Mechanics of Materials Solutions Chapter11 Probs47 57

11.47 A W530 × 92 structural steel wideflange shape [E = 200 GPa; I = 554 × 106 mm4] is loaded and supported as shown in Fig. P11.47. 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. P11.47

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: Pb( L2 − b 2 ) Pa ( L2 − a 2 ) θA = − and θC = 6 LEI 6 LEI Values: P = 150 kN, L = 10 m, a = 6 m, b = 4 m Calculation:

θA = − θC =

Pb( L2 − b 2 ) (150 kN)(4 m) 840 kN-m 2 ⎡⎣ (10 m) 2 − (4 m) 2 ⎤⎦ = − =− 6 LEI 6(10 m)EI EI

Pa( L2 − a 2 ) (150 kN)(6 m) 960 kN-m 2 ⎡⎣(10 m) 2 − (6 m) 2 ⎤⎦ = = 6 LEI 6(10 m)EI EI

Determine the slopes at A and C caused by moment reaction MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML θA = − and θC = 3EI 6 EI Values: M = MA, L = 10 m

Calculation: ML M (10 m) (3.333333 m)M A θA = − =− A = − 3EI 3EI EI

θC =

ML M A (10 m) (1.666667 m)M A = = 6 EI 6 EI EI

Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Determine the slopes at A and C caused by moment reaction MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML θA = − and θC = 6 EI 3EI Values: M = MC, L = 10 m

Calculation: ML M (10 m) (1.666667 m)M C θA = − =− C = − 6 EI 6 EI EI

θC =

ML M C (10 m) (3.333333 m)M C = = 3EI 3EI EI

Compatibility equation for slope at A: 840 kN-m 2 (3.333333 m)M A (1.666667 m)M C − − − =0 EI EI EI Compatibility equation for slope at C: 960 kN-m 2 (1.666667 m)M A (3.333333 m)M C + + =0 EI EI EI

(a)

(b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as: (3.333333 m)M A + (1.666667 m)M C = −840 kN-m 2 (1.666667 m)M A + (3.333333 m)M C = −960 kN-m 2 and solved simultaneously for MA and MC: M A = −144 kN-m = 144 kN-m (ccw)

M C = −216 kN-m = 216 kN-m (cw)

Ans. Ans.

Equilibrium equations for entire beam: ΣM A = − M A + M C − (150 kN)(6 m) + RC (10 m) = 0 (150 kN)(6 m) + (−144 kN-m) − (−216 kN-m) ∴ RC = 10 m = 97.2 kN = 97.2 kN ↑ ΣFy = RA + RC − 150 kN = 0

Ans.

∴ RA = 150 kN − 97.2 kN = 52.8 kN = 52.8 kN ↑

Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 554 × 106 mm 4

d = 533 mm S = 2,080 × 103 mm3 Maximum bending moment magnitude Mmax = 216 kN-m (at C) Bending stresses at maximum moment (216 kN-m)(533 mm/2)(1,000) 2 σx = 554 × 106 mm 4 Ans. = 103.9 MPa or using the tabulated value for the section modulus: (216 kN-m)(1,000) 2 σx = 2,080 × 103 mm3

= 103.8 MPa

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: Pab 2 vB = − ( L − a2 − b2 ) 6 LEI Values: P = 150 kN, L = 10 m, a = 6 m, b=4m

Calculation: vB = −

Pab 2 (150 kN)(6 m)(4 m) 2,880 kN-m3 ⎡⎣(10 m) 2 − (6 m) 2 − (4 m) 2 ⎤⎦ = − ( L − a 2 − b2 ) = − 6 LEI 6(10 m)EI EI

Determine the deflection at B caused by MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI Values: M = −144 kN-m, L = 10 m, x = 6 m


Calculation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−144 kN-m)(6 m) 806.4 kN-m3 2 2 ⎡⎣ 2(10 m) − 3(10 m)(6 m) + (6 m) ⎤⎦ = =− 6(10 m)EI EI

Determine the deflection at B caused by MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI Values: M = −216 kN-m, L = 10 m, x = 4 m

Calculation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI (−216 kN-m)(4 m) 1,382.4 kN-m3 2 2 ⎡⎣ 2(10 m) − 3(10 m)(4 m) + (4 m) ⎤⎦ = =− 6(10 m)EI EI

Beam deflection vB: 2,880 kN-m3 806.4 kN-m3 1,382.4 kN-m3 + + vB = − EI EI EI 3 3 691.2 kN-m 691.2 kN-m =− =− = −0.006238 m = 6.24 mm ↓ 110,800 kN-m 2 EI

Ans.


