Mechanics of Materials Solutions Chapter10 Probs1 20

10.1 For the loading shown, use the doubleintegration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam. Fig. P10.1

Solution Integration of moment equation: d 2v EI 2 = M ( x) = − M 0 dx dv EI = − M 0 x + C1 dx M 0 x2 EI v = − + C1 x + C2 2

(a) (b)

Boundary conditions: dv x=0 = 0 at dx at v=0 x=0 Evaluate constants: From Eq. (a), C1 = 0. From Eq. (b), C2 = 0 (a) Elastic curve equation: M x2 M x2 EI v = − 0 ∴v=− 0 2 2 EI (b) Deflection at the free end: M 0 ( L) 2 M 0 L2 vB = − = − 2 EI 2 EI (c) Slope at the free end: dv M ( L) M L = θB = − 0 = − 0 dx B EI EI

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10.2 For the loading shown, use the doubleintegration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.

Fig. P10.2

Solution Integration of moment equation: d 2v wx 2 EI 2 = M ( x) = − 2 dx

dv wx3 =− + C1 dx 6 wx 4 EI v = − + C1 x + C2 24

EI

(a) (b)

Boundary conditions: dv = 0 at x=L dx v=0 at x=L Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w( L)3 wL3 EI (0) = − + C1 ∴ C1 = 6 6 Substitute x = L and v = 0 into Eq. (b) to determine C2: w( L) 4 wL4 wL4 wL4 EI (0) = − + C1 ( L) + C2 = − + + C2 ∴ C2 = − 24 24 6 8 (a) Elastic curve equation: wx 4 wL3 x wL4 w ⎡ x 4 − 4 L3 x + 3L4 ⎤⎦ EI v = − + − ∴v=− 24 6 8 24 EI ⎣ (b) Deflection at the free end: w 3wL4 wL4 ⎡⎣ −(0) 4 + 4 L3 (0) − 3L4 ⎤⎦ = − vA = = − 24 EI 24 EI 8 EI (c) Slope at the free end: dv w(0)3 wL3 wL3 = θA = − + = dx A 6 EI 6 EI 6 EI

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10.3 For the loading shown, use the doubleintegration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.

Fig. P10.3

Solution Integration of moment equation: d 2v w x3 EI 2 = M ( x) = − 0 dx 6L 4 dv wx EI = − 0 + C1 dx 24 L w0 x 5 EI v = − + C1 x + C2 120 L

(a) (b)

Boundary conditions: dv = 0 at x=L dx v=0 at x=L Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w0 ( L) 4 w0 L3 EI (0) = − + C1 ∴ C1 = 24 L 24 Substitute x = L and v = 0 into Eq. (b) to determine C2: w ( L)5 w L5 w L3 EI (0) = − 0 + C1 ( L) + C2 = − 0 + 0 ( L) + C2 120 L 120 L 24 4 4 4 wL wL wL ∴ C2 = 0 − 0 = − 0 120 24 30 (a) Elastic curve equation: w x 5 w L3 w L4 EI v = − 0 + 0 x − 0 120 L 24 30

∴v = −

(b) Deflection at the free end: w0 w L4 ⎡⎣(0)5 − 5 L4 (0) + 4 L5 ⎤⎦ = − 0 vA = − 120 L EI 30 EI (c) Slope at the free end: dv w (0) 4 w0 L3 w L3 = θA = − 0 + = 0 dx A 24 L EI 24 EI 24 EI

w0 ⎡ x 5 − 5 L4 x + 4 L5 ⎤⎦ 120 L EI ⎣

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10.4 For the beam and loading shown in Fig. P10.4, use the double-integration method to determine (a) the equation of the elastic curve for segment AB of the beam, (b) the deflection at B, and (c) the slope at A. Assume that EI is constant for the beam.

Fig. P10.4

Solution Integration of moment equation: d 2v P EI 2 = M ( x) = x dx 2 2 dv Px EI = + C1 dx 4

EI v =

(a)

Px 3 + C1 x + C2 12

(b)

Boundary conditions: at v=0 x=0

dv =0 dx

at

x=

L 2

Evaluate constants: Substitute x = L/2 and dv/dx = 0 into Eq. (a) to determine C1: P ( L / 2) 2 PL2 EI (0) = + C1 ∴ C1 = − 4 16 Substitute x = 0 and v = 0 into Eq. (b) to determine C2: P (0)3 PL2 (0) EI (0) = − + C2 ∴ C2 = 0 12 16 (a) Elastic curve equation: Px 3 PL2 x EI v = − 12 16

∴v=−

Px ⎡⎣3L2 − 4 x 2 ⎤⎦ 48EI

(b) Deflection at B: 2 P( L / 2) ⎡ 2 PL3 ⎛L⎞ ⎤ vB = − 3 L 4 − = − ⎢ ⎜ ⎟ ⎥ 48EI ⎢⎣ 48EI ⎝ 2 ⎠ ⎥⎦ (c) Slope at A: dv P (0) 2 PL2 PL2 = θA = − = − dx A 4 EI 16 EI 16 EI

(0 ≤ x ≤

L ) 2

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10.5 For the beam and loading shown in Fig. P10.5, use the double-integration method to determine (a) the equation of the elastic curve for the beam, (b) the slope at A, (c) the slope at B, and (d) the deflection at midspan. Assume that EI is constant for the beam.

