Mechanics of Materials Solutions Chapter10 Probs21 28

10.21 For the beam and loading shown in Fig. P10.21, integrate the load distribution to determine (a) the equation of the elastic curve for the beam, and (b) the maximum deflection for the beam. Assume that EI is constant for the beam. Fig. P10.21

Solution Integrate the load distribution: d 4v wx EI 4 = − 0 dx L 3 d v w0 x 2 EI 3 = − + C1 dx 2L d 2v w x3 EI 2 = − 0 + C1 x + C2 dx 6L dv w x4 C x2 EI = − 0 + 1 + C2 x + C3 dx 24 L 2 5 wx C x3 C x 2 EI v = − 0 + 1 + 2 + C3 x + C4 120 L 6 2 Boundary conditions and evaluate constants: d 3v at x = 0, V = EI 3 = 0 dx d 2v at x = 0, M = EI 2 = 0 dx dv w ( L) 4 at x = L, =0 − 0 + C3 = 0 dx 24 L w ( L)5 w0 L3 ( L) at x = L, v = 0 − 0 + + C4 = 0 120 L 24 (a) Elastic curve equation: w0 x 5 w0 L3 x w0 L4 + − EI v = − 120 L 24 30

∴v=−

∴ C1 = 0 ∴ C2 = 0 w0 L3 24 w L4 ∴ C4 = − 0 30

∴ C3 =

w0 ⎡⎣ x 5 − 5 L4 x + 4 L5 ⎤⎦ 120 LEI

(b) Maximum deflection: w0 w0 L4 5 4 5 ⎡ ⎤ (0) − 5L (0) + 4 L ⎦ = − vmax = − 120 LEI ⎣ 30 EI

Ans.

Ans.

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10.22 For the beam and loading shown in Fig. P10.22, integrate the load distribution to determine (a) the equation of the elastic curve for the beam, and (b) the deflection midway between the supports. Assume that EI is constant for the beam. Fig. P10.22

Solution Integrate the load distribution: d 4v wx EI 4 = − 0 dx L 3 d v w x2 EI 3 = − 0 + C1 dx 2L 2 d v w x3 EI 2 = − 0 + C1 x + C2 dx 6L dv w x4 C x2 EI = − 0 + 1 + C2 x + C3 dx 24 L 2 5 w0 x C1 x 3 C2 x 2 EI v = − + + + C3 x + C4 120 L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx d 2v w ( L )3 at x = L, M = EI 2 = 0 − 0 + C1 ( L) = 0 dx 6L at x = 0, v = 0 at x = L, v = 0

−

w0 ( L)5 w0 Lx3 + + C3 x = 0 120 L 36

(a) Elastic curve equation: w x 5 w Lx3 7 w0 L3 x EI v = − 0 + 0 − 120 L 36 360

∴v=−

∴ C2 = 0 w0 L 6 ∴ C4 = 0 ∴ C1 =

∴ C3 = −

7 w0 L3 360

w0 ⎡⎣3 x 5 − 10 L2 x 3 + 7 L4 x ⎤⎦ 360 LEI

(b) Maximum deflection: 5w L4 w0 ⎡⎣3( L / 2)5 − 10 L2 ( L / 2)3 + 7 L4 ( L / 2) ⎤⎦ = − 0 vmax = − 360 LEI 768 EI

Ans.

Ans.

