Resistencia De Materiales

PROBLEM 7.45
Two short angle sections CE and DF are bolted to the uniform beam AB of weight 3.33 kN, and the assembly is temporarily supported by the vertical cables EG and FH as shown. A second beam resting on beam AB at I exerts a downward force of 3 kN on AB. Knowing that a = 0.3 m and neglecting the weight of the ngle sections, (a) draw the shear and bending-moment diagrams for beam AB, (b)determine the maximum absolute values of the shear and bending moment in the beam.

SOLUTION FBD angle CE:

(a) By symmetry:

T =

3.33 kN + 3 kN = 3.165 kN 2 PC = T = 3.165 kN M C = 0.3165 kN ⋅ m

ΣFy = 0: T − PC = 0

ΣM C = 0: M C − ( 0.1 m )( 3.165 kN ) = 0 By symmetry:

PD = 3.165 kN; M D = 0.3165 kN ⋅ m

Along AC:

ΣFy = 0: − x (1.11 kN/m ) − V = 0 V = − (1.11 kN/m ) x V =−1.332 kN at C

( x = 1.2 m )

ΣM J = 0: M + M = ( 0.555 kN/m ) x 2

x (1.11 kN/m ) x = 0 2 M = − 0.7992 kN ⋅ m at C

Along CI:

ΣFy = 0: − (1.11 kN/m ) x + 3.165 kN − V = 0 V = 3.165 kN − (1.11 kN/m ) x ΣM k = 0: M + (1.11 kN/m ) x − ( x − 1.2 m )( 3.165 kN ) − ( 0.3165 kN ⋅ m ) = 0 V = 1.5 kN at I

( x = 1.5 m )

PROBLEM 7.45 CONTINUED
M = 3.4815 kN ⋅ m + ( 3.165 kN ) x − ( 0.555kN/m ) x 2 M = − 0.4827 kN ⋅ m at C M = 0.01725 kN ⋅ m at I

Note: At I, the downward 3 kN force will reduce the shear V by 3 kN, from +1.5 kN to –1.5 kN, with no change in M. From I to B, the diagram can be completed by symmetry. (b) From diagrams:
V M
max

= 1.833 kN at C and D = 799 N ⋅ m at C and D

max

PROBLEM 7.46
Solve Prob. 7.45 when a = 0.6 m.

SOLUTION FBD angle CE:

(a)By symmetry:

T =

3.33 kN + 3 kN = 3.165 kN 2 PC = T = 3.165 kN M C = 0.3165 kN ⋅ m

ΣFy = 0: T − PC = 0

ΣM C = 0: M C − ( 0.1 m )( 3.165 kN ) = 0

By symmetry:
Along AC:

PD = 3.165 kN

M D = 0.3165 kN ⋅ m

ΣFy = 0: − (1.11 kN/m ) x − V = 0 V = − (1.11 kN/m ) x V = − 0.999 kN at C

( x = 0.9 m )

ΣM J = 0: M + M = − ( 0.555 kN/m ) x 2

x (1.11 kN/m ) x = 0 2 M = − 0.44955kN ⋅ m at C

Along CI:

ΣFy = 0: − x (1.11 kN/m ) + 3.165 kN − V = 0 V = 3.165 kN − (1.11 kN/m ) x V = 1.5 kN at I ΣM K = 0: M − 0.3165 kN ⋅ m + ( x − 0.9 m )( 3.165 kN ) + x (1.11 kN/m ) x = 0 2 V = 2.166 kN at C

( x = 1.5 m )

PROBLEM 7.46 CONTINUED
M = −2.532 kN ⋅ m + ( 3.165 kN ) x − ( 0.555 kN/m ) x 2 M = − 0.13305 kN ⋅ m at C M = 0.96675 kN ⋅ m at I

Note: At I, the downward 3kN force will reduce the shear V by 3 kN, from +1.5 kN to –1.5 kN, with no change in M. From I to B, the diagram can be completed by symmetry. (b) From diagrams:
V
max

= 2.17 kN at C and D M
max

= 967 N ⋅ m at I

PROBLEM 7.47
Draw the shear and bending-moment diagrams for the beam AB, and determine the shear and bending moment (a) just to the left of C, (b) just to the right of C.SOLUTION FBD CD:
ΣFy = 0: −1.2 kN + C y = 0 ΣM C = 0: ( 0.4 m )(1.2 kN ) − M C = 0
C y = 1.2 kN

M C = 0.48 kN ⋅ m

FBD Beam:
ΣM A = 0: (1.2 m ) B + 0.48 kN ⋅ m − ( 0.8 m )(1.2 kN ) = 0

B = 0.4 kN
ΣFy = 0: Ay − 1.2 kN + 0.4 kN = 0

A y = 0.8 kN

Along AC:

ΣFy = 0: 0.8 kN − V = 0 ΣM J = 0: M − x ( 0.8 kN ) = 0

V = 0.8 kN M = ( 0.8 kN ) x

M = 0.64 kN ⋅ m at x = 0.8 m

AlongCB:

ΣFy = 0: V + 0.4 kN = 0 ΣM K = 0: x1 ( 0.4 kN ) − M = 0

V = −0.4 kN M = ( 0.4 kN ) x1

M = 0.16 kN ⋅ m at x1 = 0.4 m

(a)

Just left of C:

V = 800 N M = 640 N ⋅ m

(b)

Just right of C:

V = −400 N M = 160.0 N ⋅ m

PROBLEM 7.48
Draw the shear and bending-moment diagrams for the beam AB, and determine the maximum absolute values of the shear and bending moment.SOLUTION FBD angle:
ΣFy = 0: Fy − 600 N = 0 Fy = 600 N M = 180 N ⋅ m

ΣM Base = 0: M − ( 0.3 m )( 600 N ) = 0

All three angles are the same.
FBD Beam:
ΣM A = 0: (1.8 m ) B − 3 (180 N ⋅ m ) − ( 0.3 m + 0.9 m + 1.5 m )( 600 N ) = 0

B = 1200 N
ΣFy = 0: Ay − 3 ( 600 N ) + 1200 N = 0
A y = 600 N

Along AC:

ΣFy = 0: 600 N − V = 0

V = 600 N

ΣM J = 0: M − x ( 600 N ) = 0 M = ( 600 N...