Fig. P10.5

Solution Beam FBD: ΣFy = Ay + By = 0

∴ Ay = − By ΣM A = B y L − M 0 = 0 ∴ By =

M0 L

and

Ay = −

M0 L

Moment equation:

ΣM a − a = M ( x) − Ay x − M 0 = M ( x) + ∴ M ( x) = M 0 −

M0 x − M0 = 0 L

M0x L

Integration of moment equation: d 2v M x EI 2 = M ( x) = M 0 − 0 dx L 2 dv M x EI = M 0 x − 0 + C1 dx 2L 2 M x M x3 EI v = 0 − 0 + C1 x + C2 2 6L

(a) (b)

Boundary conditions: v=0 at x=0 v=0 at x=L Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: M 0 (0) 2 M 0 (0)3 EI (0) = − + C1 (0) + C2 2 6L Substitute x = L and v = 0 into Eq. (b) to determine C1:

∴ C2 = 0

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M 0 ( L ) 2 M 0 ( L )3 EI (0) = − + C1 ( L) 2 6L M L M L M L ∴ C1 = 0 − 0 = − 0 6 2 3 (a) Elastic curve equation: M 0 x 2 M 0 x3 M 0 Lx − − EI v = 2 6L 3

∴v=−

M0x 2 ⎡⎣ x − 3Lx + 2 L2 ⎤⎦ 6 L EI

(b) Slope at A: dv M (0) 2 M 0 L M L = θ A = M 0 (0) − 0 − = − 0 dx A 2 L EI 3EI 3EI (c) Slope at B: dv M 0 ( L) M 0 ( L) 2 M 0 L M 0 M L = θB = − − = [ 6 L − 3L − 2 L ] = 0 dx B EI 2 L EI 3EI 6 EI 6 EI (d) Deflection at midspan: 2 ⎤ M 0 ( L / 2) ⎡⎛ L ⎞ M 0 L2 ⎛L⎞ 2 − + = − vx = L / 2 = − 3 L 2 L ⎢⎜ ⎟ ⎥ ⎜ ⎟ 6 L EI ⎢⎣⎝ 2 ⎠ 16 EI ⎝2⎠ ⎥⎦

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10.6 For the beam and loading shown in Fig. P10.6, use the double-integration method to determine (a) the equation of the elastic curve for the beam, (b) the maximum deflection, and (c) the slope at A. Assume that EI is constant for the beam.

Fig. P10.6

Solution Moment equation:

wLx wx 2 + =0 2 2 wx 2 wLx ∴ M ( x) = − + 2 2 Integration of moment equation: d 2v wx 2 wLx EI 2 = M ( x) = − + dx 2 2 3 2 dv wx wLx EI =− + + C1 6 4 dx wx 4 wLx 3 EI v = − + + C1 x + C2 24 12 ΣM a − a = M ( x ) −

(a) (b)

Boundary conditions: v=0 at x=0 v=0

at

x=L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: w(0) 4 wL(0)3 EI (0) = − + + C1 (0) + C2 ∴ C2 = 0 24 12 Substitute x = L and v = 0 into Eq. (b) to determine C1: w( L) 4 wL( L)3 w( L) 4 w( L) 4 wL3 EI (0) = − + + C1 ( L) ∴ C1 = − =− 24 12 24 L 12 L 24 (a) Elastic curve equation: wx 4 wLx 3 wL3 x wx ⎡ x 3 − 2 Lx 2 + L3 ⎤⎦ + − ∴v=− EI v = − 24 12 24 24 EI ⎣ (b) Maximum deflection: At x = L/2: 3 2 ⎤ w( L / 2) ⎡⎛ L ⎞ wL ⎡ L3 L3 5wL4 ⎛L⎞ 2 2⎤ vmax = − − 2 L + L = − − + L = − ⎢⎜ ⎟ ⎥ ⎢ ⎥ ⎜ ⎟ 24 EI ⎢⎣⎝ 2 ⎠ 48 EI ⎣ 8 2 384 EI ⎝2⎠ ⎥⎦ ⎦ (c) Slope at A: dv w(0)3 wL(0) 2 wL3 wL3 = θA = − + − = − dx A 6 4 24 24

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10.7 For the beam and loading shown in Fig. P10.7, use the double-integration method to determine (a) the equation of the elastic curve for segment AB of the beam, (b) the deflection midway between the two supports, (c) the slope at A, and (d) the slope at B. Assume that EI is constant for the beam.

Fig. P10.7

Solution Beam FBD: ΣFy = Ay + By − P = 0

3L P=0 2 3P ∴ By = and 2

ΣM A = B y L −

Ay = −

P 2

Moment equation:

P x=0 2 Integration of moment equation: d 2v Px EI 2 = M ( x) = − dx 2 2 dv Px EI =− + C1 dx 4 Px3 EI v = − + C1 x + C2 12 ΣM a − a = M ( x ) +

∴ M ( x) = −

Px 2

(a) (b)

Boundary conditions: v=0 at x=0 v=0

at

x=L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: P(0)3 EI (0) = − + C1 (0) + C2 ∴ C2 = 0 12 Substitute x = L and v = 0 into Eq. (b) to determine C1: P ( L )3 PL2 EI (0) = − + C1 ( L) ∴ C1 = 12 12

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(a) Elastic curve equation for segment AB of the beam: Px3 PL2 x Px ⎡⎣ L2 − x 2 ⎤⎦ + ∴v= EI v = − 12 12 12 EI (b) Deflection at midspan: 2 P( L / 2) ⎡ 2 ⎛ L ⎞ ⎤ PL ⎡ 3L ⎤ PL3 = vx = L / 2 = ⎢L − ⎜ ⎟ ⎥ = ⎢ ⎥ 12 EI ⎢⎣ ⎝ 2 ⎠ ⎥⎦ 24 EI ⎣ 4 ⎦ 32 EI (c) Slope at A: dv P (0) 2 PL2 PL2 = θA = − + = dx A 4 EI 12 EI 12 EI (d) Slope at B: dv P ( L) 2 PL2 PL2 = θB = − + = − dx B 4 EI 12 EI 6 EI

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10.8 For the beam and loading shown in Fig. P10.8, use the double-integration method to determine (a) the equation of the elastic curve for segment BC of the beam, (b) the deflection midway between B and C, and (c) the slope at C. Assume that EI is constant for the beam.

