Member (1) is a two-force member that is oriented at with respect to the horizontal axis:
30 in.
tan 0.75 36.870
40 in.
From a FBD of rigid structure ABC, the following equilibrium equations can be written:
1cos 36.870 0
(56 in.)sin 15,000 lb
36. (C)
F 870
Ans.
(b) The normal stress magnitude in member (1) is
1
Therefore, its factor of safety with respect to the 36,000 psi yield stress is
1
2 2 2 2
( 12,000 lb) ( 2,700 lb) 12,300 lb
x y
C C C Ans.
(d) The allowable shear stress for the pins at C is
allow
60,000 psi
20,000 psi
FS 3.0
U
The double-shear pin connection at C must support a load of 12,300 lb. The minimum shear area AV required to support this load is
2
allow
12,300 lb
0.6150 in.
20,000 psi
V
A C
The pin diameter can be computed from
2 2
pin pin
(2 surfaces) 0.6150 in. 0.625717 in. 0.626 in.
4d d
Ans.
P4.17 A rectangular steel plate is used as an axial member to support a dead load of 70 kips and a live load of 110 kips. The yield strength of the steel is 50 ksi.
(a) Use the ASD method to determine the minimum cross-sectional area required for the axial member if a factor of safety of 1.67 with respect to yielding is required.
(b) Use the LRFD method to determine the minimum cross-sectional area required for the axial member based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.
Solution
(a) The service load on the axial member is 70 kips 110 kips 180 kips P D L
The allowable normal stress is
allow
The minimum cross-sectional area required to support the service load is
min
allow
180 kips 2
29.940 ks 6.01 in
i .
A P
Ans.
(b) The ultimate load for LRFD is
1.2 1.6 1.2(70 kips) 1.6(110 kips) 260 kips
U D L
The design equation for an axial member subjected to tension can be written in LRFD as
t YA U
Consequently, the minimum cross-sectional area required for the tension member is
2
P4.18 A 20-mm-thick steel plate will be used as an axial member to support a dead load of 150 kN and a live load of 220 kN. The yield strength of the steel is 250 MPa.
(a) Use the ASD method to determine the minimum plate width b required for the axial member if a factor of safety of 1.67 with respect to yielding is required.
(b) Use the LRFD method to determine the minimum plate width b required for the axial member based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.
Solution
(a) The service load on the axial member is 150 kN 220 kN 370 kN P D L
The allowable normal stress is
allow
The minimum cross-sectional area required to support the service load is
2
Since the plate is 20-mm-thick, the minimum plate width is therefore:
2
(b) The ultimate load for LRFD is
1.2 1.6 1.2(150 kN) 1.6(220 kN) 532 kN
U D L
The design equation for an axial member subjected to tension can be written in LRFD as
t YA U
Consequently, the minimum cross-sectional area required for the tension member is
2
Since the plate is 20-mm-thick, the minimum plate width is therefore:
2
P4.19 A round steel tie rod is used as a tension member to support a dead load of 30 kips and a live load of 15 kips. The yield strength of the steel is 46 ksi.
(a) Use the ASD method to determine the minimum diameter required for the tie rod if a factor of safety of 2.0 with respect to yielding is required.
(b) Use the LRFD method to determine the minimum diameter required for the tie rod based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.
Solution
(a) The service load on the axial member is 30 kips 15 kips 45 kips P D L
The allowable normal stress is
allow
The minimum cross-sectional area required to support the service load is
2
and thus, the minimum tie rod diameter is
2 2
min 1.956522 in. min 1.578 in
4d d .
Ans.
(b) The ultimate load for LRFD is
1.2 1.6 1.2(30 kips) 1.6(15 kips) 60 kips
U D L
The design equation for an axial member subjected to tension can be written in LRFD as
t YA U
Consequently, the minimum cross-sectional area required for the tension member is
2
and thus, the minimum tie rod diameter is
2 2
min 1.449275 in. min 1.358 in
4d d .
Ans.
P4.20 A round steel tie rod is used as a tension member to support a dead load of 190 kN and a live load of 220 kN. The yield strength of the steel is 320 MPa.
(a) Use the ASD method to determine the minimum diameter D required for the tie rod if a factor of safety of 2.0 with respect to yielding is required.
