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7.7 For the simply supported beam subjected to the loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Report the maximum bending moment and its location.

Fig. P7.7

Solution Beam equilibrium:

(50 kN)(3 m) (75 kN)(6 m) (10 m) 0

60 kN

50 kN 75 kN 0

65 kN

A y

y

y y y

y

M DD

F A DA

Σ = − − + =

∴ =

Σ = + − − =

∴ =

Section a-a: For the interval 0 ≤ x < 3 m:

65 kN 0 65 kN

(65 kN) 0 (65 kN)y y

a a y

F A V V V

M A x M x M M x−

Σ = − = − = ∴ =

Σ = − + = − + = ∴ =

Section b-b: For the interval 3 m ≤ x < 6 m:

50 kN 65 kN 50 kN 0

15 kNy yF A V V

V

Σ = − − = − − =

∴ =

(50 kN)( 3 m)

(65 kN) (50 kN)( 3 m) 0

(15 kN) 150 kN-m

b b yM A x x Mx x M

M x

−Σ = − + − +

= − + − + =

∴ = +

Section c-c: For the interval 6 m ≤ x < 10 m:

50 kN 75 kN

65 kN 50 kN 75 kN 0

60 kN

y yF A VV

V

Σ = − − −

= − − − =

∴ = −

(50 kN)( 3 m) (75 kN)( 6 m)

(65 kN) (50 kN)( 3 m) (75 kN)( 6 m) 0

(60 kN) 600 kN-m

c c yM A x x x Mx x x M

M x

−Σ = + − + − +

= − + − + − + =

∴ = − +

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(b) Shear-force and bending-moment diagrams

(c) Maximum bending moment and its location Mmax = 240 kN-m @ x = 6 m

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7.8 For the simply supported beam subjected to theloading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam.(Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Report the maximum positive bending moment,the maximum negative bending moment, and their respective locations.

Fig. P7.8

Solution Beam equilibrium:

(20 kN)(2 m) (60 kN)(6 m) (8 m) 0

40 kN

20 kN 60 kN 0

40 kN

B y

y

y y y

y

M DD

F B DB

Σ = − + =

∴ =

Σ = + − − =

∴ =

Section a-a: For the interval 0 ≤ x < 2 m:

20 kN 0 20 kN

(20 kN) 0 (20 kN)y

a a

F V V

M x M M x−

Σ = − − = ∴ = −

Σ = + = ∴ = −

Section b-b: For the interval 2 m ≤ x < 8 m:

20 kN 20 kN 40 kN 0

20 kN(20 kN) ( 2 m)

(20 kN) (40 kN)( 2 m) 0

(20 kN) 80 kN-m

y y

b b y

F B V V

VM x B x M

x x M

M x

−

Σ = − + − = − + − =

∴ =Σ = − − +

= − − + =

∴ = −

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Section c-c: For the interval 8 m ≤ x < 10 m:

20 kN 60 kN

20 kN 40 kN 60 kN 0

40 kN

y yF B VV

V

Σ = − + − −

= − + − − =

∴ = −

(20 kN) ( 2 m) (60 kN)( 8 m)

(20 kN) (40 kN)( 2 m) (60 kN)( 8 m) 0

(40 kN) 400 kN-m

c c yM x B x x Mx x x M

M x

−Σ = − − + − +

= − − + − + =

∴ = − +

(b) Shear-force and bending-moment diagrams

(c) Maximum bending moment and its location Mmax-positive = 80 kN-m @ x = 8 m Mmax-negative = –40 kN-m @ x = 2 m

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7.9 For the simply supported beam subjected to theloading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam.(Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Report the maximum positive bending moment,the maximum negative bending moment, and theirrespective locations.

Fig. P7.9

Solution Beam equilibrium:

(7 kips/ft)(30 ft)(15 ft) (21 ft) 0

150 kips

(7 kips/ft)(30 ft) 0

60 kips

C y

y

y y y

y

M BB

F B CC

Σ = − =

∴ =

Σ = + − =

∴ =

Section a-a: For the interval 0 ≤ x < 9 ft:

2

(7 kips/ft) 0 (7 kips/ft)

(7 kips/ft)(7 kips/ft)( ) 02 2

y

a a

F x V V x

xM x M M x−

Σ = − − = ∴ = −

⎛ ⎞Σ = + = ∴ = −⎜ ⎟⎝ ⎠

Section b-b: For the interval 9 ft ≤ x < 30 ft:

(7 kips/ft) (7 kips/ft) 150 kips 0

(7 kips/ft) 150 kipsy yF x B V x V

V x

Σ = − + − = − + − =

∴ = − +

2

(7 kips/ft)( ) ( 9 ft)2

(7 kips/ft)( ) (150 kips)( 9 ft) 02

(7 kips/ft) (150 kips) 1,350 kips2

b b yxM x B x M

xx x M

M x x

−⎛ ⎞Σ = − − +⎜ ⎟⎝ ⎠⎛ ⎞= − − + =⎜ ⎟⎝ ⎠

∴ = − + −

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(b) Shear-force and bending-moment diagrams

