Mecanica De Los Solidos

PROBLEM 7.24
A rod of weight W and uniform cross section is bent into the circular arc of radius r shown. Determine the bending moment at point J when θ = 120°.

SOLUTION
(a) FBD Rod: ΣFx = 0: Ax = 0 ΣM B = 0: rAy + 2r W 2r 2W − =0 π 3 π 3 2W 3π

Ay =

FBD AJ: Note:

α =

60° π = 30° = 2 6 60 2W = 270 9

Weight of segment = W F = r sin α =

α

r 3r sin 30° = π /6 π 2W 2W + ( r −r sin 30° ) −M =0 9 3π

ΣM J = 0: ( r cos α − r sin 30° )
2W 9

M =

 3r 3 r  3 1 3r  1  − + − +   = Wr   π 2   3π 2 2π  9 3π    

M = 0.1788Wr

PROBLEM 7.25
A quarter-circular rod of weight W and uniform cross section is supported as shown. Determine the bending moment at point J when

θ = 30o.

SOLUTION
FBD Rod:

ΣFx = 0: A x = 0 ΣM B = 0: 2r

π

W − rAy= 0

Ay =

2W

π

α = 15°, weight of segment = W
FBD AJ: r = r sin α = 2W

30° W = 90° 3

α

r sin15° = 0.9886r π /12 W cos 30° − F = 0 3

ΣFy′ = 0:

π
F=

cos 30° −

W 3  2 1  −  2 π 3

2W  W  ΣM 0 = M + r  F − + r cos15° =0 π  3  
M = 0.0557 Wr

PROBLEM 7.26
A quarter-circular rod of weight W and uniform cross section is supported as shown. Determinethe bending moment at point J when

θ = 30o.

SOLUTION
FBD Rod:

ΣM A = 0: rB −
B=

2r

π

W =0

2W

π

FBD BJ:

α = 15° =
r = r

π
12

π /12

sin15° = 0.98862r 30° W = 90° 3

Weight of segment = W ΣFy′ = 0: F −

W 2W cos 30° − sin 30° = 0 π 3

 3 1 + W F=  6 π  

ΣM 0 = 0: rF − ( r cos15° )

W −M =0 3

 3 1  cos15°  M = rW  +  −  0.98862  Wr 6   π 3  

M = 0.289Wr

PROBLEM 7.27
For the rod of Prob.7.26, determine the magnitude and location of the maximum bending moment.

SOLUTION
FBD Bar:

ΣM A = 0: rB −

2r

π

W =0

B=

2W

π

α =

θ
2

so r = r

0≤α ≤ sin α ,

π
4

α

Weight of segment = W = ΣFx′ = 0: F − F = =
FBD BJ:

2α π /2 W sin 2α = 0

4α

π

4α

π

W cos 2α −

2Wπ

2W

π
2W

( sin 2α ( sin θ

+ 2α cos 2α )

π

+ θ cos θ ) 4α W −M =0

ΣM 0 = 0: rF − ( r cos α ) M = 2

π

r  4α Wr ( sin θ + θ cos θ ) −  sin α cos α  W π α  π sin α cos α = M = 2Wr 1 1 sin 2α = sin θ 2 2 + θ cosθ − sin θ ) 2 Wrθ cosθ

But, so or

π

( sin θ
M =

π

dM 2 = Wr ( cos θ − θ sin θ ) = 0 at θ tan θ = 1 dθ π

PROBLEM 7.27 CONTINUED
θ =0.8603 rad

Solving numerically

and

M = 0.357Wr

at θ = 49.3° (Since M = 0 at both limits, this is the maximum)

PROBLEM 7.28
For the rod of Prob.7.25, determine the magnitude and location of the maximum bending moment.

SOLUTION
FBD Rod:

ΣFx = 0: Ax = 0 ΣM B = 0: 2r

π

W − rAy = 0

Ay =

2W

π

α =

θ
2

,

r =

r

α

sin α 2α 4α W = π /2 π 2W

Weightof segment = W ΣFx′ = 0: − F − F = 2W 4α

π

W cos 2α + = 2W

π

cos 2α = 0

π

(1 − 2α ) cos 2α

π

(1 − θ ) cosθ

FBD AJ:

2W  ΣM 0 = 0: M +  F − π  M = But, so 2W

4α  W =0  r + ( r cos α ) π  4αW r

π

(1 + θ cosθ

− cosθ ) r −

π

α

sin α cos α

sin α cos α = M = 2r

1 1 sin 2α = sin θ 2 2

π

W (1 − cosθ + θ cosθ − sin θ )

dM 2rW = ( sin θ −θ sin θ + cosθ − cosθ ) = 0 dθ π for dM =0 dθ Only 0 and 1 in valid range At for

(1 − θ ) sin θ

=0

θ = 0, 1, nπ ( n = 1, 2,

)

θ = 0 M = 0,
at θ = 57.3°

at θ = 1 rad M = M max = 0.1009 Wr

PROBLEM 7.29
For the beam and loading shown, (a) draw the shear and bendingmoment diagrams, (b) determine the maximum absolute values of the shear and bending moment.

SOLUTION
FBDbeam:

(a) By symmetry: Ay = D =
Along AB:

1 L ( w) 2 2

Ay = D =

wL 4

ΣFy = 0: ΣM J = 0: M − x
Along BC:

wL −V = 0 4 M =

V =

wL 4

wL =0 4

wL x (straight) 4

ΣFy = 0:

wL − wx1 − V = 0 4 wL − wx1 4 V =0 at x1 = L 4

V = straight with ΣM k = 0: M +
M =

x1 L  wL wx1 −  + x1  =0 2 4  4

w  L2 L 2   8 + 2 x1 − x1   2 

PROBLEM 7.29 CONTINUED...