(a) Determine an expression for the diameter of the truncated cone through the use of the similar triangles concept:
Next, consider equilibrium of the truncated cone:
( ) 0 ( )
Fx F x P F x P
The internal force F(x) can be expressed in terms of stress and area:
( ) ( ) ( )
A
F x
dA x A xEquating this expression to the equilibrium equation gives:
( ) ( ) ( )
From Eq. (a), the area of the cone at any location x is expressed as
2
(b) The elongation in the axial member is found from Eq. (5.5):
For the truncated cone axial member:
2 2 2
P5.15 Rigid bar ABCD is loaded and supported as shown in Figure P5.15. Bars (1) and (2) are unstressed before the load P is applied. Bar (1) is made of bronze [E = 100 GPa] and has a cross-sectional area of 520 mm2. Bar (2) is made of aluminum [E = 70 GPa] and has a cross-sectional area of 960 mm2. After the load P is applied, the force in bar (2) is found to be 25 kN (in tension).
Determine:
(a) the stresses in bars (1) and (2).
(b) the vertical deflection of point A.
(c) the load P.
FIGURE P5.15 Solution
Given that the axial force in bar (2) is 25 kN (in tension), the deformation can be computed as:
2 2
Since the pin at C is a perfect connection, the deflection of the rigid bar at C is equal to the deformation of bar (2):
2 0.297619 mm
vC
From a deformation diagram of the rigid bar, the vertical deflection of joint C is related to B by similar triangles:
Note that a downward displacement at B causes contraction (and hence, compression) in bar (1).
(a) The normal stress in aluminum bar (2) can be computed from the known force in the bar:
2
The normal stress in bronze bar (1) can be computed from its contraction:
1 1 1 1
( 0.952381 mm)(100,000 N/mm )
158.730159 MPa
(b) From a deformation diagram of the rigid bar, the vertical deflection of joint A is related to joint B by similar triangles:
2.0 m 0.5 m
2.0 m
4 4(0.297619 mm) 1.190476 mm 1.190 m
0.5 m m
equilibrium equation for the sum of moments about D can be used to determine the load P:
1 2
P5.16 In Figure P5.16, aluminum [E = 70 GPa] links (1) and (2) support rigid beam ABC. Link (1) has a cross-sectional area of 300 mm2 and link (2) has a cross-sectional area of 450 mm2. For an applied load of P = 55 kN, determine the rigid beam deflection at point B.
FIGURE P5.16 Solution
From a FBD of the rigid beam, write two equilibrium equations:
and backsubstitute into Eq. (a) to obtain F1:
1 55 kN 2 55 kN 35 kN 20 kN
F F
Next, determine the deformations in links (1) and (2):
1 1
Since the connections at A and C are perfect, the rigid beam deflections at these joints are equal to the deformations of links (1) and (2), respectively:
1 2.380952 mm and 2 4.444444 mm
A C
v v
The deflection at B can be determined
1, 400 mm
( )
2,200 mm 1, 400 mm
(4.444444 mm 2.380952 mm) 2.380952 mm 2,200 mm
(0.636364)(2.053492 mm) 2.3 3.6
80952 mm
= 9 mm
B C A A
v v v v
Ans.
P5.17 Rigid bar ABC is supported by bronze rod (1) and aluminum rod (2), as shown in Fig P5-17. A concentrated load P is applied to the free end of aluminum rod (3). Bronze rod (1) has an elastic modulus of E1 = 15,000 ksi and a diameter of d1 = 0.50 in. Aluminum rod (2) has an elastic modulus of E2 = 10,000 ksi and a diameter of d2 = 0.75 in. Aluminum rod (3) has a diameter of d3 = 1.0 in. The yield strength of the bronze is 48 ksi and the yield strength of the aluminum is 40 ksi.
(a) Determine the magnitude of load P that can safely be applied to the structure if a minimum factor of safety of 1.67 is required.
