Mechanics of Materials Solutions Chapter11 Probs47 57

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11.47 A W530 × 92 structural steel wide-flange shape [ E = 200 GPa; I = 554 × 10

6

mm4] is loaded and supported as shown in

Fig. P11.47. Determine:

(a) the force and moment reactions atsupports A and C.

(b) the maximum bending stress in the

beam.

(c) the deflection of the beam at B.Fig. P11.47

Solution

(a) Reactions at A and C . Choose the moment reactions at A and C as the redundants. This will leave a

simply supported beam between A and C as the released beam.

Determine the slopes at A and C caused by the 150-kN concentrated load.

[Appendix C, SS beam with concentrated load not at midspan.] Relevant equations from Appendix C:

2 2 2 2( ) ( )

and6 6 A C

Pb L b Pa L a

LEI LEI θ θ

− −

= − =

Values:

P = 150 kN, L = 10 m, a = 6 m, b = 4 m

Calculation:2 2 2

2 2

2 2 22 2

( ) (150 kN)(4 m) 840 kN-m(10 m) (4 m)

6 6(10 m)

( ) (150 kN)(6 m) 960 kN-m(10 m) (6 m)

6 6(10 m)

A

C

Pb L b

LEI EI EI

Pa L a

LEI EI EI

θ

θ

−⎡ ⎤= − = − − = −⎣ ⎦

−⎡ ⎤= = − =⎣ ⎦

Determine the slopes at A and C caused by moment reaction M A.

[Appendix C, SS beam with concentrated moment at one end.]

Relevant equations from Appendix C:

and3 6

A C

L ML

EI EI θ θ = − =

Values: M = M A, L = 10 m

Calculation:(10 m) (3.333333 m)

3 3

(10 m) (1.666667 m)

6 6

A A A

A AC

L M M

EI EI EI

ML M M

EI EI EI

θ

θ

= − = − = −

= = =





Determine the slopes at A and C caused by moment reaction M C .



and6 3

A C

L ML

EI EI θ θ = − =

Values: M = M C , L = 10 m


6 6

(10 m) (3.333333 m)

3 3

C C A

C C C

L M M

EI EI EI

ML M M

EI EI EI

θ

θ

= − = − = −

= = =

Compatibility equation for slope at A:2840 kN-m (3.333333 m) (1.666667 m)

0 A C M M

EI EI EI − − − = (a)

Compatibility equation for slope at C :2960 kN-m (1.666667 m) (3.333333 m)

0 A C M M

EI EI EI + + = (b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as:2

2

(3.333333 m) (1.666667 m) 840 kN-m

(1.666667 m) (3.333333 m) 960 kN-m

A C

A C

M M

M M

+ = −

+ = −

and solved simultaneously for M A and M C :

144 kN-m 144 kN-m (ccw) A M = − = Ans.

216 kN-m 216 kN-m (cw)C M = − = Ans.

Equilibrium equations for entire beam:

(150 kN)(6 m) (10 m) 0 A A C C

M M M RΣ = − + − + =

(150 kN)(6 m) ( 144 kN-m) ( 216 kN-m)

10 m

97.2 kN 97.2 kN

C R + − − −

∴ =

= = ↑ Ans.

150 kN 0 y A C F R RΣ = + − =

150 kN 97.2 kN 52.8 kN 52.8 kN A

R∴ = − = = ↑ Ans.





Shear-force and bending-moment diagrams

(b) Magnitude of maximum bending stress:

Section properties (from Appendix B):6 4

3 3

554 10 mm

533 mm

2, 080 10 mm

I

d

S

= ×

=

= ×

Maximum bending moment magnitude

M max = 216 kN-m (at C )

Bending stresses at maximum moment2

6 4

(216 kN-m)(533 mm/2)(1,000)

554 10 mm

103.9 MPa

xσ =×

= Ans.

or using the tabulated value for the section

modulus:2

3 3(216 kN-m)(1,000)

2,080 10 mm

103.8 MPa

xσ =×

= Ans.

(c) Beam deflection at B:

Determine the deflection at B caused by the 150-kN concentrated load.

[Appendix C, SS beam with concentrated load not at midspan.]

Relevant equation from Appendix C:

2 2 2( )6

B

Pabv L a b

LEI = − − −

Values: P = 150 kN, L = 10 m, a = 6 m,

b = 4 m

Calculation:3

2 2 2 2 2 2(150 kN)(6 m)(4 m) 2,880 kN-m( ) (10 m) (6 m) (4 m)

6 6(10 m) B

Pabv L a b

LEI EI EI ⎡ ⎤= − − − = − − − = −⎣ ⎦

Determine the deflection at B caused by M A.


Relevant equation from Appendix C:2 2(2 3 )

6 B

M xv L Lx x

LEI = − − +

Values: M = −144 kN-m, L = 10 m, x = 6 m





Calculation:

2 2

32 2

(2 3 )6

( 144 kN-m)(6 m) 806.4 kN-m2(10 m) 3(10 m)(6 m) (6 m)

6(10 m)

B

M xv L Lx x

LEI

EI EI

= − − +

−⎡ ⎤= − − + =⎣ ⎦

Determine the deflection at B caused by M C .



2 2(2 3 )6

B

M xv L Lx x

LEI = − − +

Values: M = −216 kN-m, L = 10 m, x = 4 m

Calculation:

2 2

32 2

(2 3 )

6( 216 kN-m)(4 m) 1,382.4 kN-m

2(10 m) 3(10 m)(4 m) (4 m)6(10 m)

B

M xv L Lx x

LEI

EI EI

= − − +

−⎡ ⎤= − − + =⎣ ⎦

Beam deflection v B:3 3 3

3 3

2

2,880 kN-m 806.4 kN-m 1,382.4 kN-m

691.2 kN-m 691.2 kN-m0.006238 m 6.24 mm

110,800 kN-m

Bv EI EI EI

EI

= − + +

= − = − = − = ↓ Ans.