Fig. P10.8

Solution Beam FBD: ΣM B = PL + C y (4 L) − P(5L) = 0

∴ Cy = P ΣFy = By + C y − 2 P = 0 ∴ By = P Moment equation: ΣM a − a = M ( x) − By x + P( L + x) = M ( x) − Px + P( L + x) = 0

∴ M ( x) = − PL Integration of moment equation: d 2v EI 2 = M ( x) = − PL dx dv EI = − PLx + C1 dx PLx 2 EI v = − + C1 x + C2 2

(a) (b)

Boundary conditions: v=0 at x=0 v=0

at

x = 4L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: PL(0) 2 EI (0) = − + C1 (0) + C2 ∴ C2 = 0 2 Substitute x = 4L and v = 0 into Eq. (b) to determine C1: PL(4 L) 2 EI (0) = − + C1 (4 L) ∴ C1 = 2 PL2 2

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(a) Elastic curve equation for segment BC of the beam: PLx 2 PLx EI v = − + 2 PL2 x ∴v= [4L − x] 2 2 EI (b) Deflection at midspan: PL(2 L) 2 PL3 vx = L / 2 = [ 4 L − (2 L)] = 2 EI EI (c) Slope at C: dv PL(4 L) 2 PL2 2 PL2 = θC = − + = − dx C EI EI EI

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10.9 For the beam and loading shown in Fig. P10.9, use the double-integration method to determine (a) the equation of the elastic curve for segment AB of the beam, (b) the deflection midway between A and B, and (c) the slope at B. Assume that EI is constant for the beam.

Fig. P10.9

Solution Beam FBD:

wL2 ⎛ 5L ⎞ − P ⎜ ⎟ + By L = 0 2 ⎝ 4 ⎠ wL 5P ∴ By = + 2 4 ΣFy = Ay + By − wL − P = 0

ΣM A = −

∴ Ay =

wL P − 2 4

Moment equation:

wx 2 wx 2 wLx Px − Ay x = M ( x) + − + =0 2 2 2 4 wx 2 wLx Px ∴ M ( x) = − + − 2 2 4 Integration of moment equation: d 2v wx 2 wLx Px EI 2 = M ( x) = − + − dx 2 2 4 3 2 2 dv wx wLx Px EI =− + − + C1 dx 6 4 8 wx 4 wLx 3 Px 3 EI v = − + − + C1 x + C2 24 12 24 ΣM a − a = M ( x ) +

(a) (b)

Boundary conditions: v=0 at x=0 v=0

at

x=L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: w(0) 4 wL(0)3 P(0)3 EI (0) = − + − + C1 (0) + C2 24 12 24 Substitute x = L and v = 0 into Eq. (b) to determine C1: w( L) 4 wL( L)3 P( L)3 EI (0) = − + − + C1 ( L) 24 12 24

∴ C2 = 0

∴ C1 = −

wL3 PL2 + 24 24

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(a) Elastic curve equation for segment AB of the beam: wx 4 wLx 3 Px 3 wL3 x PL2 x EI v = − + − − + 24 12 24 24 24 wx Px ⎡⎣ x 3 − 2 Lx 2 + L3 ⎤⎦ − ⎡ x 2 − L2 ⎤⎦ ∴v=− 24 EI 24 EI ⎣

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(b) Deflection at midspan: 3 2 ⎤ P( L / 2) ⎡⎛ L ⎞2 ⎤ w( L / 2) ⎡⎛ L ⎞ ⎛L⎞ 3 2 vx = L / 2 = − ⎢⎜ ⎟ − 2 L ⎜ ⎟ + L ⎥ − ⎢⎜ ⎟ − L ⎥ 24 EI ⎣⎢⎝ 2 ⎠ 24 EI ⎢⎣⎝ 2 ⎠ ⎝2⎠ ⎦⎥ ⎦⎥

= −

5wL4 PL3 + 384 EI 64 EI

(c) Slope at B: dv w( L)3 wL( L) 2 P( L) 2 wL3 PL2 wL3 PL2 = θB = − + − − + = − dx B 6 EI 4 EI 8EI 24 EI 24 EI 24 EI 12 EI

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10.10 For the beam and loading shown in Fig. P10.10, use the double-integration method to determine (a) the equation of the elastic curve for segment AC of the beam, (b) the deflection at B, and (c) the slope at A. Assume that EI is constant for the beam.

Fig. P10.10

Solution Beam FBD: 3L + C y (2 L) = 0 2 9 wL ∴Cy = 4 ΣFy = Ay + C y − w(3L) = 0

ΣM A = − w(3L)

∴ Ay = 3wL −

9 wL 3wL = 4 4

Moment equation: 3wL ⎛x⎞ ⎛x⎞ ΣM a − a = M ( x) − Ay x + wx ⎜ ⎟ = M ( x) − x + wx ⎜ ⎟ = 0 4 ⎝2⎠ ⎝2⎠ wx 2 3wLx ∴ M ( x) = − + 2 4

Integration of moment equation: d 2v wx 2 3wLx EI 2 = M ( x) = + dx 2 4 3 2 dv wx 3wLx EI =− + + C1 dx 6 8 wx 4 3wLx 3 EI v = − + + C1 x + C2 24 24

(a) (b)

Boundary conditions: v=0 at x=0 v=0

at

x = 2L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2: w(0) 4 3wL(0)3 EI (0) = − + + C1 (0) + C2 24 24 Substitute x = 2L and v = 0 into Eq. (b) to determine C1:

∴ C2 = 0

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w(2 L) 4 3wL(2 L)3 + + C1 (2 L) 24 24 8wL3 12 wL3 wL3 ∴ C1 = − =− 24 24 6

EI (0) = −

(a) Elastic curve equation for segment AC of the beam: wx 4 3wLx 3 wL3 x wx 3 ⎡⎣ x − 3Lx 2 + 4 L3 ⎤⎦ EI v = − + − =− 24 24 6 24 wx ⎡⎣ x 3 − 3Lx 2 + 4 L3 ⎤⎦ ∴v=− 24 EI (b) Deflection at B: w( L) wL4 ⎡⎣( L)3 − 3L( L) 2 + 4 L3 ⎤⎦ = − vB = − 24 EI 12 EI (c) Slope at A: dv w(0)3 3wL(0) 2 wL3 wL3 = θA = − + − = − dx A 6 EI 8 EI 6 EI 6 EI

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10.11 For the simply supported steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Fig. P10.11, use the double-integration method to determine the deflection at B. Assume L = 4 m, P = 60 kN, and w = 40 kN/m.