(c) Maximum bending moment and its location Mmax-positive = 257.14 kip-ft @ x = 21.43 ft Mmax-negative = –283.50 kip-ft @ x = 9 ft

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7.10 For the cantilever beam and loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam.(Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. Fig. P7.10

Solution Beam equilibrium:

(4 kips/ft)(8 ft) 0

32 kips

(4 kips/ft)(8 ft)(12 ft) 0384 kip-ft

y y

y

C C

C

F CC

M MM

Σ = + =

∴ = −

Σ = − + =∴ =

Section a-a: For the interval 0 ≤ x < 8 ft:

2

(4 kips/ft) 0 (4 kips/ft)

4 kips/ft(4 kips/ft)( ) 02 2

y

a a

F x V V x

xM x M M x−

Σ = − = ∴ =

⎛ ⎞Σ = − + = ∴ =⎜ ⎟⎝ ⎠

Section b-b: For the interval 8 ft ≤ x < 16 ft:

(4 kips/ft)(8 ft) 0

32 kips

(4 kips/ft)(8 ft)( 4 ft) 0

(32 kips) 128 kip-ft

y

b b

F V

V

M x M

M x−

Σ = − =

∴ =

Σ = − − + =

∴ = −

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(b) Shear-force and bending-moment diagrams

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7.11 For the simply supported beam subjected to the loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Report the maximum bending moment and its location.

Fig. P7.11

Solution Beam equilibrium:

(42 kips)(10 ft) (6 kips/ft)(20 ft)(20 ft)

(30 ft) 0

94 kips

A

y

y

MCC

Σ = − −+ =

∴ =

42 kips (6 kips/ft)(20 ft) 0

68 kipsy y y

y

F A CA

Σ = + − − =

∴ =

Section a-a: For the interval 0 ≤ x < 10 ft:

68 kips 0 68 kips

(68 kips) 0 (68 kips)

y y

a a y

F A V V V

M A x M x M M x−

Σ = − = − = ∴ =

Σ = − + = − + = ∴ =

Section b-b:

For the interval 10 ft ≤ x < 30 ft:

42 kips (6 kips/ft)( 10 ft)

68 kips 42 kips (6 kips/ft)( 10 ft) 0

(6 kips/ft) 86 kips

y yF A x Vx V

V x

Σ = − − − −

= − − − − =

∴ = − +

2

2

10 ft(42 kips)( 10 ft) (6 kips/ft)( 10 ft)2

6 kips/ft(68 kips) (42 kips)( 10 ft) ( 10 ft) 02

3 86 120 kip-ft

b b yxM A x x x M

x x x M

M x x

−−⎛ ⎞Σ = − + − + − +⎜ ⎟

⎝ ⎠⎛ ⎞= − + − + − + =⎜ ⎟⎝ ⎠

∴ = − + +

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(b) Shear-force and bending-moment diagrams

(c) Maximum bending moment and its location Mmax = 736.33 kip-ft @ x = 14.33 ft

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7.12 For the simply supported beam subjected to the loading shown, (a) Derive equations for the shear force V and the bending moment M for any location in the beam. (Place the origin at point A.) (b) Plot the shear-force and bending-moment diagrams for the beam using the derived functions. (c) Report the maximum positive bending moment, the maximum negative bending moment, and their respective locations.

Fig. P7.12

Solution Beam equilibrium:

180 kN-m (9 m) (36 kN)(12 m) 0

28 kN

36 kN 0

8 kN

A y

y

y y y

y

M CC

F A CA

Σ = + − =

∴ =

Σ = + − =

∴ =

Section a-a: For the interval 0 ≤ x < 4 m:

8 kN 0 8 kN

(8 kN) 0 (8 kN)y y

a a y

F A V V V

M A x M x M M x−

Σ = − = − = ∴ =

Σ = − + = − + = ∴ =

Section b-b: For the interval 4 m ≤ x < 9 m:

8 kN 0 8 kN

180 kN-m

(8 kN) 180 kN-m 0

(8 kN) 180 kN-m

y y

x y

F A V V VM A x M

x M

M x

Σ = − = − = ∴ =

Σ = − + +

= − + + =

∴ = −

Section c-c:

For the interval 9 m ≤ x < 12 m:

8 kN 28 kN 0

36 kNy y yF A C V V

V

Σ = + − = + − =

∴ =

( 9 m) 180 kN-m

(8 kN) (28 kN)( 9 m) 180 kN-m 0

(36 kN) 432 kN-m

c c y yM A x C x Mx x M

M x

−Σ = − − − + +

= − − − + + =

∴ = −

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(b) Shear-force and bending-moment diagrams

(c) Maximum bending moment and its location Mmax-positive = 32 kN-m @ x = 4 m Mmax-negative = –148 kN-m @ x = 4 m