11.48 A W530 × 92 structural steel wide-flange shape [ E = 200 GPa; I = 554 × 10

6

mm4] is loaded and supported as shown in

Fig. P11.48. Determine:(a) the force and moment reactions at

supports A and C.

(b) the maximum bending stress in the beam.

(c) the deflection of the beam at B.

Fig. P11.48

Solution

(a) Reactions at A and C . Choose the moment reactions at A and C as the redundants. This will leave asimply supported beam between A and C as the released beam.

Determine the slopes at A and C caused by the 80 kN/m uniformly distributed load.


Relevant equations from Appendix C:2

2

22 2

(2 ) and

24

(2 )24

A

C

wa L a

LEI wa

L a LEI

θ

θ

= − −

= −

Values: w = 80 kN/m, L = 9 m, a = 4.5 m

Calculation:

[ ]2 2 2

22

2 2 2

2 2 2 2

(80 kN/m)(4.5 m) 1,366.875 kN-m(2 ) 2(9 m) (4.5 m)

24 24(9 m)

(80 kN/m)(4.5 m) 1,063.125 kN-m(2 ) 2(9 m) (4.5 m)

24 24(9 m)

A

C

wa L a

LEI EI EI

wa L a

LEI EI EI

θ

θ

= − − = − − = −

⎡ ⎤= − = − =⎣ ⎦

Determine the slopes at A and C caused by moment reaction M A.

[Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C:

and3 6

A C

L ML

EI EI θ θ = − =

Values:

M = M A, L = 9 m

Calculation:(9 m) (3 m)

3 3

(9 m) (1.5 m)

6 6

A A A

A AC

L M M

EI EI EI

ML M M

EI EI EI

θ

θ

= − = − = −

= = =





Determine the slopes at A and C caused by moment reaction M C .



and6 3

A C

L ML

EI EI θ θ = − =

Values: M = M C , L = 9 m


6 6

(9 m) (3 m)

3 3

C C A

C C C

L M M

EI EI EI

ML M M

EI EI EI

θ

θ

= − = − = −

= = =

Compatibility equation for slope at A:21,366.875 kN-m (3 m) (1.5 m)

0 A C M M

EI EI EI − − − = (a)

Compatibility equation for slope at C :21,063.125 kN-m (1.5 m) (3 m)

0 A C M M

EI EI EI + + = (b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as:2

2

(3 m) (1.5 m) 1,366.875 kN-m

(1.5 m) (3 m) 1,063.125 kN-m

A C

A C

M M

M M

+ = −

+ = −

and solved simultaneously for M A and M C :

371.25 kN-m 371.25 kN-m (ccw) A M = − = Ans.

168.75 kN-m 168.8 kN-m (cw)C M = − = Ans.


(80 kN/m)(4.5 m)(2.25 m) (9 m) 0 A A C C M M M RΣ = − + − + =

(80 kN/m)(4.5 m)(2.25 m) ( 371.25 kN-m) ( 168.75 kN-m)

9 m

67.5 kN 67.5 kN

C R + − − −

∴ =

= = ↑ Ans.

(80 kN/m)(4.5 m) 0 y A C F R RΣ = + − =

(80 kN/m)(4.5 m) 67.5 kN 292.5 kN 293 kN A R∴ = − = = ↑ Ans.






(b) Magnitude of maximum bending stress:

Section properties (from Appendix B):6 4

3 3

554 10 mm

533 mm

2, 080 10 mm

I

d

S

= ×

=

= ×


M max = 371.25 kN-m (at A)

Bending stresses at maximum moment2

6 4

(371.25 kN-m)(533 mm/2)(1,000)

554 10 mm

178.6 MPa

xσ =×

= Ans.


modulus: 2

3 3

(371.25 kN-m)(1,000)

2,080 10 mm

178.5 MPa

xσ =×

= Ans.

(c) Beam deflection at B:

Determine the deflection at B caused by the 80 kN/m uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load over a portion of the span.] Relevant equation from Appendix C:3

2 2(4 7 3 )24

B

wav L aL a

LEI = − − +

Values: w = 80 kN/m, L = 9 m, a = 4.5 m

Calculation:3

2 2

3 3

2 2

(4 7 3 )24

(80 kN/m)(4.5 m) 3,417.1875 kN-m4(9 m) 7(4.5 m)(9 m) 3(4.5 m)

24(9 m)

B

wav L aL a

LEI

EI EI

= − − +

⎡ ⎤= − − + = −⎣ ⎦





Determine the deflection at B caused by M A.



2 2(2 3 )6

B

M xv L Lx x

LEI = − − +

Values: M = −371.25 kN-m, L = 9 m,

x = 4.5 m

Calculation:

2 2

32 2

(2 3 )6

( 371.25 kN-m)(4.5 m) 1,879.4531 kN-m2(9 m) 3(9 m)(4.5 m) (4.5 m)

6(9 m)

B

M xv L Lx x

LEI

EI EI

= − − +

−⎡ ⎤= − − + =⎣ ⎦

Determine the deflection at B caused by M C .