11.48 A W530 × 92 structural steel wideflange shape [E = 200 GPa; I = 554 × 106 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 a simply 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. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equations from Appendix C: wa 2 θA = − (2 L − a) 2 and 24 LEI

θC =

wa 2 (2 L2 − a 2 ) 24 LEI

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

θA = − θC =

wa 2 (80 kN/m)(4.5 m) 2 1,366.875 kN-m 2 2 (2 L − a) 2 = − 2(9 m) (4.5 m) − = − [ ] 24 LEI 24(9 m)EI EI

wa 2 (80 kN/m)(4.5 m) 2 1,063.125 kN-m 2 ⎡⎣ 2(9 m) 2 − (4.5 m) 2 ⎤⎦ = (2 L2 − a 2 ) = 24 LEI 24(9 m)EI EI

Determine the slopes at A and C caused by moment reaction MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML θA = − and θC = 3EI 6 EI Values: M = MA, L = 9 m

Calculation: ML M (9 m) (3 m)M A θA = − =− A = − 3EI 3EI EI

θC =

ML M A (9 m) (1.5 m)M A = = 6 EI 6 EI EI


Determine the slopes at A and C caused by moment reaction MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equations from Appendix C: ML ML θA = − and θC = 6 EI 3EI Values: M = MC, L = 9 m

Calculation: ML M (9 m) (1.5 m)M C θA = − =− C = − 6 EI 6 EI EI

θC =

ML M C (9 m) (3 m)M C = = 3EI 3EI EI

Compatibility equation for slope at A: 1,366.875 kN-m 2 (3 m)M A (1.5 m)M C − − − =0 EI EI EI Compatibility equation for slope at C: 1,063.125 kN-m 2 (1.5 m)M A (3 m)M C + + =0 EI EI EI

(a)

(b)

Solve Equations (a) and (b). Equations (a) and (b) can be rewritten as: (3 m)M A + (1.5 m)M C = −1,366.875 kN-m 2 (1.5 m)M A + (3 m)M C = −1,063.125 kN-m 2 and solved simultaneously for MA and MC: M A = −371.25 kN-m = 371.25 kN-m (ccw)

M C = −168.75 kN-m = 168.8 kN-m (cw)

Ans. Ans.

Equilibrium equations for entire beam: ΣM A = − M A + M C − (80 kN/m)(4.5 m)(2.25 m) + RC (9 m) = 0 (80 kN/m)(4.5 m)(2.25 m) + (−371.25 kN-m) − (−168.75 kN-m) ∴ RC = 9m = 67.5 kN = 67.5 kN ↑

Ans.

ΣFy = RA + RC − (80 kN/m)(4.5 m) = 0

∴ RA = (80 kN/m)(4.5 m) − 67.5 kN = 292.5 kN = 293 kN ↑

Ans.


Shear-force and bending-moment diagrams (b) Magnitude of maximum bending stress: Section properties (from Appendix B): I = 554 × 106 mm 4 d = 533 mm

S = 2,080 × 103 mm3 Maximum bending moment magnitude Mmax = 371.25 kN-m (at A) Bending stresses at maximum moment (371.25 kN-m)(533 mm/2)(1,000) 2 σx = 554 × 106 mm 4 = 178.6 MPa Ans. or using the tabulated value for the section modulus: (371.25 kN-m)(1,000) 2 σx = 2,080 × 103 mm3

= 178.5 MPa

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: wa 3 vB = − (4 L2 − 7aL + 3a 2 ) 24 LEI Values: w = 80 kN/m, L = 9 m, a = 4.5 m

Calculation: wa 3 vB = − (4 L2 − 7 aL + 3a 2 ) 24 LEI (80 kN/m)(4.5 m)3 3, 417.1875 kN-m3 2 2 ⎡ ⎤ =− ⎣ 4(9 m) − 7(4.5 m)(9 m) + 3(4.5 m) ⎦ = − EI 24(9 m)EI


Determine the deflection at B caused by MA. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI Values: M = −371.25 kN-m, L = 9 m, x = 4.5 m

Calculation: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI =−

(−371.25 kN-m)(4.5 m) 1,879.4531 kN-m3 ⎡⎣ 2(9 m) 2 − 3(9 m)(4.5 m) + (4.5 m) 2 ⎤⎦ = EI 6(9 m)EI

Determine the deflection at B caused by MC. [Appendix C, SS beam with concentrated moment at one end.] Relevant equation from Appendix C: Mx vB = − (2 L2 − 3Lx + x 2 ) 6 LEI Values: M = −168.75 kN-m, L = 9 m, x = 4.5 m