Fig. P10.11

Solution Beam FBD: ⎛L⎞ ⎛L⎞ ΣM A = − wL ⎜ ⎟ − P ⎜ ⎟ + C y ( L) = 0 ⎝2⎠ ⎝2⎠ wL P ∴Cy = + 2 2 ΣFy = Ay + C y − w( L) − P = 0 ∴ Ay =

wL P + 2 2

Moment equation: wx 2 wx 2 wLx Px ΣM a − a = M ( x ) + − Ay x = M ( x) + − − =0 2 2 2 2 wx 2 wLx Px ∴ M ( x) = − + + 2 2 2 Integration of moment equation: d 2v wx 2 wLx Px EI 2 = M ( x) = − + + dx 2 2 2 3 2 2 dv wx wLx Px EI =− + + + C1 dx 6 4 4 wx 4 wLx 3 Px3 EI v = − + + + C1 x + C2 24 12 12

(a) (b)

Boundary conditions: v=0 at x=0 dv L = 0 at x= dx 2 Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L/2 and dv/dx = 0 into Eq. (b) to determine C1:

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w( L / 2)3 wL( L / 2) 2 P ( L / 2) 2 EI (0) = − + + + C1 6 4 4 wL3 wL3 PL2 wL3 PL2 ∴ C1 = − − =− − 48 16 16 24 16 Elastic curve equation: wx 4 wLx 3 Px3 wL3 x PL2 x + + − − EI v = − 24 12 12 24 16 wx Px ⎡⎣ x 3 − 2 Lx 2 + L3 ⎤⎦ − ⎡⎣3L2 − 4 x 2 ⎤⎦ ∴v=− 24 EI 48 EI Deflection at B: At x = L/2: 5wL4 PL3 vB = − − 384 EI 48 EI

Let E = 200 GPa, I = 129 × 106 mm4, w = 40 kN/m, P = 60 kN, and L = 4 m. 5(40 N/mm)(4,000 mm) 4 (60,000 N)(4,000 mm)3 vB = − − 384(200,000 N/mm 2 )(129 × 106 mm 4 ) 48(200,000 N/mm 2 )(129 × 106 mm 4 ) = −5.1680 mm − 3.1008 mm = −8.27 mm

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10.12 For the cantilever steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Fig. P10.12, use the double-integration method to determine the deflection at A. Assume L = 2.5 m, P = 50 kN, and w = 30 kN/m.

Fig. P10.12

Solution Moment equation:

ΣM a − a = M ( x ) +

wx 2 + Px = 0 2

∴ M ( x) = −

wx 2 − Px 2

Integration of moment equation: d 2v wx 2 EI 2 = M ( x) = − − Px dx 2 dv wx 3 Px 2 EI =− − + C1 dx 6 2 wx 4 Px 3 EI v = − − + C1 x + C2 24 6

(a) (b)

Boundary conditions: v=0 at x=L dv = 0 at x=L dx Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w( L)3 P( L) 2 wL3 PL2 EI (0) = − − + C1 ∴ C1 = + 6 2 6 2 Substitute x = L and v = 0 into Eq. (b) to determine C2: w( L) 4 P( L)3 wL3 PL2 wL4 wL4 PL3 PL3 ( L) + ( L ) + C2 = − EI (0) = − − + + − + + C2 24 6 6 2 24 6 6 2 wL4 PL3 ∴ C2 = − − 8 3 Elastic curve equation: wx 4 wL3 x wL4 Px 3 PL2 x PL3 EI v = − + − − + − 24 6 8 6 2 3 w P ⎡⎣ x 4 − 4 L3 x + 3L4 ⎤⎦ − ⎡⎣ x 3 − 3L2 x + 2 L3 ⎤⎦ ∴v = − 24 EI 6 EI

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Deflection at A: w P 3wL4 PL3 ⎡⎣ −(0) 4 + 4 L3 (0) − 3L4 ⎤⎦ − ⎡⎣(0)3 − 3L2 (0) + 2 L3 ⎤⎦ = − − vA = 24 EI 6 EI 24 EI 3EI Let E = 200 GPa, I = 129 × 106 mm4, w = 30 kN/m, P = 50 kN, and L = 2.5 m. 3(30 N/mm)(2,500 mm) 4 (50,000 N)(2,500 mm)3 vA = − − 24(200,000 N/mm 2 )(129 × 106 mm 4 ) 3(200,000 N/mm 2 )(129 × 106 mm 4 ) = −5.6777 mm − 10.0937 mm = −15.77 mm

Ans.

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10.13 For the cantilever steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Fig. P10.13, use the double-integration method to determine the deflection at B. Assume L = 3 m, M0 = 70 kN-m, and w = 15 kN/m.

Fig. P10.13

Solution Moment equation:

w( L − x) 2 − M0 = 0 2 w( L − x) 2 ∴ M ( x) = − − M0 2

ΣM a − a = − M ( x ) −

Integration of moment equation: d 2v w( L − x) 2 w wL2 wx 2 − M 0 = − ⎡⎣ L2 − 2 Lx + x 2 ⎤⎦ − M 0 = − + wLx − − M0 EI 2 = M ( x) = − dx 2 2 2 2 dv wL2 x wLx 2 wx 3 EI =− + − − M 0 x + C1 dx 2 2 6 wL2 x 2 wLx3 wx 4 M 0 x 2 EI v = − + − − + C1 x + C2 4 6 24 2

(a) (b)

Boundary conditions: v=0 at x=0 dv = 0 at x=0 dx Evaluate constants: Substitute x = 0 and dv/dx = 0 into Eq. (a) to determine C1 = 0. Next, substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Elastic curve equation: wL2 x 2 wLx 3 wx 4 M 0 x 2 EI v = − + − − 4 6 24 2 w M x2 ⎡⎣ x 4 − 4 Lx 3 + 6 L2 x 2 ⎤⎦ − 0 ∴v = − 24 EI 2 EI Deflection at B: w M ( L) 2 wL4 M 0 L2 ⎡⎣( L) 4 − 4 L( L)3 + 6 L2 ( L) 2 ⎤⎦ − 0 =− − vB = − 24 EI 2 EI 8 EI 2 EI

Let E = 200 GPa, I = 129 × 106 mm4, w = 15 kN/m, M0 = 70 kN-m, and L = 3 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.