Relevant equation from Appendix C:2 2(2 3 )

6 B

M xv L Lx x

LEI = − − +

Values: M = −168.75 kN-m, L = 9 m,

x = 4.5 m

Calculation:

2 2

32 2

(2 3 )6

( 168.75 kN-m)(4.5 m) 854.2969 kN-m2(9 m) 3(9 m)(4.5 m) (4.5 m)6(9 m)

B

M xv L Lx x

LEI

EI EI

= − − +

− ⎡ ⎤= − − + =⎣ ⎦

Beam deflection v B:3 3 3

3 3

2

3,417.1875 kN-m 1,879.4531 kN-m 854.2969 kN-m

683.4375 kN-m 683.4375 kN-m0.006168 m 6.17 mm

110,800 kN-m

Bv EI EI EI

EI

= − + +

= − = − = − = ↓ Ans.





11.49 A timber [ E = 1,800 ksi] beam isloaded and supported as shown in Fig.

P11.49. The cross section of the timber beam

is 4-in. wide and 8-in. deep. The beam issupported at B by a ½-in.-diameter steel [ E =

30,000 ksi] rod, which has no load before the

distributed load is applied to the beam. After

a distributed load of 900 lb/ft is applied to

the beam, determine:(a) the force carried by the steel rod.

(b) the maximum bending stress in thetimber beam.


Fig. P11.49

Solution

Section properties:3

4

beam beam

2 2

1 1

(4 in.)(8 in.)Beam: 170.6667 in. 1,800,000 psi

12Rod (1): (0.50 in.) 0.1963495 in. 30,000,000 psi

4

I E

A E π

= = =

= = =

(a) Force carried by the steel rod.

The reaction force from rod (1) will be

taken as the redundant, leaving a simplysupported beam between A and C as the

released beam.

For this analysis, a tension force is

assumed to exist in axial member (1).

Beam free-body diagram

Downward deflection of wood beam at B due to 900 lb/ft uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load.]

Relevant equation from Appendix C:45

384 B

wLv

EI = −

Values: w = 900 lb/ft, L = 14 ft, EI = 307.2 ×10

6 lb-in.

2

Calculation:4 4 3

6 2

5 5(900 lb/ft)(14 ft) (12 in./ft)2.532305 in.

384 384(307.2 10 lb-in. ) B

wLv

EI = − = − = −

×





Upward deflection of wood beam at B due to force F 1 in rod (1).

[Appendix C, SS beam with concentrated load at midspan.]


48 B

PLv

EI = −

Values:

P = − F 1, L = 14 ft, EI = 307.2 ×106 lb-in.

2

Calculation:3 3 3

6116 2

( )(14 ft) (12 in./ft)(321.5625 10 in./lb)

48 48(307.2 10 lb-in. ) B

PL F v F

EI

−−= − = − = ×

×

Elongation of steel rod (1) due to force F 1.

61 11 1 12 6

1 1

(16 ft)(12 in./ft)(32.59493 10 in./lb)

(0.1963495 in. )(30 10 psi)

F Le F F

A E

−= = = ××

The deflection of the wood beam at B will not equal zero in this instance because the rod that supports

the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore, 6

1 1(32.59493 10 in./lb) Bv e F −= − = − ×

Compatibility equation at B:

The sum of the downward deflection caused by the uniformly distributed load and the upward deflectioncaused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will

produce a downward (i.e., negative) deflection of the wood beam at B.

6 6

1 1

1 6

2.532305 in. (321.5625 10 in./lb) (32.59493 10 in./lb)

2.532305 in.7,150.22 lb 7,150 lb (T)

354.1574 10 in./lb

F F

F

− −

−

− + × = − ×

∴ = = =×

Ans.





(b) Determine maximum bending stress in

wood beam:


M max = 4,125 lb-ft = 49,500 lb-in.

Bending stress at maximum moment

4

(49,500 lb-in.)(8 in./2) 1,160 psi170.6667 in.

xσ = = Ans.

(c) Beam deflection at B.

The beam deflection at B is equal to the elongation of the steel rod:

1 11 2 6

1 1

(7,150.22 lb)(16 ft)(12 in./ft)0.233 in. 0.233 in.

(0.1963495 in. )(30 10 psi) B

F Lv e

A E = − = − = − = − = ↓

× Ans.





11.50 A W360 × 72 structural steel [ E = 200GPa] wide-flange shape is loaded and

supported as shown in Fig. P11.50. The beam

is supported at B by 20-mm-diameter solidaluminum [ E = 70 GPa] rod. After a

concentrated load of 40 kN is applied to the tip

of the cantilever, determine:

(a) the force produced in the aluminum rod.

(b) the maximum bending stress in the beam.(c) the deflection of the beam at B.

Fig. P11.50

Solution

Section properties:6 4

beam beam

3 3

beam

2 21 1

W360 × 72: 201 10 mm 351 mm 200,000 MPa

1,150 10 mm

Rod (1): (20 mm) 314.159266 mm 70,000 MPa4

I d E

S

A E π

= × = =

= ×

= = =

(a) Force in the aluminum rod.


taken as the redundant, leaving a cantilever beam as the released beam.



Downward deflection of W360 × 72 beam at B due to 40-kN concentrated load.

[Appendix C, Cantilever beam with concentrated load at tip.]


(3 ) (elastic curve)6

B

Pxv L x

EI = − −

Values: P = 40 kN, L = 5 m, x = 3.6 m, EI = 40,200 kN-m

2

Calculation:

[ ]2 2

3

2

(40 kN)(3.6 m)(3 ) 3(5 m) (3.6 m) 24.50149 10 m

6 6(40,200 kN-m ) B

Pxv L x

EI

−= − − = − − = − ×





Upward deflection of W360 × 72 beam at B due to force F 1 in rod (1).