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10.23 For the beam and loading shown in Fig. P10.23, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection at the left end of the beam, and (c) the support reactions RB and MB. Assume that EI is constant for the beam. Fig. P10.23

Solution Integrate the load distribution: d 4v w0 x 3 EI 4 = − 3 dx L 3 d v w x4 EI 3 = − 0 3 + C1 dx 4L 2 d v w x5 EI 2 = − 0 3 + C1 x + C2 dx 20 L dv w x6 C x2 EI = − 0 3 + 1 + C2 x + C3 dx 120 L 2 7 w0 x C1 x3 C2 x 2 EI v = − + + + C3 x + C4 840 L3 6 2 Boundary conditions and evaluate constants: d 3v at x = 0, V = EI 3 = 0 dx d 2v at x = 0, M = EI 2 = 0 dx dv w ( L)6 at x = L, =0 − 0 3 + C3 = 0 dx 120 L w ( L)7 w L3 ( L) at x = L, v = 0 − 0 3 + 0 + C4 = 0 840 L 120 (a) Elastic curve equation: w x 7 7 w L3 x 6 w0 L4 w0 ⎡ x7 − ∴v=− EI v = − 0 3 + 0 840 L 840 840 840 L3 EI ⎣ (b) Deflection at left end of beam: w0 6 w0 L7 6 w0 L4 7 6 7 ⎡ ⎤ vmax = − − L + L = − = − (0) 7 (0) 6 ⎦ 840 L3 EI ⎣ 840 L3 EI 140 EI (c) Support reactions RB and MB: d 3v w0 ( L) 4 wL VB = EI 3 =− =− 0 3 dx x = L 4L 4 M B = EI

d 2v dx 2

=− x=L

w0 ( L)5 w0 L2 = − 20 L3 20

∴ RB =

w0 L ↑ 4

∴ MB =

w0 L2 (cw) 20

∴ C1 = 0 ∴ C2 = 0 w0 L3 120 6 w L4 ∴ C4 = − 0 840

∴ C3 =

− 7 L6 x + 6 L7 ⎤⎦

Ans.

Ans.

Ans. Ans.

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10.24 For the beam and loading shown in Fig. P10.24, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, and (c) the support reactions RA and RB. Assume that EI is constant for the beam. Fig. P10.24

Solution Integrate the load distribution: d 4v w x3 EI 4 = − 0 3 dx L 3 d v w x4 EI 3 = − 0 3 + C1 dx 4L 2 d v w x5 EI 2 = − 0 3 + C1 x + C2 dx 20 L dv w0 x 6 C1 x 2 EI =− + + C2 x + C3 dx 120 L3 2 w x 7 C x3 C x 2 EI v = − 0 3 + 1 + 2 + C3 x + C4 840 L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx d 2v w ( L )5 at x = L, M = EI 2 = 0 − 0 3 + C1 ( L) = 0 dx 20 L at x = 0, v = 0

∴ C2 = 0 w0 L 20 ∴ C4 = 0

∴ C1 =

w0 ( L)7 w0 L( L)3 6w0 L3 at x = L, v = 0 − + + C3 ( L) = 0 ∴ C3 = − 840 L3 120 840 (a) Elastic curve equation: w0 x 7 w0 Lx 3 6 w0 L3 x w0 ⎡⎣ x 7 − 7 L6 x 3 + 6 L6 x ⎤⎦ EI v = − + − ∴v=− 3 3 840 L 120 840 840 L EI (b) Deflection midway between the supports: 3 ⎡⎛ L ⎞7 ⎤ w0 13w0 L4 6⎛ L⎞ 6⎛ L⎞ vx = L / 2 = − ⎢⎜ ⎟ − 7 L ⎜ ⎟ + 6 L ⎜ ⎟ ⎥ = − 840 L3 EI ⎣⎢⎝ 2 ⎠ 5120 EI ⎝2⎠ ⎝ 2 ⎠ ⎦⎥ (c) Support reactions RA and RB: d 3v w0 (0) 4 w0 L w0 L VA = EI 3 =− + = dx x = 0 4 L3 20 20 VB = EI

d 3v dx 3

=− x=L

w0 ( L) 4 w0 L 4w L + =− 0 3 4L 20 20

Ans.

Ans.

∴ RA =

w0 L ↑ 20

Ans.

∴ RB =

w0 L ↑ 5

Ans.

