. (Hint: Draw the bending-moment diagram and equate the absolute values of the largest positive and negative bending moments obtained.)

SOLUTION By symmetry, A  B 0: 8 8 0 8 0.5 y F A W B A B W           Bending moment at C: 0: (8 0.5 )(1) 0 (8 0.5 ) kN m C C C M W M M W          Bending moment at D: 0: (8 0.5 )(2) (8)(1) 0 (8 ) kN m D D D M W M M W           Equate: 8MDMC W   8 0.5W W 10.67 kN  (a) 3 max 10.6667 kN 2.6667 kN m 2.6667 kN m 2.6667.10 N m | | 2.6667 kN m C D W M M M          

For W310 23.8 rolled-steel shape,

3 3 6 3 280 10 mm 280 10 m x S      (b) max 3 6 max 6 | | 2.6667 10 9.52 10 Pa 280 10 x M S        max 9.52 MPa 

(35)

B A C D a 8 ft 5 ft W14 ⫻ 22 10 kips 5 kips

### PROBLEM 5.28

Determine (a) the distance a for which the maximum absolute value of the bending moment in the beam is as small as possible, (b) the corresponding maximum normal stress due to bending. (See hint of Prob. 5.27.)

SOLUTION Reaction at B: 0: 5 (8)(10) 13 0 1 (80 5 ) 18 C B B M a R R a        Bending moment at D: 0: 5 0 5 5 (80 5 ) 13 D D B D B M M R M R a         Bending moment at C: 0 5 0 5 C C C M a M M a      Equate: 5 5 (80 5 ) 13 C D M M a a     4.4444 ft a  ( )a a 4.44 ft  Then MCMD (5)(4.4444)  22.222 kip ft max

| |M 22.222 kip ft  266.67 kip in. For W14 22 rolled-steel section, S 29.0 in3

Normal stress: 266.67 9.20 ksi

29.0 M

S

(36)

B A a C D P Q 12 mm 18 mm 500 mm 500 mm

### PROBLEM 5.29

Knowing that PQ 480 N, determine (a) the distance a for which the absolute value of the bending moment in the beam is as small as possible, (b) the corresponding maximum normal stress due to bending. (See hint of Prob. 5.27.)

SOLUTION 480 N 480 N PQReaction at A: 0: 480( 0.5) 480(1 ) 0 720 960 N D M Aa a a A a              Bending moment at C: 0: 0.5 0 360 0.5 480 N m C C C M A M M A a           Bending moment at D: 0: 480(1 ) 0 480(1 ) N m D D D M M a M a           (a) Equate: MD MC 480(1 a) 480 360 a      0.86603 a  m 866 a  mm  128.62 A  N MC 64.31 N m MD  64.31 N m

(b) For rectangular section, 1 2 6 Sbh 2 3 9 3 1 (12)(13) 648 mm 648 10 m 6 S      6 max max 9 | | 64.31 99.2 10 Pa 6.48 10 M S       max 99.2 MPa 

(37)

B A a C D P Q 12 mm 18 mm 500 mm 500 mm

### PROBLEM 5.30

Solve Prob. 5.29, assuming that P  480 Nand 320 N.QPROBLEM 5.29 Knowing that PQ 480 N, determine (a) the distance a for which the absolute value of the bending moment in the beam is as small as possible, (b) the corresponding maximum normal stress due to bending. (See hint of Prob. 5.27.) SOLUTION 480 N 320 N PQReaction at A: 0: 480( 0.5) 320(1 ) 0 560 800 N D M Aa a a A a             Bending moment at C: 0: 0.5 0 280 0.5 400 N m C C C M A M M A a             Bending moment at D: 0: 320(1 ) 0 ( 320 320 ) N m D D D M M a M a           (a) Equate: MD MC 320 320a 400 280 a      2 320a 80a280 0 a 0.81873 m, 1.06873 m

Reject negative root. a 819 mm 

116.014 N C 58.007 N m D 58.006 N m

AM   M   

(b) For rectangular section, 1 2

6 S bh 2 3 9 3 1 (12)(18) 648 mm 648 10 m 6 S      6 max max 9 | | 58.0065 89.5 10 Pa 648 10 M S       max 89.5 MPa 

(38)

Hinge 18 ft B a C 4 kips/ft W14 ⫻ 68 A

### PROBLEM 5.31

Determine (a) the distance a for which the absolute value of the bending moment in the beam is as small as possible, (b) the corresponding maximum normal stress due to bending. (See hint of Prob. 5.27.)