3 B

PLv

EI = −

Values:

P = − F 1, L = 3.6 m, EI = 40,200 kN-m2

Calculation:3 3

6112

( )(3.6 m)(386.8657 10 m/kN)

3 3(40,200 kN-m ) B

PL F v F

EI

−−= − = − = ×

Elongation of aluminum rod (1) due to force F 1.

61 11 1 12 2

1 1

(3 m)(136.4185 10 m/kN)

(314.159266 mm )(70,000 N/mm )(1 kN/1,000 N)

F Le F F

A E

−= = = ×

The deflection of the W360 × 72 beam at B will not equal zero in this instance because the rod that

supports the beam at B will elongate, thus permitting the W360 × 72 beam to deflect downward.

Therefore, 6

1 1(136.4185 10 m/kN) Bv e F −= − = − ×


The sum of the downward deflection caused by the 40-kN concentrated load and the upward deflection

caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the

aluminum rod will produce a downward (i.e., negative) deflection of the W360 × 72 beam at B.

3 6 6

1 1

3

1 6

24.50149 10 m (386.8657 10 m/kN) (136.4185 10 m/kN)

24.50149 10 m 46.82254 kN 46.8 kN (T)523.2842 10 m/kN

F F

F

− − −

−

−

− × + × = − ×

×∴ = = =×

Ans.





(b) Determine maximum bending stress in the

W360 × 72 beam:


M max = 56 kN-m (at B)


2

6 4

(56 kN-m)(351 mm/2)(1,000)

201 10 mm

48.9 MPa

xσ =×

= Ans.

or using the tabulated value for the sectionmodulus of the W360 × 72 beam:

2

3 3

(56 kN-m)(1,000)48.7 MPa

1,150 10 mm x

σ = =×

Ans.


The beam deflection at B is equal to the elongation of the aluminum rod:

1 11 2 2

1 1

(46,822.54 N)(3,000 mm)6.38746 mm 6.39 mm

(314.159266 mm )(70,000 N/mm ) B

F Lv e

A E = − = − = − = − = ↓ Ans.





11.51 A W18 × 55 structural steel [ E = 29,000ksi] wide-flange shape is loaded and supported

as shown in Fig. P11.51. The beam is supported

at C by a ¾-in.-diameter aluminum [ E = 10,000ksi] rod, which has no load before the

distributed load is applied to the beam. After a

distributed load of 4 kips/ft is applied to the

beam, determine:

(a) the force carried by the aluminum rod.(b) the maximum bending stress in the steel

beam.(c) the deflection of the beam at C .

Fig. P11.51

Solution


beam beam

3

beam

2 2

1 1

Beam: 890 in. 18.1 in. 29,000 ksi

98.3 in.

Rod (1): (0.75 in.) 0.4417865 in. 10,000 ksi4

I d E

S

A E π

= = =

=

= = =

(a) Force in the aluminum rod.

The reaction force from rod (1) will betaken as the redundant, leaving a

cantilever beam as the released beam.



Downward deflection of W18 × 55 beam at C due to 4 kips/ft uniformly distributed load.

[Appendix C, Cantilever beam with uniformly distributed load.]

Relevant equations from Appendix C:4 3

and8 6

B B

wL wLv

EI EI θ = − = −

Values: w = 4 kips/ft, L = 11 ft, EI = 25.81×10

6 kip-in.

2

Calculation:4 4 3

6 2

3 3 23

6 2

(4 kips/ft)(11 ft) (12 in./ft)0.490113 in.

8 8(25.81 10 kip-in. )

(4 kips/ft)(11 ft) (12 in./ft)4.950639 10 rad

6 6(25.81 10 kip-in. )

0.490113 in. (5 ft)(12 in./ft)( 4.9

B

B

C

wLv

EI

wL

EI

v

θ −

= − = − = −×

= − = − = − ××

= − + − 350639 10 rad) 0.787152 in.−× = −





Upward deflection of W18 × 55 beam at C due to force F 1 in rod (1).



3C

PLv

EI = −

Values:

P = − F 1, L = 16 ft, EI = 25.81×106 kip-in.

2

Calculation:3 3 3

3116 2

( )(16 ft) (12 in./ft)(91.41015 10 in./kip)

3 3(25.81 10 kip-in. )C

PL F v F

EI

−−= − = − = ×

×


31 11 1 12

1 1

(14 ft)(12 in./ft)(38.02742 10 in./kip)

(0.4417865 in. )(10,000 ksi)

F Le F F

A E

−= = = ×

The deflection of the W18 × 55 beam at C will not equal zero in this instance because the rod that

supports the beam at C elongates, thus permitting the W18 × 55 beam to deflect downward. Therefore, 3

1 1(38.02742 10 in./kip)C v e F

−= − = − ×

Compatibility equation at C :

The sum of the downward deflection caused by the uniformly distributed load and the upward deflectioncaused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of the

aluminum rod will produce a downward (i.e., negative) deflection of the W18 × 55 beam at C .

3 3

1 1

1 3

0.787152 in. (91.41015 10 in./kip) (38.02742 10 in./kip)

0.787152 in.6.081323 kips 6.08 kips (T)

129.4376 10 in./kip

F F

F

− −

−

− + × = − ×

∴ = = =×

Ans.






W18 × 55 beam:


M max = 144.70 kip-ft (at A)


4

(144.70 kip-ft)(18.1 in./2)(12 in./ft)

890 in.

17.66 ksi

xσ =

= Ans.

or using the tabulated value for the sectionmodulus of the W18 × 55 beam:

3

(144.70 kip-ft)(12 in./ft)

98.3 in.