Calculation: Mx (2 L2 − 3Lx + x 2 ) vB = − 6 LEI =−

(−168.75 kN-m)(4.5 m) 854.2969 kN-m3 ⎡⎣ 2(9 m) 2 − 3(9 m)(4.5 m) + (4.5 m) 2 ⎤⎦ = 6(9 m)EI EI

Beam deflection vB: 3, 417.1875 kN-m3 1,879.4531 kN-m3 854.2969 kN-m3 vB = − + + EI EI EI 3 3 683.4375 kN-m 683.4375 kN-m =− =− = −0.006168 m = 6.17 mm ↓ 110,800 kN-m 2 EI

Ans.


11.49 A timber [E = 1,800 ksi] beam is loaded 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 is supported 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 the timber beam. (c) the deflection of the beam at B.

Fig. P11.49

Solution Section properties:

Beam:

I beam =

Rod (1):

A1 =

π 4

(4 in.)(8 in.)3 = 170.6667 in.4 12

Ebeam = 1,800,000 psi

(0.50 in.) 2 = 0.1963495 in.2

E1 = 30,000,000 psi

(a) Force carried by the steel rod. The reaction force from rod (1) will be taken as the redundant, leaving a simply supported 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: 5wL4 vB = − 384 EI Values: w = 900 lb/ft, L = 14 ft, EI = 307.2 ×106 lb-in.2

Calculation: 5wL4 5(900 lb/ft)(14 ft)4 (12 in./ft)3 vB = − =− = −2.532305 in. 384 EI 384(307.2 × 106 lb-in.2 )


Upward deflection of wood beam at B due to force F1 in rod (1). [Appendix C, SS beam with concentrated load at midspan.] Relevant equation from Appendix C: PL3 vB = − 48 EI Values: P = −F1, L = 14 ft, EI = 307.2 ×106 lb-in.2

Calculation: PL3 (− F1 )(14 ft)3 (12 in./ft)3 vB = − =− = (321.5625 × 10−6 in./lb)F1 6 2 48EI 48(307.2 × 10 lb-in. ) Elongation of steel rod (1) due to force F1. FL (16 ft)(12 in./ft) e1 = 1 1 = F1 = (32.59493 × 10−6 in./lb)F1 2 6 A1E1 (0.1963495 in. )(30 × 10 psi) 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,

vB = −e1 = −(32.59493 × 10−6 in./lb)F1 Compatibility equation at B: 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.

−2.532305 in. + (321.5625 × 10−6 in./lb)F1 = −(32.59493 × 10−6 in./lb)F1 ∴ F1 =

2.532305 in. = 7,150.22 lb = 7,150 lb (T) 354.1574 × 10−6 in./lb

Ans.


(b) Determine maximum bending stress in wood beam: Maximum bending moment magnitude Mmax = 4,125 lb-ft = 49,500 lb-in. Bending stress at maximum moment (49,500 lb-in.)(8 in./2) σx = = 1,160 psi 170.6667 in.4

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the steel rod: FL (7,150.22 lb)(16 ft)(12 in./ft) vB = −e1 = − 1 1 = − = −0.233 in. = 0.233 in. ↓ A1E1 (0.1963495 in.2 )(30 × 106 psi)

Ans.

Ans.


11.50 A W360 × 72 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Fig. P11.50. The beam is supported at B by 20-mm-diameter solid aluminum [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: W360 × 72: I beam = 201 × 106 mm 4

d = 351 mm

Ebeam = 200,000 MPa

S beam = 1,150 × 103 mm3 Rod (1):

A1 =

π 4

(20 mm) 2 = 314.159266 mm 2

E1 = 70,000 MPa

(a) Force in the aluminum rod. The reaction force from rod (1) will be taken as the redundant, leaving a cantilever beam as the released beam.

For this analysis, a tension force is assumed to exist in axial member (1).

Downward deflection of W360 × 72 beam at B due to 40-kN concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: Px 2 vB = − (3L − x) (elastic curve) 6 EI Values: P = 40 kN, L = 5 m, x = 3.6 m, EI = 40,200 kN-m2

Calculation: Px 2 (40 kN)(3.6 m) 2 vB = − (3L − x) = − [3(5 m) − (3.6 m)] = −24.50149 × 10−3 m 6 EI 6(40, 200 kN-m 2 )


Upward deflection of W360 × 72 beam at B due to force F1 in rod (1). [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −F1, L = 3.6 m, EI = 40,200 kN-m2