(15 N/mm)(3,000 mm) 4 (70 kN-m)(1,000 N/kN)(1,000 mm/m)(3,000 mm)2 vB = − − 8(200,000 N/mm 2 )(129 × 106 mm 4 ) 2(200,000 N/mm 2 )(129 × 106 mm 4 ) = −5.8866 mm − 12.2093 mm = −18.10 mm

Ans.

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10.14 For the cantilever steel beam [E = 200 GPa; I = 129 × 106 mm4] shown in Fig. P10.14, use the double-integration method to determine the deflection at A. Assume L = 2.5 m, P = 50 kN-m, and w0 = 90 kN/m.

Fig. P10.14

Solution Moment equation: ΣM a − a = M ( x ) +

w0 x ⎛ x ⎞ ( x) ⎜ ⎟ + Px = 0 2L ⎝3⎠

w0 x 3 ∴ M ( x) = − − Px 6L

Integration of moment equation: d 2v w0 x 3 EI 2 = M ( x) = − − Px dx 6L dv w x 4 Px 2 EI =− 0 − + C1 dx 24 L 2 w x 5 Px 3 EI v = − 0 − + C1 x + C2 120 L 6

(a) (b)

Boundary conditions: v=0 at x=L dv = 0 at x=L dx Evaluate constants: Substitute x = L and dv/dx = 0 into Eq. (a) to determine C1: w0 ( L) 4 P( L) 2 w0 L3 PL2 EI (0) = − − + C1 ∴ C1 = + 24 L 2 24 2 Substitute x = L and v = 0 into Eq. (b) to determine C2: w ( L)5 P ( L)3 w0 L3 PL2 w L4 w L4 PL3 PL3 ( L) + ( L ) + C2 = − 0 + 0 − EI (0) = − 0 − + + + C2 120 L 6 24 2 120 24 6 2 w L4 PL3 ∴ C2 = − 0 − 30 3 Elastic curve equation: w x 5 Px 3 w0 L3 PL2 w L4 PL3 + EI v = − 0 − x+ x− 0 − 120 L 6 24 2 30 3 w0 P ⎡⎣ x 5 − 5 L4 x + 4 L5 ⎤⎦ − ⎡ x 3 − 3L2 x + 2 L3 ⎤⎦ ∴v = − 120 L EI 6 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.

Deflection at A: Let E = 200 GPa, I = 129 × 106 mm4, w0 = 90 kN/m, P = 50 kN, and L = 2.5 m. w0 P ⎡⎣(0)5 − 5 L4 (0) + 4 L5 ⎤⎦ − ⎡⎣(0)3 − 3L2 (0) + 2 L3 ⎤⎦ vA = − 120 L EI 6 EI w0 L4 PL3 − 30 EI 3EI (90 N/mm)(2,500 mm) 4 (50,000 N)(2,500 mm)3 =− − 30(200,000 N/mm 2 )(129 × 106 mm 4 ) 3(200,000 N/mm 2 )(129 × 106 mm 4 ) = −4.5422 mm − 10.0937 mm =−

= −14.64 mm

Ans.

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10.15 For the beam and loading shown in Fig. P10.15, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.

Fig. P10.15

Solution Beam FBD: ΣM A = − M A −

w0 L ⎛ 2 L ⎞ ⎜ ⎟=0 2 ⎝ 3 ⎠

w0 L2 3 wL ΣFy = Ay − 0 = 0 2 w0 L ∴ Ay = 2 ∴MA = −

Moment equation:

ΣM a − a = M ( x ) − M A + = M ( x) +

w0 x ⎛ x ⎞ ( x) ⎜ ⎟ − Ay x 2L ⎝3⎠

w0 L2 w0 x ⎛ x ⎞ w0 L + ( x) ⎜ ⎟ − ( x) = 0 3 2L 2 ⎝3⎠

∴ M ( x) = −

w0 x 3 w0 Lx w0 L2 + − 6L 2 3

Integration of moment equation: d 2v w x 3 w Lx w L2 EI 2 = M ( x) = − 0 + 0 − 0 dx 6L 2 3 4 2 2 dv wx w Lx wLx EI =− 0 + 0 − 0 + C1 dx 24 L 4 3 w x 5 w Lx 3 w0 L2 x 2 EI v = − 0 + 0 − + C1 x + C2 120 L 12 6

(a) (b)

Boundary conditions: v=0 at x=0 dv = 0 at x=0 dx Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq. (b) to determine C1 = 0. 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.

(a) Elastic curve equation: w x 5 w Lx 3 w0 L2 x 2 EI v = − 0 + 0 − 120 L 12 6

∴v=−

w0 ⎡ x 5 − 10 L2 x3 + 20 L3 x 2 ⎤⎦ 120 L EI ⎣

(b) Deflection at the free end: w0 11w0 L5 ⎡⎣( L)5 − 10 L2 ( L)3 + 20 L3 ( L) 2 ⎤⎦ = − vB = − 120 L EI 120 L EI (c) Slope at the free end: dv w0 ( L) 4 w0 L( L) 2 w0 L2 ( L) w0 L3 6w0 L3 8w0 L3 w0 L3 = θB = − + − =− + − = − dx B 24 L 4 3 24 24 24 8EI

Ans.

Ans.

Ans.

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10.16 For the beam and loading shown in Fig. P10.16, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam AB, (b) the deflection at the free end, and (c) the slope at the free end. Assume that EI is constant for each beam.