SOLUTION For W14 68, Sx 103 in3 Let (18b  a) ft Segment BC: By symmetry, VBC 0: 4 0 2 y B B F V C b V b       2 2 0: (4 ) 0 2 2 2 2 lb ft J B B x M V x x M M V x x bx x                 2 2 2 max 1 2 0 2 1 1 2 2 m m dM b x x b dx M b b b        Segment AB: 2 ( ) 0: 4( ) 2 ( ) 0 2( ) 2 ( ) K B a x M a x V a x M M a x b a x               max 2 2 max | | occurs at 0. | | 2 2 2 2 (18 ) 36 M x M a ab a a a a         

(a) Equate the two values of |Mmax|:

###  

2 2 2 2 2 1 2 1 1 1 36 (18 ) 162 18 2 2 2 1 54 162 0 54 (54) (4) (162) 2 a b a a a a a a             54a  50.91183.0883 ft 3.09 a  ft 

(b) | |Mmax 36a 111.179 kip ft 1334.15 kip in.

2 max | | 1334.15 12.95 kips/in 103 x M S     12.95 m  ksi 

(39)

B d A

L ⫽ 10 ft

### PROBLEM 5.32

A solid steel rod of diameter d is supported as shown. Knowing that for steel  490 lb/ft ,3 determine the smallest diameter d that can be used if the normal stress due to bending is not to exceed 4 ksi.

SOLUTION

Let W  total weight.

2 4 WAL  d LReaction at A: 1 2 AW

Bending moment at center of beam:

2 2 0: 0 2 2 2 4 8 32 C W L W L M M WL Md L                    

For circular cross section,

1

### 

2 cd 4, 3 3 4 4 32 I I c S c d c        Normal stress: 2 2 2 32 3 32 d L M L S d d         Solving for d, 2 L d    Data: 3 3 3 2 10 ft (12)(10) 120 in. 490 490 lb/ft 0.28356 lb/in 12 4 ksi 4000 lb/in L           2 (120) (0.28356) 4000 d  1.021 d  in. 

(40)

B b b A C D 1.2 m 1.2 m 1.2 m

### PROBLEM 5.33

A solid steel bar has a square cross section of side b and is supported as shown. Knowing that for steel   7860 kg / m ,3 determine the dimension b for which the maximum normal stress due to bending is (a) 10 MPa, (b) 50 MPa.

SOLUTION

Weight density:   g Let L  total length of beam.

2 WAL g b L gReactions at C and D: 2 W CD  Bending moment at C: 0: 0 6 3 18 C L W M M WL M           

Bending moment at center of beam:

0: 0 4 2 6 2 24 E L W L W WL M       M M                 2 2 max| | 18 18 WL b L g M   

For a square section, 1 3

6 Sb Normal stress: 2 2 2 3 | | /18 3 /6 M b L g L g S b b       Solve for b: 2 3 L g b    Data: L 3.6 m  7860 kg/m3 g 9.81 m/s2 (a)  10 10 Pa 6 (b)  50 10 Pa 6 (a) 2 3 6 (3.6) (7860)(9.81) 33.3 10 m (3)(10 10 ) b      33.3 b  mm  (b) 2 3 6 (3.6) (7860)(9.81) 6.66 10 m (3)(50 10 ) b      6.66 b  mm 

(41)

B w

A

L

### PROBLEM 5.34

Using the method of Sec. 5.2, solve Prob. 5.1a.

PROBLEM 5.1 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION 0: 0 2 2 B L wL M AL wL A        0: 0 2 2 A L wL M BL wL B       dV w dx   0 x A VV  

### 

wdx  wx A VVwxAwx 2 wL V  wxdM V dx  0 0 2 2 2 2 x x A wL M M V dx wx dx wLx wx         

### 

2 2 2 A wLx wx MM   ( 2) 2 w MLxx

Maximum M occurs at 1, where 2 x  0 dM V dx   | |max 82 wL M  

(42)

B P C A L b a

### PROBLEM 5.35

Using the method of Sec. 5.2, solve Prob. 5.2a.