17.66 ksi

xσ =

= Ans.

(c) Beam deflection at C .

The beam deflection at C is equal to the elongation of the aluminum rod:

1 11 2

1 1

(6.081323 kips)(14 ft)(12 in./ft)0.231257 in. 0.231 in.

(0.4417865 in. )(10,000 ksi)C

F Lv e

A E = − = − = − = − = ↓ Ans.





11.52 A W250 × 32.7 structural steel [ E =200 GPa] wide-flange shape is loaded and

supported as shown in Fig. P11.52. A

uniformly distributed load of 16 kN/m isapplied to the beam, causing the roller

support at B to settle downward (i.e.,

displace downward) by 15 mm. Determine:

(a) the reactions at supports A, B, and C.

(b) the maximum bending stress in the beam.

Fig. P11.52

Solution

Section properties:6 4 3 3250 32.7 : 49.1 10 mm 259 mm 380 10 mmW I d S × = × = = ×

(a) Reactions at A, B, and C . Choose the reaction force at B as the redundant; therefore, the released

beam is simply supported between A and C .

Consider downward deflection of simply supported beam at B due to uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load over a portion of the span.]


3 2 3( 2 ) (elastic curve)24

B

wxv L Lx x

EI = − − +

Values: w = 16 kN/m, L = 10 m, x = 4 m

Calculation:

3 2 3

33 2 3

( 2 )24

(16 kN/m)(4 m) 1,984 kN-m(10 m) 2(10 m)(4 m) (4 m)

24

B

wxv L Lx x

EI

EI EI

= − − +

⎡ ⎤= − − + = −⎣ ⎦

Consider upward deflection of simply supported beam at B due to concentrated load R B.



2 2 2( )6

B

Pabv L a b

LEI = − − −

Values:

P = − R B, L = 10 m, a = 4 m, b = 6 m

Calculation:

2 2 2

32 2 2

( )6

( )(4 m)(6 m) (19.2 m )(10 m) (4 m) (6 m)

6(10 m)

B

B B

Pabv L a b

LEI

R R

EI EI

= − − −

−⎡ ⎤= − − − =⎣ ⎦





Compatibility equation for deflection at B:2 6 4 12 2 3 2

3 3

3 2 3

3

(200,000 N/mm )(49.1 10 mm ) 9.82 10 N-mm 9.82 10 kN-m

1,984 kN-m (19.2 m )0.015 m

( 0.015 m)(9.82 10 kN-m ) 1,984 kN-m95.6615 kN 95.7 kN

19.2 m

B

B

EI

R

EI EI

R

= × = × = ×

− + = −

− × +∴ = = = ↑ Ans.


(4 m) (10 m) (16 kN/m)(10 m)(5 m) 0 A B C M R RΣ = + − =

(16 kN/m)(10 m)(5 m) (95.6615 kN)(4 m)

10 m

41.7354 kN 41.7 kN

C R −

∴ =

= = ↑ Ans.

(16 kN/m)(10 m) 0 y A B C F R R RΣ = + + − =

(16 kN/m)(10 m) 95.6615 kN 41.7354 kN

22.6031 kN 22.6 kN

A R∴ = − −

= = ↑ Ans.



W250 × 32.7 beam:


M max = 54.4327 kN-m

Bending stress at maximum moment2

6 4

(54.4327 kN-m)(259 mm/2)(1,000)

49.1 10 mm

143.6 MPa

xσ =×

= Ans.


modulus of the W250 × 32.7 beam:2

3 3

(54.4327 kN-m)(1,000)

380 10 mm

143.2 MPa

xσ = ×

= Ans.





11.53 A W10 × 22 structural steel [ E =29,000 ksi] wide-flange shape is loaded

and supported as shown in Fig. P11.53.

The beam is supported at C by a timber [ E = 1,800 ksi] post having a cross-

sectional area of 16 in.2. After a

concentrated load of 10 kips is applied to

the beam, determine:

(a) the reactions at supports A and C. (b) the maximum bending stress in the

beam.(c) the deflection of the beam at C .

Fig. P11.53

Solution


beam beam

3

beam

1 1

W10 22: 118 in. 10.2 in. 29,000 ksi

23.2 in.

Post (1): 16 in. 1,800 ksi

I d E

S

A E

× = = =

=

= =

(a) Reactions at supports A and C.

The reaction force from post (1) will be

taken as the redundant, leaving a

cantilever beam as the released beam.

To be consistent with earlier signconventions (e.g., Chapter 5), we will

assume that the force in the axial

member is tension (even thoughintuitively we recognize that the post

must be in compression).Beam free-body diagram

Downward deflection of W10 × 22 beam at C due to 10-kip concentrated load.


Relevant equations from Appendix C:3 2

and3 2

B B

PL PLv

EI EI θ = − = −

Values: P = 10 kips, L = 14 ft, EI = 3,422,000 kip-in.

2

Calculation:3 3 3

2

2 2 2

2

(10 kips)(14 ft) (12 in./ft)4.618773 in.

3 3(3,422,000 kip-in. )

(10 kips)(14 ft) (12 in./ft)0.0412390 rad

2 2(3,422,000 kip-in. )

4.618773 in. (0.0412390 rad)(6 ft)(12 in./f

B

B

C

PLv

EI

PL

EI

v

θ

= − = − = −

= − = − = −

= − − t) 7.587984 in.= −





Since we are assuming that a tension force exists in post (1), the tension force from the post will

cause a downward deflection of W10 × 22 beam at C .