Fig. P10.16

Solution Beam FBD: ΣM A = − M A −

w0 L ⎛ L ⎞ ⎜ ⎟=0 2 ⎝3⎠

w0 L2 6 wL ΣFy = Ay − 0 = 0 2 w0 L ∴ Ay = 2 ∴MA = −

Moment equation:

ΣM a − a = − M ( x ) −

w0 ( L − x)3 =0 2L 3

w0 ( L − x )3 6L w = − 0 ( L3 − 3L2 x + 3Lx 2 − x 3 ) 6L w L2 w Lx w x 2 w x 3 =− 0 + 0 − 0 + 0 6 2 2 6L

M ( x) = −

Integration of moment equation: d 2v w x3 w x 2 w Lx w L2 EI 2 = M ( x) = 0 − 0 + 0 − 0 dx 6L 2 2 6 4 3 2 2 dv w0 x wx w Lx wLx EI = − 0 + 0 − 0 + C1 dx 24 L 6 4 6 w0 x 5 w0 x 4 w0 Lx3 w0 L2 x 2 EI v = − + − + C1 x + C2 120 L 24 12 12

(a) (b)

Boundary conditions: v=0 at x=0 dv = 0 at x=0 dx

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Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq. (b) to determine C1 = 0. (a) Elastic curve equation: w x 5 w x 4 w Lx3 w0 L2 x 2 − EI v = 0 − 0 + 0 120 L 24 12 12 w0 ⎡⎣ x5 − 5Lx 4 + 10 L2 x3 − 10 L3 x 2 ⎤⎦ ∴v= 120 L EI (b) Deflection at the free end: w0 4 w0 L5 w0 L4 5 4 2 3 3 2 ⎡( L) − 5 L( L) + 10 L ( L) − 10 L ( L) ⎤⎦ = − vB = = − 120 L EI ⎣ 120 L EI 30 EI (c) Slope at the free end: dv w ( L) 4 w0 ( L)3 w0 L( L) 2 w0 L2 ( L) w L3 = θB = 0 − + − = − 0 dx B 24 L EI 6 EI 4 EI 6 EI 24 EI

Ans.

Ans.

Ans.

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10.17 For the beam and loading shown in Fig. P10.17, use the double-integration method to determine (a) the equation of the elastic curve for the cantilever beam, (b) the deflection at B, (c) the deflection at the free end, and (d) the slope at the free end. Assume that EI is constant for the beam.

Fig. P10.17

Solution Beam FBD: wL ⎛ L L ⎞ ⎜ + ⎟=0 2 ⎝2 4⎠ 3wL2 ∴MA = − 8 wL ΣFy = Ay − =0 2 wL ∴ Ay = 2

ΣM A = − M A −

Consider beam segment AB (0 ≤ x ≤ L/2) Moment equation:

ΣM a − a = M ( x) − M A − Ay x = M ( x) + ∴ M ( x) = −

3wL2 wL x=0 − 8 2

3wL2 wLx + 8 2

Integration of moment equation: d 2v 3wL2 wLx EI 2 = M ( x) = − + dx 8 2 2 2 dv 3wL x wLx EI =− + + C1 dx 8 4 3wL2 x 2 wLx 3 EI v = − + + C1 x + C2 16 12

(a) (b)

Boundary conditions: v=0 at x=0 dv = 0 at x=0 dx Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 0 and dv/dx = 0 into Eq. (b) to determine C1 = 0. 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.

Elastic curve equation for beam segment AB: 3wL2 x 2 wLx3 + EI v = − 16 12

∴v=−

wLx 2 [9 L − 4 x ] 48 EI

(0 ≤ x ≤ L / 2)

Slope at B: Let x = L/2 dv 3wL2 ( L / 2) wL( L / 2) 2 wL3 = θB = − + =− dx B 8EI 4 EI 8EI Deflection at B: Let x = L/2 wL( L / 2) 2 ⎡ 7 wL4 ⎛ L ⎞⎤ − = − vB = − 9 L 4 ⎜ ⎟⎥ 48EI ⎢⎣ 192 EI ⎝ 2 ⎠⎦

Consider beam segment BC (L/2 ≤ x ≤ L) Moment equation:

ΣM b − b

w⎛ L⎞ = M ( x) − M A − Ay x + ⎜ x − ⎟ 2⎝ 2⎠

2

2

= M ( x) +

w⎛ L⎞ 3wL2 wL − x+ ⎜x− ⎟ =0 8 2 2⎝ 2⎠ 2

w⎛ L ⎞ wL 3wL2 ∴ M ( x) = − ⎜ x − ⎟ + x− 2⎝ 2⎠ 2 8 2 2 wx wL =− + wLx − 2 2

Integration of moment equation: d 2v wx 2 wL2 EI 2 = M ( x) = − + wLx − dx 2 2 3 2 2 dv wx wLx wL x EI =− + − + C3 dx 6 2 2 wx 4 wLx3 wL2 x 2 EI v = − + − + C3 x + C4 24 6 4 Continuity conditions: 7 wL4 v=− at 192 EI dv wL3 =− at dx 8 EI

(c) (d)

L 2 L x= 2 x=

Evaluate constants: Substitute the slope continuity condition into Eq. (c) for x = L/2 and solve for C3: 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.

dv w( L / 2)3 wL( L / 2) 2 wL2 ( L / 2) wL3 =− + − + C3 = − EI dx 6 2 2 8 3 wL ∴ C3 = 48 Next, substitute the deflection continuity condition into Eq. (d) for x = L/2 and solve for C4 w( L / 2) 4 wL( L / 2)3 wL2 ( L / 2) 2 wL3 7 wL4 EI v = − + − + ( L / 2) + C4 = − 24 6 4 48 192 4 wL ∴ C4 = − 384 Elastic curve equation for beam segment BC: wx 4 wLx3 wL2 x 2 wL3 x wL4 EI v = − + − − − 24 6 4 48 384 w ⎡⎣16 x 4 − 64 Lx 3 + 96 L2 x 2 − 8 L3 x + L4 ⎤⎦ ∴v=− 384 EI

( L / 2 ≤ x ≤ L)

(a) Elastic curve equations for entire beam: wLx 2 v=− (0 ≤ x ≤ L / 2) [9 L − 4 x ] 48 EI

v=−

w ⎡16 x 4 − 64 Lx 3 + 96 L2 x 2 − 8L3 x + L4 ⎤⎦ 384 EI ⎣

Ans.