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10.25 For the beam and loading shown in Fig. P10.25, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection at the left end of the beam, and (c) the support reactions RB and MB. Assume that EI is constant for the beam.

Fig. P10.25

Solution Integrate the load distribution: πx d 4v EI 4 = − w0 cos dx 2L 3 d v 2w L πx + C1 EI 3 = − 0 sin dx π 2L d 2v 4 w L2 πx EI 2 = 02 cos + C1 x + C2 dx π 2L dv 8w0 L3 π x C1 x 2 EI = sin + + C2 x + C3 dx π3 2L 2 16w0 L4 π x C1 x3 C2 x 2 EI v = − cos + + + C3 x + C4 π4 2L 6 2 Boundary conditions and evaluate constants: d 3v at x = 0, V = EI 3 = 0 ∴ C1 = 0 dx d 2v 4 w0 L2 π (0) 4 w0 L2 at x = 0, M = EI 2 = 0 cos + C = 0 ∴ C = − 2 2 dx π2 2L π2 dv 8w0 L3 π ( L) 4w0 L2 ( L) 4w0 L3 at x = L, =0 sin − + C3 = 0 ∴ C3 = − 3 (2 − π ) dx π3 2L π2 π 4 2 2 3 16 w0 L π ( L) 4 w0 L ( L) 4w0 L ( L) − − − at x = L, v = 0 cos (2 − π ) + C4 = 0 4 π 2L 2π 2 π3 2 w L4 ∴ C4 = 03 (4 − π )

π

(a) Elastic curve equation: 16 w0 L4 π x 4 w0 L2 x 2 4w0 L3 2 w0 L4 − − EI v = − cos (2 − π ) + (4 − π ) π4 2L 2π 2 π3 π3 ∴v=−

w0 2π 4 EI

πx ⎡ ⎤ 4 2 2 2 3 4 ⎢32 L cos 2 L + 4π L x + 8π L x(2 − π ) − 4π L (4 − π ) ⎥ ⎣ ⎦

Ans.

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(b) Deflection at left end of beam: w ⎡ π (0) ⎤ v A = − 40 ⎢32 L4 cos + 4π 2 L2 (0) 2 + 8π L3 (0)(2 − π ) − 4π L4 (4 − π ) ⎥ 2π EI ⎣ 2L ⎦ =−

w0 w0 L4 4 4 ⎡ ⎤ 32 L − 4 π L (4 − π ) = − [32 − 4π (4 − π )] ⎦ 2π 4 EI ⎣ 2π 4 EI

w0 L4 = −0.1089 EI

Ans.

(c) Support reactions RB and MB: d 3v 2w L π ( L) 2w L = − 0 sin =− 0 VB = EI 3 π π dx x = L 2L d 2v M B = EI 2 dx

= x=L

4 w0 L2

π2

cos

π ( L) 2L

−

4 w0 L2

π2

∴ RB =

=−

4w0 L2

π2

2 w0 L

∴ MB =

π

↑

4w0 L2

π2

Ans. (cw)

Ans.

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10.26 For the beam and loading shown in Fig. P10.26, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, (c) the slope at the left end of the beam, and (d) the support reactions RA and RB. Assume that EI is constant for the beam. Fig. P10.26

Solution Integrate the load distribution: d 4v πx EI 4 = − w0 sin dx L 3 d v wL πx EI 3 = 0 cos + C1 dx π L πx d 2v w L2 EI 2 = 0 2 sin + C1 x + C2 π dx L π x C1 x 2 dv w0 L3 EI = − 3 cos + + C2 x + C3 π dx L 2 π x C1 x3 C2 x 2 w0 L4 EI v = − 4 sin + + + C3 x + C4 π L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx d 2v w0 L2 π ( L) at x = L, M = EI 2 = 0 sin + C1 ( L) = 0 2 dx π L at x = 0, v = 0 at x = L, v = 0 (a) Elastic curve equation: w0 L4 πx EI v = − 4 sin L π

−

w0 L4

π

4

sin

π ( L) L

∴ C2 = 0 ∴ C1 = 0 ∴ C4 = 0

+ C3 ( L) = 0

∴ C3 = 0

w0 L4 πx ∴ v = − 4 sin L π EI

Ans.