3 B

PLv

EI = −

Values: P = F 1, L = 20 ft, EI = 3,422,000 kip-in.

2

Calculation:3 3 3

112

( )(20 ft) (12 in./ft)(1.346581 in./kip)

3 3(3,422,000 kip-in. ) B

PL F v F

EI = − = − = −

Axial elongation of post (1) due to force F 1.

1 11 1 12

1 1

(12 ft)(12 in./ft)(0.005 in./kip)

(16 in. )(1,800 ksi)

F Le F F

A E = = =

The deflection of the steel beam will not equal zero at C in this instance because the post deforms. If we

are consistent and assume that there is tension in the post, then the steel beam must deflect upward at C. Therefore,

1 1(0.005 in./kip)C v e F = =

Compatibility equation at C :

The sum of the downward deflection caused by the 10-kip concentrated load and the upward deflection

caused by tension in post (1) must equal the elongation of the post.

1

1

7.587984 in. (1.346581 in./kip) (0.005 in./kip)

7.587984 in.5.614154 kips 5.61 kips (C)

1.351582 in./kip C

F F

F R

− − =

∴ = = − = =−

Ans.

Equilibrium equations for entire beam:1(20 ft) (10 kips)(14 ft) 0 A A M M F Σ = − − − =

( 5.614154 kips)(20 ft) (10 kips)(14 ft) 27.7169 kN-m 27.7 kN-m (ccw) A M ∴ = − − − = − = Ans.

1 (10 kips) 0 y A F R F Σ = − − =

10 kips ( 5.614154 kips) 4.3858 kips 4.39 kips A

R∴ = + − = = ↑ Ans.






beam:


M max = 33.685 kip-ft = 404.218 kip-in. (at B)


4

(404.218 kip-in.)(10.2 in./2)

118 in.

17.47 ksi

xσ =

= Ans.


modulus of the W10 × 22 beam:

3

404.218 kip-in.17.42 ksi

23.2 in. xσ = = Ans.

(c) Beam deflection at C .

The beam deflection at C is equal to the elongation of the post:

1 11 2

1 1

( 5.614154 kips)(12 ft)(12 in./ft)0.0281 in. 0.0281 in.

(16 in. )(1,800 ksi) B

F Lv e

A E

−= = = = − = ↓ Ans.





11.54 A timber [ E = 12 GPa] beam is loadedand supported as shown in Fig. P11.54. The

cross section of the timber beam is 100-mm.

wide and 300-mm deep. The beam issupported at B by a 12-mm-diameter steel [ E

= 200 GPa] rod, which has no load before the

distributed load is applied to the beam. Aftera distributed load of 7 kN/m is applied to the

beam, determine:(a) the force carried by the steel rod.

(b) the maximum bending stress in the timber beam.


Fig. P11.54

Solution


6 4

beam beam

2 2

1 1

(100 mm)(300 mm)Beam: 225 10 mm 12,000 MPa

12Rod (1): (12 mm) 113.097336 mm 200,000 MPa

4

I E

A E π

= = × =

= = =

(a) Force carried by the steel rod.


taken as the redundant, leaving a simplysupported beam between A and C as the

released beam.

For this analysis, a tension force is assumed

to exist in axial member (1).

Downward deflection of wood beam at B due to 7 kN/m uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load.]


3 2 3( 2 ) (elastic curve)24

B

wxv L Lx x

EI = − − +

Values: w = 7 kN/m, L = 6 m, x = 4 m, EI = 2,700 kN-m

2

Calculation:

3 2 3

3 2 3 3

2

( 2 )24

(7 kN/m)(4 m)(6 m) 2(6 m)(4 m) (4 m) 38.02469 10 m

24(2,700 kN-m )

B

wxv L Lx x

EI

−

= − − +

⎡ ⎤= − − + = − ×⎣ ⎦








2 2 2( )6

B

Pabv L a b

LEI = − − −

Values: P = − F 1, L = 6 m, a = 4 m, b = 2 m,

EI = 2,700 kN-m2

Calculation:

2 2 2

2 2 2 3112

( )6

( )(4 m)(2 m)(6 m) (4 m) (2 m) (1.316872 10 m/kN)

6(6 m)(2,700 kN-m )

B

Pabv L a b

LEI

F F −

= − − −

−⎡ ⎤= − − − = ×⎣ ⎦

Elongation of steel rod (1) due to force F 1.

61 11 1 12 2

1 1

(5 m)(221.0485 10 m/kN)

(113.097336 mm )(200,000 N/mm )(1 kN/1,000 N)

F Le F F

A E

−= = = ×

The deflection of the wood beam at B will not equal zero in this instance because the rod that supportsthe beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore,

6

1 1(221.0485 10 m/kN) Bv e F −= − = − ×


The sum of the downward deflection caused by the uniformly distributed load and the upward deflection

caused by the force in rod (1) must equal the elongation of the steel rod. Elongation of the steel rod will

produce a downward (i.e., negative) deflection of the wood beam at B.

3 3 6

1 1

3

1 3

38.02469 10 m (1.316872 10 m/kN) (221.0485 10 m/kN)38.02469 10 m

24.72474 kN 24.7 kN (T)1.537921 10 m/kN

F F

F

− − −

−

−

− × + × = − ××

∴ = = =×

Ans.






wood beam:


M max = 11.6270 kN-m


2

6 4

(11.6270 kN-m)(300 mm/2)(1,000)

225 10 mm

7.75 MPa

xσ =×

= Ans.