PROBLEM 5.2 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION 0: 0 C Pb M LA bP A L      0: 0 A Pa M LC aP C L      At ,AV A Pb M 0 L    A to B: 0   x a 0 0 x 0 w

### 

wdx  0 A VVV Pb L   0 0 a a B A Pb Pba M M V dx dx L L  

### 

B Pba M L   At B +, V A P Pb P Pa L L       B to C: a   x L 0 ax 0 w

### 

wdx  0 C B VVV Pa L    ( ) 0 L C B a C B Pa Pab M M V dx L a L L

Pab Pba Pab

M M L L L            

### 

max | |M Pab L  

(43)

B w0

A

L

### PROBLEM 5.36

Using the method of Sec. 5.2, solve Prob. 5.3a.

PROBLEM 5.3 For the beam and loading shown, (a) draw the shear

and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves.

SOLUTION

Free body diagram for determining reactions.

Reactions: 0 0 0 : 0 2 2  yA  Aw L w L F V V 0 2 0 : 0 2 3             A A w L L M M 2 0 3   A w L M 0 0 2 0 , 2 , 3AA  x w L w L w w V M L 0     dV w x w dx L 0 0 2 0 2A  

### 

x   w x w x V V dx L L 2 0 0 2 2  w Lw x V L  2 2 2   oo dM w L w x V dx L 2 0 0 0 0 2 2 x x A w L w x M M V dx dx L          

### 

3 0 0 2 6  w Lxw x L 2 3 0 0 0 3 2 6   w Lw Lw x M x L

(44)

B

w

L A

### PROBLEM 5.37

Using the method of Sec. 5.2, solve Prob. 5.4a.

PROBLEM 5.4 For the beam and loading shown, (a) draw the shear

and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves.

SOLUTION 0: 0 y A A F V wL V wL      2 0: ( ) 0 2 2 A A L wL M M wL   M            dV w dx   0  A  

### 

x   V V wdx wx VwLwx dM V wL wx dx    2 0( ) 2A  

### 

x    wx M M wL wx dx wLx 2 2 2 2  wL   wx M wLx  max 2 max 2   V wL wL M

(45)

B P P C A a a

### PROBLEM 5.38

Using the method of Sec. 5.2, solve Prob. 5.5a.

PROBLEM 5.5 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION At A+: VA   P Over AB: dV w 0 dx    A dM V V P dx      M  PxC 1 0 at 0 0 MxCM  Px At point B: xa M  Pa At point B+: 2V   P P  P Over BC: dV w 0 dx    2 dM V P dx     2 2 M   PxC At B: xa M  Pa 2 2 2 Pa  PaC CPa 2 M   PxPa At C: 2xa M  3Pa 

(46)

D A B a a C L w w

### PROBLEM 5.39

Using the method of Sec. 5.2, solve Prob. 5.6a.

PROBLEM 5.6 For the beam and loading shown, (a) draw the shear and bending-moment diagrams, (b) determine the equations of the shear and bending-moment curves. SOLUTION Reactions: ADwa A to B: 0   x a ww , A VAwa MA  0 0  A  

### 

x   V V wdx wx ( ) Vw a   x dM V wa wx dx    0 0( ) x x A MM

V dx

### 

wawx dx 2 1 2 Mwaxwx  2 1 0 2 B B VMwa B to C: a    x L a V   0 0 dM V dx   0 x B a MM

### 

V dxB MM 1 2 2 Mwa       

(47)

PROBLEM 5.39 (Continued) L   a x L C to D: [VVC  

### 

L ax w dx  w x(La )] [ )]    V w L x a  [ ( )] x x C L a L a MM

V dx

### 

w xLa dx 2 2 2 2 2 2 ( ) 2 ( ) ( ) ( ) 2 2 ( ) ( ) 2 2 x L a x w L a x x L a w L a x L a x L a w L a x                                    2 2 2 1 ( ) ( ) 2 2 2 x L a Mwaw  La x     

(48)

B

A C D E

3 kN 2 kN 5 kN 2 kN

0.3 m 0.3 m 0.3 m 0.4 m

### PROBLEM 5.40

Using the method of Sec. 5.2, solve Prob. 5.7.