Calculation: PL3 (− F1 )(3.6 m)3 vB = − =− = (386.8657 × 10−6 m/kN)F1 3EI 3(40, 200 kN-m 2 ) Elongation of aluminum rod (1) due to force F1. FL (3 m) e1 = 1 1 = F1 = (136.4185 × 10−6 m/kN)F1 2 2 A1E1 (314.159266 mm )(70,000 N/mm )(1 kN/1,000 N) 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,

vB = −e1 = −(136.4185 × 10−6 m/kN)F1 Compatibility equation at B: 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. −24.50149 × 10−3 m + (386.8657 × 10−6 m/kN)F1 = −(136.4185 × 10−6 m/kN)F1 ∴ F1 =

24.50149 × 10−3 m = 46.82254 kN = 46.8 kN (T) 523.2842 × 10−6 m/kN

Ans.


(b) Determine maximum bending stress in the W360 × 72 beam: Maximum bending moment magnitude Mmax = 56 kN-m (at B) Bending stress at maximum moment (56 kN-m)(351 mm/2)(1,000) 2 σx = 201 × 106 mm 4 = 48.9 MPa Ans. or using the tabulated value for the section modulus of the W360 × 72 beam: (56 kN-m)(1,000) 2 σx = = 48.7 MPa Ans. 1,150 × 103 mm3

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the aluminum rod: FL (46,822.54 N)(3,000 mm) vB = −e1 = − 1 1 = − = −6.38746 mm = 6.39 mm ↓ A1E1 (314.159266 mm 2 )(70,000 N/mm 2 )

Ans.


11.51 A W18 × 55 structural steel [E = 29,000 ksi] 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 = 10,000 ksi] 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 Section properties: Beam: I beam = 890 in.4

d = 18.1 in.

Ebeam = 29,000 ksi

3

S beam = 98.3 in. Rod (1):

A1 =

π 4

(0.75 in.) 2 = 0.4417865 in.2

E1 = 10,000 ksi

(a) Force in the aluminum rod. The reaction force from rod (1) will be taken as the redundant, leaving a cantilever beam as the released beam.

For this analysis, a tension force is assumed to exist in axial member (1). 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: wL4 wL3 vB = − and θ B = − 8EI 6 EI Values: w = 4 kips/ft, L = 11 ft, EI = 25.81×106 kip-in.2

Calculation: wL4 (4 kips/ft)(11 ft)4 (12 in./ft)3 vB = − =− = −0.490113 in. 8EI 8(25.81 × 106 kip-in.2 )

θB = −

wL3 (4 kips/ft)(11 ft)3 (12 in./ft) 2 =− = −4.950639 × 10−3 rad 6 2 6 EI 6(25.81 × 10 kip-in. )

vC = −0.490113 in. + (5 ft)(12 in./ft)( − 4.950639 × 10−3 rad) = −0.787152 in.


Upward deflection of W18 × 55 beam at C due to force F1 in rod (1). [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vC = − 3EI Values: P = −F1, L = 16 ft, EI = 25.81×106 kip-in.2

Calculation: PL3 (− F1 )(16 ft)3 (12 in./ft)3 vC = − =− = (91.41015 × 10−3 in./kip)F1 6 2 3EI 3(25.81 × 10 kip-in. ) Elongation of aluminum rod (1) due to force F1. FL (14 ft)(12 in./ft) e1 = 1 1 = F1 = (38.02742 × 10−3 in./kip)F1 A1E1 (0.4417865 in.2 )(10,000 ksi) 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,

vC = −e1 = −(38.02742 × 10−3 in./kip)F1 Compatibility equation at C: 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 aluminum rod. Elongation of the aluminum rod will produce a downward (i.e., negative) deflection of the W18 × 55 beam at C.

−0.787152 in. + (91.41015 × 10−3 in./kip)F1 = −(38.02742 × 10−3 in./kip)F1 ∴ F1 =

0.787152 in. = 6.081323 kips = 6.08 kips (T) 129.4376 × 10−3 in./kip

Ans.


(b) Determine maximum bending stress in W18 × 55 beam: Maximum bending moment magnitude Mmax = 144.70 kip-ft (at A) Bending stress at maximum moment (144.70 kip-ft)(18.1 in./2)(12 in./ft) σx = 890 in.4 = 17.66 ksi Ans. or using the tabulated value for the section modulus of the W18 × 55 beam: (144.70 kip-ft)(12 in./ft) σx = 98.3 in.3

= 17.66 ksi

(c) Beam deflection at C. The beam deflection at C is equal to the elongation of the aluminum rod: FL (6.081323 kips)(14 ft)(12 in./ft) vC = −e1 = − 1 1 = − = −0.231257 in. = 0.231 in. ↓ A1E1 (0.4417865 in.2 )(10,000 ksi)

Ans.