The beam deflection at B is equal to the elongation of the steel rod:

1 11

1 1

2 2

(24.72474 kN)(5 m)(1,000 N/kN)(1,000 mm/m)

(113.097336 mm )(200,000 N/mm )

5.465367 mm 5.47 mm

B

F Lv e

A E = − = −

= −

= − = ↓ Ans.






supported as shown in Fig. P11.55. The

beam is supported at B by a timber [ E = 12

GPa] post having a cross-sectional area o20,000 mm

2. After a uniformly distributed

load of 50 kN/m is applied to the beam,

determine:

(a) the reactions at supports A, B, and C. (b) the maximum bending stress in the

beam.(c) the deflection of the beam at B.

Fig. P11.55

Solution

Section properties:6 4

beam beam

3 3

beam

21 1

W360 × 72: 201 10 mm 351 mm 200,000 MPa

1,150 10 mm

Post (1): 20,000 mm 12,000 MPa

I d E

S

A E

= × = =

= ×

= =

(a) Reactions at supports A, B, and C .

The reaction force from post (1) will betaken as the redundant, leaving a simply

supported beam as the released beam. To

be consistent with earlier sign conventions(e.g., Chapter 5), we will assume that the

force in the axial member is tension (even

though intuitively we recognize that the post must be in compression).

Downward deflection of W360 × 72 beam at B due to 50 kN/m uniformly distributed load.

[Appendix C, SS beam with uniformly distributed load over a portion of the span.]


2 2(4 7 3 )24

B

wav L aL a

LEI = − − +

Values:

P = 50 kN/m, L = 13 m, a = 6 m, EI = 40,200 kN-m

2

Calculation:3

2 2

32 2 3

2

(4 7 3 )24

(50 kN/m)(6 m)4(13 m) 7(6 m)(13 m) 3(6 m) 204.937 10 m

24(13 m)(40,200 kN-m )

B

wav L aL a

LEI

−

= − − +

⎡ ⎤= − − + = − ×⎣ ⎦





Since we are assuming that a tension force exists in post (1), the tension force from the post will

cause a downward deflection of W360 × 72 beam at B.



2 2 2( )6

B

Pabv L a b

LEI = − − −

Values: P = F

1, L = 13 m, a = 6 m, b = 7 m,

EI = 40,200 kN-m2

Calculation:

2 2 2

2 2 2 3112

( )6

( )(6 m)(7 m)(13 m) (6 m) (7 m) (1.12514 10 m/kN)

6(13 m)(40,200 kN-m )

B

Pabv L a b

LEI

F F −

= − − −

⎡ ⎤= − − − = − ×⎣ ⎦

Elongation of wood post (1) due to force F 1.

61 11 1 12 2

1 1

(5 m) (20.8333 10 m/kN)(20,000 mm )(12,000 N/mm )(1 kN/1,000 N)

F Le F F A E

−= = = ×

The deflection of the W360 × 72 beam at B will not equal zero in this instance because the post deforms.If we are consistent and assume that there is tension in the post, then the steel beam must deflect upward

at B. Therefore, 6

1 1(20.8333 10 m/kN) Bv e F −= = ×

Compatibility equation at B:The sum of the downward deflection caused by the 50 kN/m uniformly distributed load and the

downward deflection caused by the force in post (1) must equal the elongation of the wood post.

3 3 6

1 1

3

1 3

204.937 10 m (1.12514 10 m/kN) (20.8333 10 m/kN)

204.937 10 m178.832 kN 178.8 kN (C)

1.14598 10 m/kN B

F F

F R

− − −

−

−

− × − × = ×

×∴ = = − = =

− × Ans.


1(6 m) (13 m) (50 kN/m)(6 m)(3 m) 0 A C M F RΣ = − + − =

(50 kN/m)(6 m)(3 m) ( 178.832 kN)(6 m)13.3071 kN 13.31 kN

13 mC R

+ −∴ = = − = ↓ Ans.





1 (50 kN/m)(6 m) 0 y A C F R F RΣ = − + − =

(50 kN/m)(6 m) ( 178.832 kN) ( 13.3071 kN) 134.475 kN 134.5 kN A

R∴ = + − − − = = ↑ Ans.


W360 × 72 beam:


M max = 180.836 kN-m


6 4

(180.836 kN-m)(351 mm/2)(1,000)

201 10 mm

157.9 MPa

xσ =×

= Ans.


modulus of the W360 × 72 beam:2

3 3

(180.836 kN-m)(1,000)

1,150 10 mm

157.2 MPa

xσ =×

= Ans.


The beam deflection at B is equal to the deformation of the wood post:

1 11 2 2

1 1

( 178.832 kN)(5 m)(1,000 N/kN)(1,000 mm/m)

(20,000 mm )(12,000 N/mm )

3.7257 mm 3.73 mm

B

F Lv e

A E

−= = =

= − = ↓ Ans.





11.56 A timber [ E = 1,800 ksi] beam is loadedand supported as shown in Fig. P11.56. The

cross section of the timber beam is 4-in. wide

and 8-in. deep. The beam is supported at B by a¾-in.-diameter aluminum [ E = 10,000 ksi] rod,

which has no load before the distributed load is

applied to the beam. After a distributed load o

800 lb/ft is applied to the beam, determine:

(a) the force carried by the aluminum rod.(b) the maximum bending stress in the timber

beam.(c) the deflection of the beam at B.

Fig. P11.56

Solution


4

beam beam

2 2

1 1

(4 in.)(8 in.)Beam: 170.6667 in. 1,800,000 psi

12

Rod (1): (0.75 in.) 0.441786 in. 10,000,000 psi4

I E

A E π

= = =

= = =

(a) Force carried by the aluminum rod.