PROBLEM 5.7 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Free body diagram for determining reactions. Reactions: 0: 3 kN 2 kN 5 kN 2 kN 0 y A F V        2 kNVA   0: (3 kN)(0.3 m) (2 kN)(0.6 m) (5 kN)(0.9 m) (2 kN)(1.3 m) 0 MAMA      0.2 kN mMA  

Between concentrated loads and the vertical reaction, the scope of the shear diagram is , i.e., the shear is constant. Thus, the area under the shear diagram is equal to the change in bending moment.

A to C: 2 kN C A 0.6 C 0.4 kN V M M   M   C to D: 1 kN 0.3 0.1 kN m   DC   D    V M M M D to E: 3 kN 0.9 0.8 kN m   ED   E    V M M M E to B: 2 kN B E 0.8 B 0 (Checks) V   MM   M  

(49)

100 lb 250 lb 100 lb 10 in. 25 in. 20 in. 15 in. A B C D E

### PROBLEM 5.41

Using the method of Sec. 5.2, solve Prob. 5.8.

PROBLEM 5.8 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Free body diagram for determining reactions. Reactions:

0 : 100 lb 250 lb 100 lb 0

FYVCVE    

450 lbVCVE  

0 : (45 in.) (100 lb)(15 in.) (250 lb)(20 in.) (100 lb)(55 in.) 0

MCVE    

200 lbVE

250 lb VC

Between concentrated loads and the vertical reaction, the scope of the shear diagram is , i.e., the shear is constant. Thus, the area under the shear diagram is equal to he change in bending moment. A to C: 100 lb, C A 1500, C 1500 lb in. V   MM   M    C to D: 150 lb 3000, 1500 lb in.  DC   D    V M M M D to E: 100 lb, 2500, 1000 lb in.   ED   E    V M M M E to B: 100 lb, 1000, 0 (Checks)  BE   BV M M M

(50)

B A C D 25 kN/m 40 kN 40 kN 0.6 m 1.8 m 0.6 m

### PROBLEM 5.42

Using the method of Sec. 5.2, solve Prob. 5.9.

PROBLEM 5.9 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Free body diagram to determine reactions: 0: (3.0 m) 45 kN(1.5 m) (40 kN)(0.6 m) (40 kN)(2.4 m) 0 A B M V       62.5 kNVB   0: 40 kN 45 kN 40 kN 62.5 kN 0 y A F V        62.5 kNVA

Change in bending moment is equal to area under shear diagram. A to C: (62.5 kN)(0.6 m) 37.5 kN m C to E: 1(0.9 m)(22.5 kN) 10.125 kN m 2   E to D: 1(0.9 m)( 22.5 kN) 10.125 kN m 2     D to B: ( 62.5 kN)(0.6 m)  37.5 kN m   

(51)

B A C D 2.5 kips/ft 15 kips 6 ft 3 ft 6 ft

### PROBLEM 5.43

Using the method of Sec. 5.2, solve Prob. 5.10.

PROBLEM 5.10 Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION

Reactions at supports A and B:

0: 15( ) (12)(6)(2.5) (6)(15) 0 B A M R       18 kipsRA   0: 15 (3)(6)(2.5) (9)(15) 0 A B M R      12 kipsRB

Areas under shear diagram:

A to C: (6)(3) 1(6)(15) 63 kip ft 2    C to D: (3)(3) 9 kip ft D to B: (6)( 12)  72 kip ft Bending moments: 0 A M  0 63 63 kip ft C M     63 9MD    72 kip ft 72MB  72  0

(52)

0.5 m 4 kN 1 m 1 m 0.5 m 4 kN A E D C B F

### PROBLEM 5.44

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION 0: 3 (1)(4) (0.5)(4) 0 2 kN B M A         A 0: 3 (2)(4) (2.5)(4) 0 6 kN A M B        B Shear diagram: A to C: 2 kNVC to D: 2V    4 2 kN D to B: 2V     4 6 kN

Areas of shear diagram:

A to C: (1)(2)

### 

V dx   2 kN m C to D: (1)( 2)

### 

V dx    2 kN m D to E: (1)( 6)

### 

V dx    6 kN m Bending moments: 0 A M  0 2 2 kN m 2 4 6 kN m 6 2 4 kN m 4 2 6 kN m 6 6 0 C C D D B M M M M M                        (a) Vmax 6.00 kN  (b) M max 6.00 kN m 

(53)

300 N 300 N 200 mm 75 mm 200 mm 200 mm A C D B F E

### PROBLEM 5.45

Draw the shear and bending-moment diagrams for the beam and loading shown, and determine the maximum absolute value (a) of the shear, (b) of the bending moment.