( L / 2 ≤ x ≤ L)

(b) Deflection at B: 7 wL4 vB = − 192 EI (c) Deflection at free end of cantilever: w 41wL4 ⎡⎣16( L) 4 − 64 L( L)3 + 96 L2 ( L) 2 − 8L3 ( L) + L4 ⎤⎦ = − vC = − 384 EI 384 EI

Ans.

Ans.

Ans.

(d) Slope at free end of cantilever: dv 8w( L)3 24 wL( L) 2 24 wL2 ( L) wL3 7 wL3 EI =− + − + =− dx 48 48 48 48 48

dv 7 wL3 ∴ = θC = − dx C 48EI

Ans.

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10.18 For the beam and loading shown in Fig. P10.18, use the double-integration method to determine (a) the equation of the elastic curve for the beam, and (b) the deflection at B. Assume that EI is constant for the beam.

Fig. P10.18

Solution Beam FBD:

wL ⎛ L ⎞ ⎜ ⎟=0 2 ⎝4⎠ wL ∴Cy = 8 wL ΣFy = Ay + C y − =0 2 3wL ∴ Ay = 8

ΣM A = C y ( L ) −

Consider beam segment AB (0 ≤ x ≤ L/2) Moment equation:

wx 2 wx 2 3wL x=0 − Ay x = M ( x) + − 2 2 8 wx 2 3wLx ∴ M ( x) = − + 2 8

ΣM a − a = M ( x ) +

Integration of moment equation: d 2v wx 2 3wLx EI 2 = M ( x) = − + dx 2 8 3 2 dv wx 3wLx EI =− + + C1 dx 6 16 wx 4 wLx 3 EI v = − + + C1 x + C2 24 16

(a) (b)

Boundary conditions: v=0 at x=0 Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Slope at B: Let x = L/2 in Eq. (a). dv w( L / 2)3 3wL( L / 2) 2 wL3 3wL3 5wL3 EI = EIθ B = − + + C1 = − + + C1 = + C1 dx B 6 16 48 64 192

(c)

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Deflection at B: Let x = L/2 in Eq. (b). w( L / 2) 4 wL( L / 2)3 wL4 wL4 C1L wL4 C1L EI vB = − + + C1 ( L / 2) = − + + = + 24 16 384 128 2 192 2

(d)

Consider beam segment BC (L/2 ≤ x ≤ L) Moment equation: wL ( L − x) = 0 8 wL wL2 wLx ∴ M ( x) = ( L − x) = − 8 8 8

ΣM b − b = − M ( x ) + C y ( L − x ) = − M ( x ) +

Integration of moment equation: d 2v wLx wL2 EI 2 = M ( x) = − + dx 8 8 dv wLx 2 wL2 x EI =− + + C3 dx 16 8 wLx 3 wL2 x 2 EI v = − + + C3 x + C4 48 16

(e) (f)

Boundary conditions: v=0 at x=L Evaluate constants: Substitute x = L and v = 0 into Eq. (f) to find wL( L)3 wL2 ( L) 2 + + C3 ( L) + C4 EI (0) = − 48 16

wL4 ∴ C3 L + C4 = − 24 Slope at B: Let x = L/2 in Eq. (e). dv wL( L / 2) 2 wL2 ( L / 2) wL3 wL3 3wL3 EI = EI θ B = − + + C3 = − + + C3 = + C3 dx B 16 8 64 16 64 Deflection at B: Let x = L/2 in Eq. (f). wL( L / 2)3 wL2 ( L / 2) 2 5wL4 C3 L EI vB = − + + C3 ( L / 2) + C4 = + + C4 48 16 384 2 Continuity conditions: Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h): 5wL3 3wL3 + C1 = + C3 192 64

(g)

(h)

(i)

(j)

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Further, the deflection at B must be the same for both segments; therefore, equate Eqs. (d) and (i): wL4 C1L 5wL4 C3 L + = + + C4 192 2 384 2

(k)

Evaluate constants: Solve Eqs. (g), (j), and (k) simultaneously to determine the values of constants C1, C3, and C4: 9wL3 17 wL3 wL4 C1 = − C3 = − C4 = 384 384 384 (a) Elastic curve equation for beam segment AB: wx 4 wLx3 9 wL3 x + − EI v = − 24 16 384 wx ⎡⎣16 x 3 − 24 Lx 2 + 9 L3 ⎤⎦ ∴v=− 384 EI (a) Elastic curve equation for beam segment BC: wLx 3 wL2 x 2 17 wL3 x wL4 + − + EI v = − 48 16 384 384 wL ⎡⎣8 x 3 − 24 Lx 2 + 17 L2 x − L3 ⎤⎦ ∴v=− 384 EI (b) Deflection at B: wL4 9 wL4 5wL4 EI vB = − =− 192 768 768

(0 ≤ x ≤ L / 2)

Ans.

( L / 2 ≤ x ≤ L)

Ans.

5wL4 ∴ vB = − 768 EI

Ans.

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10.19 For the beam and loading shown in Fig. P10.19, use the double-integration method to determine (a) the equation of the elastic curve for the entire beam, (b) the deflection at C, and (c) the slope at B. Assume that EI is constant for the beam.