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 is applied 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: W 250 × 32.7 :

I = 49.1 × 106 mm 4

d = 259 mm

S = 380 × 103 mm3

(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.] Relevant equation from Appendix C: wx vB = − ( L3 − 2 Lx 2 + x 3 ) (elastic curve) 24 EI Values: w = 16 kN/m, L = 10 m, x = 4 m Calculation: wx vB = − ( L3 − 2 Lx 2 + x 3 ) 24 EI =−

(16 kN/m)(4 m) 1,984 kN-m3 ⎡⎣ (10 m)3 − 2(10 m)(4 m) 2 + (4 m)3 ⎤⎦ = − EI 24 EI

Consider upward deflection of simply supported beam at B due to concentrated load RB. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a2 − b2 ) 6 LEI Values: P = −RB, L = 10 m, a = 4 m, b = 6 m

Calculation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI (− RB )(4 m)(6 m) (19.2 m3 ) RB 2 2 2 ⎡ ⎤ =− ⎣(10 m) − (4 m) − (6 m) ⎦ = EI 6(10 m) EI Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Compatibility equation for deflection at B: EI = (200,000 N/mm 2 )(49.1 × 106 mm 4 ) = 9.82 × 1012 N-mm 2 = 9.82 × 103 kN-m 2 −

1,984 kN-m3 (19.2 m3 ) RB + = −0.015 m EI EI (−0.015 m)(9.82 × 103 kN-m 2 ) + 1,984 kN-m3 ∴ RB = = 95.6615 kN = 95.7 kN ↑ 19.2 m3

Ans.

Equilibrium equations for entire beam: ΣM A = RB (4 m) + RC (10 m) − (16 kN/m)(10 m)(5 m) = 0 (16 kN/m)(10 m)(5 m) − (95.6615 kN)(4 m) ∴ RC = 10 m = 41.7354 kN = 41.7 kN ↑

Ans.

ΣFy = RA + RB + RC − (16 kN/m)(10 m) = 0

∴ RA = (16 kN/m)(10 m) − 95.6615 kN − 41.7354 kN = 22.6031 kN = 22.6 kN ↑

Ans.

Shear-force and bending-moment diagrams (b) Determine maximum bending stress in W250 × 32.7 beam: Maximum bending moment magnitude Mmax = 54.4327 kN-m Bending stress at maximum moment (54.4327 kN-m)(259 mm/2)(1,000) 2 σx = 49.1 × 106 mm 4 = 143.6 MPa Ans. or using the tabulated value for the section modulus of the W250 × 32.7 beam: (54.4327 kN-m)(1,000) 2 σx = 380 × 103 mm3 = 143.2 MPa 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 crosssectional 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 Section properties: W10 × 22: I beam = 118 in.4

d = 10.2 in.

Ebeam = 29,000 ksi

3

Sbeam = 23.2 in. Post (1):

A1 = 16 in.

E1 = 1,800 ksi

(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 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). Beam free-body diagram Downward deflection of W10 × 22 beam at C due to 10-kip concentrated load. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equations from Appendix C: PL3 PL2 vB = − and θ B = − 3EI 2 EI Values: P = 10 kips, L = 14 ft, EI = 3,422,000 kip-in.2

Calculation: PL3 (10 kips)(14 ft)3 (12 in./ft)3 =− = −4.618773 in. vB = − 3EI 3(3, 422,000 kip-in.2 )

θB = −

PL2 (10 kips)(14 ft) 2 (12 in./ft) 2 =− = −0.0412390 rad 2 EI 2(3, 422,000 kip-in.2 )

vC = −4.618773 in. − (0.0412390 rad)(6 ft)(12 in./ft) = −7.587984 in. Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

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. [Appendix C, Cantilever beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = F1, L = 20 ft, EI = 3,422,000 kip-in.2

Calculation: PL3 ( F1 )(20 ft)3 (12 in./ft)3 vB = − =− = −(1.346581 in./kip)F1 3EI 3(3, 422,000 kip-in.2 ) Axial elongation of post (1) due to force F1. FL (12 ft)(12 in./ft) e1 = 1 1 = F1 = (0.005 in./kip)F1 A1E1 (16 in.2 )(1,800 ksi) 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, vC = e1 = (0.005 in./kip)F1 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. −7.587984 in. − (1.346581 in./kip)F1 = (0.005 in./kip)F ∴ F1 =

7.587984 in. = −5.614154 kips = 5.61 kips (C) = RC −1.351582 in./kip

Ans.