SOLUTION 3 0: 0.075 (0.2)(300) (0.6)(300) 0 3.2 10 N A EF EF M F F        3 0: 0 3.2 10 N x x EF x F A F A       0: 300 300 0 600 N y y y F A A       Couple at D: (0.075)(3.2 10 )3 240 N m D M     Shear: A to C: 600 NVC to B: 600V  300 300 N

Areas under shear diagram:

A to C: (0.2)(600)

### 

V dx  120 N m C to D: (0.2)(300)

### 

V dx   60 N m D to B: (0.2)(300)

### 

V dx   60 N m Bending moments: 0 A M  0 120 120 N m 120 60 180 N m 180 240 60 N m 60 60 0 C D D B M M M M                    Maximum 600 NV   Maximum M 180.0 N m 

(54)

B A C 3 kN/m 1.5 m 1.5 m 2.2 m 100 mm 200 mm 10 kN

### PROBLEM 5.46

Using the method of Sec. 5.2, solve Prob. 5.15.

PROBLEM 5.15 For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C. SOLUTION 0: 3 (1.5)(10) (1.1)(2.2)(3) 0 2.58 kN C M A A        0: (1.5)(10) 3 (4.1)(2.2)(3) 0 14.02 kN A M C C        Shear: A to D: 2.58 kNV D to C: 2.58 10V  7.42 kN C: 7.42 14.02V   6.60 kN B: 6.60V (2.2)(3) 0

Areas under shear diagram:

A to D: (1.5)(2.58)

### 

V dx  3.87 kN m D to C: (1.5)( 7.42)

### 

V dx    11.13 kN m C to B: 1 (2.2)(6.60) 7.26 kN m 2 V dx     

### 

Bending moments: 0 A M  0 3.87 3.87 kN m 3.87 11.13 7.26 kN m 7.26 7.26 0 D C B M M M             3 7.26 kN m 7.26 10 N m C M     

(55)

PROBLEM 5.46 (Continued)

For rectangular cross section, 1 2 1 (100)(200)2

6 6 Sbh       3 3 6 2 666.67 10 mm 666.67 10 m     Normal stress: 3 6 6 7.26 10 10.89 10 Pa 666.67 10       C M S  10.89 MPa 

(56)

750 lb B A C D 150 lb/ft 750 lb 3 in. 12 in. 4 ft 4 ft 4 ft

### PROBLEM 5.47

Using the method of Sec. 5.2, solve Prob. 5.16.

PROBLEM 5.16 For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION Reactions: CAby symmetry 0: (2)(750) (12)(150) 0 1650 lb y F A C A C         Shear: VA 1650 lb VC 1650(4)(150) 1050 lb VC 1050750300 lb VE  300(2)(150)  0

Areas under shear diagram:

A to C: 1 (1650 1050)(4) 2 V dx      

### 

5400 lb ft   C to E: 1 (300)(2) 300 lb ft 2 V dx       

### 

Bending moments: 0  A M 0 5400 5400 lb ft 5400 300 5700 lb ft         C E M M 3 5700 lb ft 68.4 10 lb in.      E M

For rectangular cross section, 1 2 1 (3)(12)2 72 in3

6 6 Sbh        Normal stress: 3 68.4 10 950 psi 72 M S      

(57)

B A a a 30 kN 50 kN 50 kN 30 kN 2 m 5 @ 0.8 m 5 4 m W310 3 52

### PROBLEM 5.48

Using the method of Sec. 5.2, solve Prob. 5.18.

PROBLEM 5.18 For the beam and loading shown, determine the maximum normal stress due to bending on section a-a.