(b) Deflection midway between the supports: w L4 π ( L / 2) w L4 = − 40 vx = L / 2 = − 40 sin L π EI π EI (c) Slope at the left end of the beam: dv w L3 π (0) w L3 = EI θ A = − 0 3 cos = − 03 EI dx A L π π

Ans.

∴ θA = −

w0 L3 π 3 EI

Ans.

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(d) Support reactions RA and RB: d 3v wL π (0) w0 L = 0 cos = VA = EI 3 π π dx x = 0 L VB = EI

d 3v dx 3

= x=L

w0 L

π

cos

π ( L) L

=−

w0 L

π

∴ RA = ∴ RB =

w0 L

π w0 L

π

↑

Ans.

↑

Ans.

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10.27 For the beam and loading shown in Fig. P10.27, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection midway between the supports, (c) the slope at the left end of the beam, and (d) the support reactions RA and RB. Assume that EI is constant for the beam. Fig. P10.27

Solution Integrate the load distribution: d 4v πx EI 4 = − w0 sin dx 2L 3 d v 2w L πx EI 3 = 0 cos + C1 dx π 2L d 2v 4 w L2 πx EI 2 = 02 sin + C1 x + C2 dx π 2L dv 8w L3 π x C1 x 2 EI = − 03 cos + + C2 x + C3 dx π 2L 2 16w0 L4 π x C1 x3 C2 x 2 EI v = − sin + + + C3 x + C4 π4 2L 6 2 Boundary conditions and evaluate constants: d 2v at x = 0, M = EI 2 = 0 dx d 2v 4 w0 L2 π ( L) at x = L, M = EI 2 = 0 sin + C1 ( L) = 0 2 dx π 2L at x = 0, v = 0 at x = L, v = 0

−

16w0 L4

π4

sin

π ( L) 2L

−

4w0 L( L)3 + C3 ( L) = 0 6π 2

∴ C2 = 0 ∴ C1 = − ∴ C4 = 0 ∴ C3 =

4w0 L

π2

2w0 L3 (24 + π 2 ) 3π 4

(a) Elastic curve equation: 16 w0 L4 π x 4 w0 Lx3 2w0 L3 x − + EI v = − sin (24 + π 2 ) π4 2L 6π 2 3π 4 ∴v=−

2 w0 3π 4 EI

πx ⎡ 4 2 3 2 3 ⎤ ⎢ 24 L sin 2 L + π Lx − (24 + π ) L x ⎥ ⎣ ⎦

Ans.

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(b) Deflection midway between the supports: 3 2w ⎡ π ( L / 2) ⎛L⎞ ⎛ L ⎞⎤ vx = L / 2 = − 4 0 ⎢ 24 L4 sin + π 2 L⎜ ⎟ − (24 + π 2 ) L3 ⎜ ⎟ ⎥ 3π EI ⎢⎣ 2L ⎝2⎠ ⎝ 2 ⎠ ⎥⎦ 2 w L4 ⎡ π π 2 (24 + π 2 ) ⎤ = − 40 ⎢ 24sin + − ⎥ 3π EI ⎣ 4 8 2 ⎦ 2 w0 L4 = − 4 (1.2694611) 3π EI = −0.0086882

w0 L4 w L4 = −0.00869 0 EI EI

Ans.

(c) Slope at the left end of the beam: dv 8w0 L3 π (0) 2w0 L(0) 2 2w0 L3 = EI θ A = − 3 cos − + EI (24 + π 2 ) 2 4 dx A π 2L π 3π =−

8w0 L3

π3

+

2w0 L3 2 ⎤ ⎡ 8 16 (24 + π 2 ) = − w0 L3 ⎢ 3 − 4 − 2 ⎥ = −0.026209w0 L3 4 3π π 3π ⎦ ⎣π

w0 L3 ∴ θ A = −0.0262 EI

Ans.