Equilibrium equations for entire beam: ΣM A = − M A − F1 (20 ft) − (10 kips)(14 ft) = 0

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

Ans.

ΣFy = RA − F1 − (10 kips) = 0

∴ RA = 10 kips + ( −5.614154 kips) = 4.3858 kips = 4.39 kips ↑

Ans.


(b) Determine maximum bending stress in the beam: Maximum bending moment magnitude Mmax = 33.685 kip-ft = 404.218 kip-in. (at B) Bending stress at maximum moment (404.218 kip-in.)(10.2 in./2) σx = 118 in.4 = 17.47 ksi Ans. or using the tabulated value for the section modulus of the W10 × 22 beam: 404.218 kip-in. Ans. σx = = 17.42 ksi 23.2 in.3

(c) Beam deflection at C. The beam deflection at C is equal to the elongation of the post: F L ( − 5.614154 kips)(12 ft)(12 in./ft) vB = e1 = 1 1 = = −0.0281 in. = 0.0281 in. ↓ A1E1 (16 in.2 )(1,800 ksi)

Ans.


11.54 A timber [E = 12 GPa] beam is loaded and supported as shown in Fig. P11.54. The cross section of the timber beam is 100-mm. wide and 300-mm deep. The beam is supported 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. After a 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. (c) the deflection of the beam at B.

Fig. P11.54


Beam:

I beam =

Rod (1):

A1 =

π 4

(100 mm)(300 mm)3 = 225 × 106 mm 4 12

Ebeam = 12,000 MPa

(12 mm) 2 = 113.097336 mm 2

E1 = 200,000 MPa

(a) Force carried by the steel rod. The reaction force from rod (1) will be taken as the redundant, leaving a simply supported 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.] Relevant equation from Appendix C: wx vB = − ( L3 − 2 Lx 2 + x3 ) (elastic curve) 24 EI Values: w = 7 kN/m, L = 6 m, x = 4 m, EI = 2,700 kN-m2

Calculation: wx vB = − ( L3 − 2 Lx 2 + x 3 ) 24 EI (7 kN/m)(4 m) ⎡(6 m)3 − 2(6 m)(4 m) 2 + (4 m)3 ⎤⎦ = −38.02469 × 10−3 m =− 24(2,700 kN-m 2 ) ⎣ Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Upward deflection of wood beam at B due to force F1 in rod (1). [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 vB = − ( L − a2 − b2 ) 6 LEI Values: P = −F1, L = 6 m, a = 4 m, b = 2 m, EI = 2,700 kN-m2

Calculation: Pab 2 ( L − a 2 − b2 ) vB = − 6 LEI (− F1 )(4 m)(2 m) ⎡(6 m) 2 − (4 m) 2 − (2 m) 2 ⎤⎦ = (1.316872 × 10−3 m/kN)F1 =− 2 ⎣ 6(6 m)(2,700 kN-m ) Elongation of steel rod (1) due to force F1. FL (5 m) e1 = 1 1 = F1 = (221.0485 × 10−6 m/kN)F1 2 2 A1E1 (113.097336 mm )(200,000 N/mm )(1 kN/1,000 N) 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,

vB = −e1 = −(221.0485 × 10−6 m/kN)F1 Compatibility equation at B: 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. −38.02469 × 10−3 m + (1.316872 × 10−3 m/kN)F1 = −(221.0485 × 10−6 m/kN)F1 ∴ F1 =

38.02469 × 10−3 m = 24.72474 kN = 24.7 kN (T) 1.537921 × 10−3 m/kN

Ans.


(b) Determine maximum bending stress in wood beam: Maximum bending moment magnitude Mmax = 11.6270 kN-m Bending stress at maximum moment (11.6270 kN-m)(300 mm/2)(1,000) 2 σx = 225 × 106 mm 4 = 7.75 MPa Ans.

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the steel rod: FL vB = −e1 = − 1 1 A1E1

=−

(24.72474 kN)(5 m)(1,000 N/kN)(1,000 mm/m) (113.097336 mm 2 )(200,000 N/mm 2 )

= −5.465367 mm = 5.47 mm ↓

Ans.