SOLUTION Reactions: By symmetry, A  B. 0: 80 kN y F   AB   Shear diagram: A to C: 80 kNVC to D: 80 30V 50 kN D to E: 50V  50  0 Areas of shear diagram:

A to C: (80)(0.8)V dx  64 kN m C to D: (50)(0.8)V dx   40 kN m D to E: 0V dx  Bending moments: 0 A M  0 64 64 kN m C M     MD  6440104 kN m ME 104 0 104 kN m 3 max 104 kN m 104 10 N m M      For W310 52, S  747 10 mm 3 3  747 10 m 6 3 Normal stress: 3 6 6 104 10 139.2 10 Pa 747 10 M S        139.2 MPa   

(58)

B A C D E F G 5 kips 5 kips 2 kips 2 kips 2 kips 6 @ 15 in. 5 90 in. S8 3 18.4

### PROBLEM 5.49

Using the method of Sec. 5.2, solve Prob. 5.20.

PROBLEM 5.20 For the beam and loading shown, determine the maximum normal stress due to bending on a transverse section at C.

SOLUTION

Use entire beam as free body. 0: B M   90 (75)(5) (60)(5) (45)(2) (30)(2) (15)(2) 0 9.5 kips A          A Shear A to C: 9.5 kipsV

Area under shear curve A to C: (15)(9.5)

142.5 kip in. V dx     0 A M  0 142.5MC   142.5 kip in. For S8 18.4, S 14.4 in3 Normal stress: 142.5 14.4 M S    9.90 ksi  

(59)

w L A B x w 5 w0 [x/L]1/2

### PROBLEM 5.50

Determine (a) the equations of the shear and bending-moment curves for the beam and loading shown, (b) the maximum absolute value of the bending moment in the beam.

SOLUTION 1 1/2 2 0 0 1/2             dV x w x w w dx L L 3/2 1 1/2 2 3   w xoV C L 0 atVxL 0 1 1 0 2 2 0 3 3   w LC Cw L 3/2 0 0 1/2 2 2 3 3         w x V w L L  5/2 0 2 0 1/2 2 2 2 3 3 5      dM w x V M C w Lx dx L 2 2 2 0 0 2 0 2 4 2 0 at 0 3 15 5        M x L C w L w L C w L 5/2 2 0 0 1/2 0 2 4 2 3 15 5   w xM w Lx w L L  2 0 max max 2 occurs at 0 5   M x M w L 

(60)

w A L B x w ⫽ w0 cos2Lx

### PROBLEM 5.51

Determine (a) the equations of the shear and bending-moment curves for the beam and loading shown, (b) the maximum absolute value of the bending moment in the beam.

SOLUTION 0 0 1 2 0 1 2 2 1 2 0 2 2 cos 2 2 sin 2 4 cos 2 0 at 0. Hence, 0. 4 0 at 0. Hence, . dV x w w dx L Lw x dM V C L dx L w x M C x C L V x C L w M x C                         (a) V  (2Lw0/sin(x L/2 )  2 0 (4 / )[1 cos( /2 )] M   L w   x L(b) M max  4w L0 2/2 

(61)

B x w w ⫽ w 0 sin A Lx L

### PROBLEM 5.52

Determine (a) the equations of the shear and bending-moment curves for the beam and loading shown, (b) the maximum absolute value of the bending moment in the beam.

SOLUTION 0 0 1 2 0 1 2 2 2 1 1 sin cos sin 0 at 0 0 0 at 0 0 0 0 dV x w w dx L w L x dM V C L dx w L x M C x C L M x C M x L C L C                         (a) V w L0 cos x L     2 0 2 sin w L x M L     0 at 2 dM L V x dx    (b) 2 0 max 2 sin 2 w L M    0 2 max 2 w L M   

(62)

B x w w ⫽ w0 A L x L

### PROBLEM 5.53

Determine (a) the equations of the shear and bending-moment curves for the beam and loading shown, (b) the maximum absolute value of the bending moment in the beam.

SOLUTION 0 2 0 1 3 0 1 2 1 2 1 6 dV x w w dx L x dM V w C L dx x M w C x C L             2 0 at 0 0 MxC  2 0 1 1 0 1 1 0 at 0 6 6 Mx L   w LC L Cw L (a) 2 2 0 0 1 1 2 6 x V w w L L    2 2 0 1 ( 3 )/ 6 Vw Lx L  3 0 0 1 1 6 6 x M w w Lx L    3 0 1 ( / ) 6 Mw Lx x L 

(b) Mmaxoccurs when 0. 2 3 2 0

m dM V L x dx     2 2 max 0 1 6 3 3 3 3 m L L L xMw      2 max 0.0642 0 Mw L 

References

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