Fig. P10.19

Solution Beam FBD: L⎞ ⎛ ΣM A = By (3L) − wL ⎜ 3L + ⎟ = 0 2⎠ ⎝ 7 wL ∴ By = 6 ΣFy = Ay + By − wL = 0 ∴ Ay = −

wL 6

Consider beam segment AB (0 ≤ x ≤ 3L) Moment equation:

⎛ wL ⎞ ΣM a − a = M ( x) − Ay x = M ( x) + ⎜ ⎟x = 0 ⎝ 6 ⎠ wLx ∴ M ( x) = − 6 Integration of moment equation: d 2v wLx EI 2 = M ( x) = − dx 6 2 dv wLx EI =− + C1 dx 12 wLx 3 EI v = − + C1 x + C2 36 Boundary conditions: v=0 at x=0

and

(a) (b)

v=0

at

x = 3L

Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = 3L and v = 0 into Eq. (b) and solve for C1: wL(3L)3 9wL3 wL3 EI (0) = − + C1 (3L) ∴ C1 = = 36 36 4

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Slope at B: Let x = 3L in Eq. (a). dv wL(3L) 2 wL3 wL3 EI = EIθ B = − + = − dx B 12 4 2

(c)

Consider beam segment BC (3L ≤ x ≤ 4L) Moment equation:

w (4 L − x) 2 = 0 2 w(4 L − x) 2 ∴ M ( x) = − 2

ΣM b − b = − M ( x ) −

Integration of moment equation: d 2v w(4 L − x) 2 EI 2 = M ( x) = − dx 2 3 dv w(4 L − x) EI = + C3 6 dx w(4 L − x) 4 EI v = − + C3 x + C4 24 Boundary conditions: v=0 at x = 3L Substitute x = 3L and v = 0 into Eq. (f) to find w(4 L − 3L) 4 wL4 + C3 (3L) + C4 = − + C3 (3L) + C4 EI (0) = − 24 24 wL4 ∴ C4 = − (3L)C3 24 Slope at B: Let x = 3L in Eq. (e). dv w(4 L − 3L)3 wL3 EI = EI θ B = + C3 = + C3 dx B 6 6 Continuity conditions: Since the slope at B must be the same for both beam segments, equate Eqs. (c) and (h): wL3 wL3 2wL3 − = + C3 C3 = − 2 6 3 Backsubstitute this result into Eq. (g) to determine C4: ⎛ −2 wL3 ⎞ 49 wL4 wL4 wL4 − (3L)C3 = − (3L) ⎜ C4 = ⎟= 24 24 24 ⎝ 3 ⎠

(e) (f)

(g)

(h)

(i)

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(a) Elastic curve equation for beam segment AB: wLx 3 9 wL3 x + EI v = − 36 36 wLx 2 ⎡ x − 9 L2 ⎤⎦ ∴v=− (0 ≤ x ≤ 3L) 36 EI ⎣ (a) Elastic curve equation for beam segment BC: w(4 L − x) 4 2 wL3 x 49 wL4 − + EI v = − 24 3 24 w ⎡⎣(4 L − x) 4 + 16 L3 x − 49 L4 ⎤⎦ ∴v=− 24 EI

Ans.

(3L ≤ x ≤ 4 L)

Ans.

(b) Deflection at C: w w 15wL4 ⎡⎣(4 L − 4 L) 4 + 16 L3 (4 L) − 49 L4 ⎤⎦ = − ⎡⎣64 L4 − 49 L4 ⎤⎦ = − vC = − 24 EI 24 EI 24 EI

5wL4 ∴ vC = − 8EI (c) Slope at B: Let x = 3L in Eq. (a). dv wL3 EI = EIθ B = − 2 dx B

Ans.

∴

dv dx

= θB = − B

wL3 2 EI

Ans.

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10.20 For the beam and loading shown in Fig. P10.20, use the double-integration method to determine (a) the equation of the elastic curve for the beam, (b) the location of the maximum deflection, and (c) the maximum beam deflection. Assume that EI is constant for the beam.

Fig. P10.20

Solution Beam FBD:

w0 L ⎛ 2 L ⎞ ⎜ ⎟=0 2 ⎝ 3 ⎠ wL ∴ By = 0 3 wL ΣFy = Ay + By − 0 = 0 2 wL ∴ Ay = 0 6

ΣM A = B y L −

Moment equation:

ΣM a − a = M ( x ) + = M ( x) + ∴ M ( x) = −

w0 x 2 ⎛ x ⎞ ⎜ ⎟ − Ay x 2L ⎝ 3 ⎠ w0 x 2 ⎛ x ⎞ w0 Lx =0 ⎜ ⎟− 2L ⎝ 3 ⎠ 6

w0 x 3 w0 Lx + 6L 6

Integration of moment equation: d 2v w x 3 w Lx EI 2 = M ( x) = − 0 + 0 dx 6L 6 4 2 dv wx w Lx EI =− 0 + 0 + C1 24 L 12 dx w0 x 5 w0 Lx 3 EI v = − + + C1 x + C2 120 L 36

(a) (b)

Boundary conditions: v=0 at x=0 v=0 at x=L

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Evaluate constants: Substitute x = 0 and v = 0 into Eq. (b) to determine C2 = 0. Next, substitute x = L and v = 0 into Eq. (b) and solve for C1: w ( L)5 w0 L( L)3 7 w L3 EI (0) = − 0 + + C1 ( L) ∴ C1 = − 0 120 L 36 360 (a) Elastic curve equation: w0 x 5 w0 Lx3 7 w0 L3 x + − EI v = − 120 L 36 360

∴v=−

w0 x ⎡⎣3 x 4 − 10 L2 x 2 + 7 L4 ⎤⎦ 360 L EI

Ans.

(b) Location of maximum deflection: The maximum deflection occurs where the beam slope is zero. Therefore, set the beam slope equation [Eq. (a)] equal to zero: dv w x 4 w Lx 2 7 w0 L3 EI =− 0 + 0 − =0 dx 24 L 12 360 Multiply by −360L/w0 to obtain: 15 x 4 − 30 L2 x 2 + 7 L4 = 0 Solve this equation numerically to obtain: x = 0.51932962236L = 0.51933L Ans. (c) Maximum beam deflection: w (0.51933L) ⎡⎣3(0.51933L) 4 − 10 L2 (0.51933L) 2 + 7 L4 ⎤⎦ vmax = − 0 360 L EI =−

w0 (0.51933) w L4 (0.0065222) w0 L4 ⎡⎣ 4.52118 L4 ⎤⎦ = − = −0.00652 0 EI EI 360 EI

Ans.

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