11.55 A W360 × 72 structural steel [E = 200 GPa] wide-flange shape is loaded and 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 of 20,000 mm2. 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: W360 × 72: I beam = 201 × 106 mm 4 3

Sbeam = 1,150 × 10 mm Post (1):

d = 351 mm

Ebeam = 200,000 MPa

3

A1 = 20,000 mm 2

E1 = 12,000 MPa

(a) Reactions at supports A, B, and C. The reaction force from post (1) will be taken 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.] Relevant equation from Appendix C: wa 3 (4 L2 − 7aL + 3a 2 ) vB = − 24 LEI Values: P = 50 kN/m, L = 13 m, a = 6 m, EI = 40,200 kN-m2

Calculation: wa 3 (4 L2 − 7aL + 3a 2 ) vB = − 24 LEI =−

(50 kN/m)(6 m)3 ⎡ 4(13 m) 2 − 7(6 m)(13 m) + 3(6 m) 2 ⎤⎦ = −204.937 × 10−3 m 2 ⎣ 24(13 m)(40, 200 kN-m )


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. [Appendix C, SS beam with concentrated load not at midspan.] Relevant equation from Appendix C: Pab 2 ( L − a2 − b2 ) vB = − 6 LEI Values: P = F1, L = 13 m, a = 6 m, b = 7 m, EI = 40,200 kN-m2

Calculation: Pab 2 vB = − ( L − a 2 − b2 ) 6 LEI ( F1 )(6 m)(7 m) ⎡(13 m) 2 − (6 m) 2 − (7 m) 2 ⎤⎦ = −(1.12514 × 10−3 m/kN)F1 =− 6(13 m)(40, 200 kN-m 2 ) ⎣ Elongation of wood post (1) due to force F1. FL (5 m) e1 = 1 1 = F1 = (20.8333 × 10−6 m/kN)F1 2 2 A1E1 (20,000 mm )(12,000 N/mm )(1 kN/1,000 N) 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,

vB = e1 = (20.8333 × 10−6 m/kN)F1 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. −204.937 × 10−3 m − (1.12514 × 10−3 m/kN)F1 = (20.8333 × 10−6 m/kN)F1 ∴ F1 =

204.937 × 10−3 m = −178.832 kN = 178.8 kN (C) = RB −1.14598 × 10−3 m/kN

Equilibrium equations for entire beam: ΣM A = − F1 (6 m) + RC (13 m) − (50 kN/m)(6 m)(3 m) = 0 (50 kN/m)(6 m)(3 m) + (−178.832 kN)(6 m) ∴ RC = = −13.3071 kN = 13.31 kN ↓ 13 m

Ans.

Ans.


ΣFy = RA − F1 + RC − (50 kN/m)(6 m) = 0

∴ RA = (50 kN/m)(6 m) + ( −178.832 kN) − (−13.3071 kN) = 134.475 kN = 134.5 kN ↑

Ans.

(b) Determine maximum bending stress in the W360 × 72 beam: Maximum bending moment magnitude Mmax = 180.836 kN-m Bending stress at maximum moment (180.836 kN-m)(351 mm/2)(1,000) 2 σx = 201 × 106 mm 4 = 157.9 MPa Ans. or using the tabulated value for the section modulus of the W360 × 72 beam: (180.836 kN-m)(1,000) 2 σx = 1,150 × 103 mm3

= 157.2 MPa

Ans.

(c) Beam deflection at B. The beam deflection at B is equal to the deformation of the wood post: F L ( − 178.832 kN)(5 m)(1,000 N/kN)(1,000 mm/m) vB = e1 = 1 1 = A1E1 (20,000 mm 2 )(12,000 N/mm 2 ) = −3.7257 mm = 3.73 mm ↓

Ans.


11.56 A timber [E = 1,800 ksi] beam is loaded and 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 a ¾-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 of 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


Beam:

I beam =

Rod (1):

A1 =

π 4

(4 in.)(8 in.)3 = 170.6667 in.4 12

Ebeam = 1,800,000 psi

(0.75 in.)2 = 0.441786 in.2

E1 = 10,000,000 psi

(a) Force carried by the aluminum rod. The reaction force from rod (1) will be taken as the redundant, leaving a cantilever 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 800 lb/ft uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = 800 lb/ft, L = 16 ft, x = 12 ft, EI = 307.2×106 lb-in.2

Calculation: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) 24 EI (800 lb/ft)(12 ft)2 (12 in./ft)3 ⎡6(16 ft) 2 − 4(16 ft)(12 ft) + (12 ft) 2 ⎤⎦ = −24.624 in. =− 24(307.2 × 106 lb-in.2 ) ⎣ Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Upward deflection of wood beam at B due to force F1 in rod (1). [Appendix C, SS beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −F1, L = 12 ft, EI = 307.2×106 lb-in.2

Calculation: PL3 vB = − 3EI (− F1 )(12 ft)3 (12 in./ft)3 =− = (3.24 × 10−3 in./lb)F1 6 2 3(307.2 × 10 lb-in. ) Elongation of aluminum rod (1) due to force F1. FL (14 ft)(12 in./ft) e1 = 1 1 = F1 = (38.02746 × 10−6 in./lb)F1 2 6 A1E1 (0.441786 in. )(10 × 10 psi) 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,

vB = −e1 = −(38.02746 × 10−6 in./lb)F1 Compatibility equation at B: 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 aluminum rod. Elongation of the aluminum rod will produce a downward (i.e., negative) deflection of the wood beam at B.