(b) Determine the deflection of point D for the load determined
Before beginning, the cross-sectional areas of the three rods can be calculated from the specified diameters:
2 2 2
1 0.196350 in. 2 0.441786 in. 3 0.785398 in.
A A A
The allowable stress of the bronze is
,bronze
and the allowable stress of the aluminum is
,alum
(a) From a FBD cut through rod (3), equilibrium requires that the internal force in rod (3) is F3 = P.
From a FBD of the rigid bar, write two equilibrium equations:
1 2 3 1 2 0
Backsubstitute this expression into Eq. (a) to obtain F1 in terms of the unknown load P:
1 2 0.6250 0.3750
F P F P P P (d)
Based on the allowable stress for the bronze, the maximum allowable force that can be supported by rod (1) is:
2
1 allow,1 1 (28.742515 ksi)(0.196350 in. ) 5.643579 kips
F A
Thus, from Eq. (f), the corresponding maximum load P is:
2.6667 1 2.6667(5.643579 kips) 15.049545 kips
P F (g)
Next, the maximum allowable force that can be supported by rod (2) based on the allowable stress for the aluminum is:
2
2 allow,2 2 (23.952096 ksi)(0.441786 in. ) 10.581711 kips
F A
which leads to a corresponding maximum load P from Eq. (e):
1.6000 2 1.6000(10.581711 kips) 16.930738 kips
P F (h)
Finally, we should also check the capacity of rod (3):
2
3 allow,3 3 (23.952096 ksi)(0.785398 in. ) 18.811931 kips
F A
which means that 18.811931 kips
P (i)
From a comparison of Eqs. (g), (h), and (i), the maximum load that may be applied to the structure is
max 15.049545 kips 15.05 kips
P Ans.
Based on this maximum load P, the internal forces in the three rods are:
1
2 3
0.3750 0.3750(15.049545 kips) 5.643579 kips 0.6250 0.6250(15.049545 kips) 9.405966 kips
15.049545 kips
(b) Next, determine the deformations in rods (1), (2), and (3):
1 1
1 2
1 1
(5.643579 kips)(6 ft)(12 in./ft)
0.137964 in.
(0.196350 in. )(15,000 ksi) F L
A E
2 2
2 2
2 2
(9.405966 kips)(8 ft)(12 in./ft)
0.204391 in.
(0.441786 in. )(10,000 ksi) F L
A E
3 3
3 2
3 3
(15.049545 kips)(3 ft)(12 in./ft)
0.068982 in.
(0.785398 in. )(10,000 ksi) F L
A E
Since the connections at A and C are perfect, the rigid bar deflections at these joints are equal to the deformations of rods (1) and (2), respectively:
1 0.137964 in. and 2 0.204391 in.
A C
v v
Solve this expression for vB: 2.5 ft
( )
4 ft 2.5 ft
(0.204391 in. 0.137964 in.) 0.137964 in.
4 ft
(0.6250)(0.066427 in.) 0.137964 in.
0.179481 in.
The deflection of joint D is equal to the deflection of the rigid bar at B plus the deformation in rod (3):
3 0.179481 in. 0.068982 in. 0.248463 in. 0.248 in.
D B
v v Ans.
(c) The pin at B has an allowable shear stress of
allow
The force tending to shear this pin is equal to the load P = 15.049545 kips. The shear area AV required for this pin is thus
Since the pin is used in a double shear connection, the shear area is equal to twice the cross-sectional area of the pin:
and so the minimum pin diameter is
2 2
pin pin
2 0.836086 in. 0.729568 in.
4d d 0.730 in.
Ans.
P5.18 The truss shown in Figure P5.18 is constructed from three aluminum alloy members, each having a cross-sectional area of A = 850 mm2 and an elastic modulus of E = 70 GPa. Assume that a = 4.0 m, b = 10.5 m, and c = 6.0 m. Calculate the horizontal displacement of roller B when the truss supports a load of P = 12 kN.
FIGURE P5.18 Solution
To determine the horizontal displacement of roller B, we need only determine the deformation in member AB. This requires calculation of the internal force in member AB.