The reaction force from rod (1) will betaken as the redundant, leaving a cantilever

beam between A and C as the released

beam.



Downward deflection of wood beam at B due to 800 lb/ft uniformly distributed load.

[Appendix C, Cantilever beam with uniformly distributed load.]


2 2(6 4 ) (elastic curve)24

B

wxv L Lx x

EI = − − +

Values: w = 800 lb/ft, L = 16 ft, x = 12 ft,

EI = 307.2×106 lb-in.

2

Calculation:2

2 2

2 32 2

6 2

(6 4 )24

(800 lb/ft)(12 ft) (12 in./ft)6(16 ft) 4(16 ft)(12 ft) (12 ft) 24.624 in.

24(307.2 10 lb-in. )

B

wxv L Lx x

EI = − − +

⎡ ⎤= − − + = −⎣ ⎦×






[Appendix C, SS beam with concentrated load at tip.]


3 B

PLv

EI = −

Values:

P = − F 1, L = 12 ft, EI = 307.2×106 lb-in.

2

Calculation:3

3 331

16 2

3

( )(12 ft) (12 in./ft)(3.24 10 in./lb)

3(307.2 10 lb-in. )

B

PLv

EI

F F −

= −

−= − = ×

×


61 11 1 12 6

1 1

(14 ft)(12 in./ft)(38.02746 10 in./lb)

(0.441786 in. )(10 10 psi)

F Le F F

A E

−= = = ××

The deflection of the wood beam at B will not equal zero in this instance because the rod that supports

the beam at B will elongate, thus permitting the wood beam to deflect downward. Therefore, 6

1 1(38.02746 10 in./lb) Bv e F −= − = − ×



caused by the force in rod (1) must equal the elongation of the aluminum rod. Elongation of thealuminum rod will produce a downward (i.e., negative) deflection of the wood beam at B.

3 6

1 1

1 3

24.624 in. (3.24 10 in./lb) (38.02746 10 in./lb)24.624 in.

7,511.835 lb 7,510 lb (T)3.278027 10 in./lb

F F

F

− −

−

− + × = − ×

∴ = = =×

Ans.






wood beam:


M max = 12,258 lb-ft


4

(12,258 lb-ft)(8 in./2)(12 in./ft)

170.6667 in.

3,447.56 psi 3,450 psi

xσ =

= = Ans.


The beam deflection at B is equal to the elongation of the aluminum rod:

1 11

1 1

2 6

(7,511.835 lb)(14 ft)(12 in./ft)

(0.441786 in. )(10 10 psi)

0.28566 in. 0.286 in.

B

F Lv e

A E = − = −

= −×

= − = ↓ Ans.






supported as shown in Fig. P11.57. A

uniformly distributed load of 70 kN/m isapplied to the beam, causing the roller support

at B to settle downward (i.e., displace

downward) by 10 mm. Determine:

(a) the reactions at supports A and B.

(b) the maximum bending stress in the beam. Fig. P11.57

Solution

Section properties:6 4 3 3W530 66: 351 10 mm 526 mm 1,340 10 mm I d S × = × = = ×

(a) Reactions at supports A and B.

The reaction force at B will be taken as the redundant, leaving a cantilever beam between A and C as thereleased beam.

Downward deflection of beam at B due to 70 kN/m uniformly distributed load.[Appendix C, Cantilever beam with uniformly distributed load.]


2 2(6 4 ) (elastic curve)24

B

wxv L Lx x

EI = − − +

Values:

w = 70 kN/m, L = 6 m, x = 4.5 m,

EI = 70,200 kN-m2

Calculation:2

2 2

22 2

2

(6 4 )24

(70 kN/m)(4.5 m)6(6 m) 4(6 m)(4.5 m) (4.5 m) 0.107903 m

24(70,200 kN-m )

B

wx

v L Lx x EI = − − +

⎡ ⎤= − − + = −⎣ ⎦

Upward deflection of beam at B due to reaction force R B.

[Appendix C, SS beam with concentrated load at tip.]


3 B

PLv

EI = −

Values: P = − R B, L = 4.5 m, EI = 70,200 kN-m2

Calculation:3

36

2

3

( )(4.5 m)(432.6923 10 m/kN)

3(70,200 kN-m )

B

B B

PLv

EI

R R−

= −

−= − = ×





caused by the reaction force R B must equal the support settlement:

6

3

6

0.107903 m (432.6923 10 m/kN) 0.010 m

97.90264 10 m226.2639 kN 226 kN

432.6923 10 m/kN

B

B

R

R

−

−

−

− + × = −

×∴ = = = ↑

×

Ans.


(4.5 m) (70 kN/m)(6 m)(3 m) 0 A A B M M RΣ = − + − =

(226.2639 kN)(4.5 m) (70 kN/m)(6 m)(3 m) 241.8125 kN-m 242 kN-m (ccw) A M ∴ = − = − = Ans.

(70 kN/m)(6 m) 0 y A B F R RΣ = + − =

(70 kN/m)(6 m) 226.2639 kN 193.7361 kN 193.7 kN A R∴ = − = = ↑ Ans.


beam:


M max = 241.8125 kN-m (at A)


6 4

(241.8125 kN-m)(526 mm/2)(1,000)

351 10 mm

181.187 MPa 181.2 MPa

xσ =×

= = Ans.


modulus:2

3 3

(241.8125 kN-m)(1,000)

1,340 10 mm

180.457 MPa 180.5 MPa

xσ =×

= = Ans.

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Mechanics of Materials Solutions Chapter11 Probs47 57