−24.624 in. + (3.24 × 10−3 in./lb)F1 = −(38.02746 × 10−6 in./lb)F1 ∴ F1 =

24.624 in. = 7,511.835 lb = 7,510 lb (T) 3.278027 × 10−3 in./lb

Ans.


(b) Determine maximum bending stress in wood beam: Maximum bending moment magnitude Mmax = 12,258 lb-ft Bending stress at maximum moment (12, 258 lb-ft)(8 in./2)(12 in./ft) σx = 170.6667 in.4 = 3, 447.56 psi = 3, 450 psi

Ans.

(c) Beam deflection at B. The beam deflection at B is equal to the elongation of the aluminum rod: FL vB = −e1 = − 1 1 A1E1

=−

(7,511.835 lb)(14 ft)(12 in./ft) (0.441786 in.2 )(10 × 106 psi)

= −0.28566 in. = 0.286 in. ↓

Ans.


11.57 A W530 × 66 structural steel [E = 200 GPa] wide-flange shape is loaded and supported as shown in Fig. P11.57. A uniformly distributed load of 70 kN/m is applied 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: W530 × 66: I = 351 × 106 mm 4

d = 526 mm

S = 1,340 × 103 mm3

(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 the released beam. Downward deflection of beam at B due to 70 kN/m uniformly distributed load. [Appendix C, Cantilever beam with uniformly distributed load.] Relevant equation from Appendix C: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) (elastic curve) 24 EI Values: w = 70 kN/m, L = 6 m, x = 4.5 m, EI = 70,200 kN-m2

Calculation: wx 2 vB = − (6 L2 − 4 Lx + x 2 ) 24 EI (70 kN/m)(4.5 m)2 ⎡6(6 m) 2 − 4(6 m)(4.5 m) + (4.5 m) 2 ⎤⎦ = −0.107903 m =− 24(70, 200 kN-m 2 ) ⎣ Upward deflection of beam at B due to reaction force RB. [Appendix C, SS beam with concentrated load at tip.] Relevant equation from Appendix C: PL3 vB = − 3EI Values: P = −RB, L = 4.5 m, EI = 70,200 kN-m2

Calculation: PL3 vB = − 3EI (− RB )(4.5 m)3 =− = (432.6923 × 10−6 m/kN)RB 2 3(70, 200 kN-m ) Excerpts from this work may be reproduced by instructors for distribution on a not-for-profit basis for testing or instructional purposes only to students enrolled in courses for which the textbook has been adopted. Any other reproduction or translation of this work beyond that permitted by Sections 107 or 108 of the 1976 United States Copyright Act without the permission of the copyright owner is unlawful.

Compatibility equation at B: The sum of the downward deflection caused by the uniformly distributed load and the upward deflection caused by the reaction force RB must equal the support settlement: −0.107903 m + (432.6923 × 10−6 m/kN)RB = −0.010 m ∴ RB =

97.90264 × 10−3 m = 226.2639 kN = 226 kN ↑ 432.6923 × 10−6 m/kN

Ans.

Equilibrium equations for entire beam: ΣM A = − M A + RB (4.5 m) − (70 kN/m)(6 m)(3 m) = 0

∴ M A = (226.2639 kN)(4.5 m) − (70 kN/m)(6 m)(3 m) = −241.8125 kN-m = 242 kN-m (ccw) Ans. ΣFy = RA + RB − (70 kN/m)(6 m) = 0

∴ RA = (70 kN/m)(6 m) − 226.2639 kN = 193.7361 kN = 193.7 kN ↑

Ans.

(b) Determine maximum bending stress in beam: Maximum bending moment magnitude Mmax = 241.8125 kN-m (at A) Bending stress at maximum moment (241.8125 kN-m)(526 mm/2)(1,000) 2 σx = 351 × 106 mm 4 Ans. = 181.187 MPa = 181.2 MPa or using the tabulated value for the section modulus: (241.8125 kN-m)(1,000) 2 σx = 1,340 × 103 mm3

= 180.457 MPa = 180.5 MPa

Ans.


Mechanics of Materials Solutions Chapter11 Probs47 57

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