(d) Support reactions RA and RB: d 3v 2w L π (0) 4w0 L 2w0 L VA = EI 3 (π − 2) = 0 cos − 2 = dx x = 0 π 2L π π2

∴ RA =

2w0 L

π2

d 3v VB = EI 3 dx ∴ RB =

(π − 2) ↑

=

2w0 L

x=L

4w0 L

π2

↑

π

cos

Ans.

π ( L) 2L

−

4w0 L

π

2

=−

4 w0 L

π2 Ans.

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10.28 For the beam and loading shown in Fig. P10.28, integrate the load distribution to determine (a) the equation of the elastic curve, (b) the deflection at the left end of the beam, and (c) the support reactions RB and MB. Assume that EI is constant for the beam.

Fig. P10.28

Solution Integrate the load distribution: πx d 4v EI 4 = − w0 sin dx 2L 3 d v 2w L πx EI 3 = 0 cos + C1 dx π 2L d 2v 4w L2 πx EI 2 = 02 sin + C1 x + C2 dx π 2L dv 8w L3 π x C1 x 2 EI = − 03 cos + + C2 x + C3 dx π 2L 2 16w0 L4 π x C1 x3 C2 x 2 EI v = − sin + + + C3 x + C4 π4 2L 6 2 Boundary conditions and evaluate constants: d 3v 2w0 L π (0) at x = 0, V = EI 3 = 0 cos + C1 = 0 dx π 2L d 2v 4w0 L2 π (0) 2w0 L(0) at x = 0, M = EI 2 = 0 sin − + C2 = 0 2 dx π 2L π dv 8w L3 π ( L) 2w0 L( L) 2 at x = L, =0 − 03 cos + + C3 = 0 dx π 2L 2π 16 w0 L4 π ( L) 2 w0 L( L)3 w0 L3 ( L) − − − + C4 = 0 at x = L, v = 0 sin π4 2L 6π π 2 w L4 ∴ C4 = 0 4 (24 − π 3 ) 3π

∴ C1 = −

2w0 L

π

∴ C2 = 0 ∴ C3 = −

w0 L3

π

(a) Elastic curve equation: 16 w0 L4 π x 2 w0 Lx3 w0 L3 x 2w0 L4 EI v = − − − + sin (24 − π 3 ) 4 4 π 2L 6π π 3π ∴v=−

w0 3π 4 EI

πx ⎡ ⎤ 4 L 48 sin + π 3 Lx 3 − 3π 3 L3 x − 2(24 − π 3 ) L4 ⎥ ⎢ 2L ⎣ ⎦

Ans.

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(b) Deflection at left end of beam: w ⎡ π (0) ⎤ v A = − 40 ⎢ 48L4 sin + π 3 L(0)3 − 3π 3 L3 (0) − 2(24 − π 3 ) L4 ⎥ 3π EI ⎣ 2L ⎦

=−

w0 w0 L4 3 4 ⎡ ⎤ L 2(24 π ) 0.0479509 − − = − ⎦ 3π 4 EI ⎣ EI

w0 L4 = −0.04795 EI

Ans.

(c) Support reactions RB and MB: d 3v 2w L π ( L) 2 w0 L 2w L = 0 cos − =− 0 VB = EI 3 dx x = L π 2L π π

M B = EI

d 2v dx 2

= x=L

∴MB = −

4w0 L2

π2

2 w0 L2

π2

sin

π ( L) 2L

(π − 2) =

−

2w0 L( L)

π

2 w0 L2

π2

=

(π − 2)

∴ RB =

4w0 L2

π2

−

(cw)

2 w0 L

π

↑

Ans.

2w0 L2

π Ans.

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