(b) Use the LRFD method to determine the minimum diameter D required for the tie rod based on yielding of the gross section using the LRFD method. Use a resistance factor of t = 0.9 and load factors of 1.2 and 1.6 for the dead and live loads, respectively.
Solution
(a) The service load on the axial member is 190 kN 220 kN 410 kN P D L
The allowable normal stress is
allow
The minimum cross-sectional area required to support the service load is
2
and thus, the minimum tie rod diameter is
2 2
min 2,562.500 mm min
4d d 57.1 mm
Ans.
(b) The ultimate load for LRFD is
1.2 1.6 1.2(190 kN) 1.6(220 kN) 580 kN
U D L
The design equation for an axial member subjected to tension can be written in LRFD as
t YA U
Consequently, the minimum cross-sectional area required for the tension member is
2
and thus, the minimum tie rod diameter is
2 2
min 2,013.889 mm min
4d d 50.6 mm
Ans.
P5.1 A steel [E = 200 GPa] rod with a circular cross section is 7.5-m long. Determine the minimum diameter required if the rod must transmit a tensile force of 50 kN without exceeding an allowable stress of 180 MPa or stretching more than 5 mm.
Solution
If the normal stress in the rod cannot exceed 180 MPa, the cross-sectional area must equal or exceed
2 2
(50 kN)(1,000 N/kN)
277.7778 mm 180 N/mm
A P
If the elongation must not exceed 5 mm, the cross-sectional area must equal or exceed
2 2
(50 kN)(1,000 N/kN)(7,500 mm)
375.0000 mm (200,000 N/mm )(5 mm)
A PL E
Therefore, the minimum cross-sectional area that may be used for the rod is Amin = 375 mm2. The corresponding rod diameter is
2 2
rod 375 mm rod 21.8510 mm 2 m
4d d 1.9 m
Ans.
P5.2 An aluminum [E = 10,000 ksi] control rod with a circular cross section must not stretch more than 0.25 in. when the tension in the rod is 2,200 lb. If the maximum allowable normal stress in the rod is 12 ksi, determine:
(a) the smallest diameter that can be used for the rod.
(b) the corresponding maximum length of the rod.
Solution
(a) If the normal stress in the rod cannot exceed 20 ksi, the cross-sectional area must equal or exceed 2.20 kips 2
0.1833 in.
12 ksi A P
The corresponding rod diameter is
2 2
rod 0.1833 in. rod 0.483
4d d in.
Ans.
(b) If the elongation must not exceed 0.25 in., the aluminum control rod having a cross-sectional area of A = 0.1833 in.2 can have a length no greater than
(0.1833 in. )(10,000 ksi)(0.25 in.)2
208.3333 in.
2.20 kips 208 in.
L AE P
Ans.
P5.3 A 12-mm-diameter steel [E = 200 GPa] rod (2) is connected to a 30-mm-wide by 8-mm-thick rectangular aluminum [E = 70 GPa] bar (1), as shown in Figure P5.3. Determine the force P required to stretch the assembly 10.0 mm.
FIGURE P5.3 Solution
The elongations in the two axial members are expressed by
1 1 2 2
The total elongation of the assembly is thus
1 1 2 2
Since the internal forces F1 and F2 are equal to external load P, this expression can be simplified to
1 2
For rectangular aluminum bar (1), the cross-sectional area is
2 1 (30 mm)(8 mm) 240 mm
A
and the cross-sectional area of steel rod (2) is
2 2
2 (12 mm) 113.0973 mm A 4
The force P required to stretch the assembly 10.0 mm is thus
1 2
P5.4 A rectangular bar of length L has a slot in the central half of its length, as shown in Figure P5.4.
The bar has width b, thickness t, and elastic modulus E. The slot has width b/3. If L = 400 mm, b = 45 mm, t = 8 mm, and E = 72 GPa, determine the overall elongation of the bar for an axial force of P = 18 kN.
FIGURE P5.4 Solution
The internal force in the bar is equal to the external load P. The unslotted portions of the bar have an area of
2
unslotted (45 mm)(8 mm) 360 mm
A
The slotted portion of the bar has an area of
2 slotted
45 mm
45 mm (8 mm) 240 mm
A 3 The overall elongation of the bar is thus
unslotted slotted
2 2 2
unslotted slotted
18,000 N 200 mm 200 mm
72,000 N/mm 360 mm 240 mm 0.347 mm
P L L
E A A
Ans.
P5.5 An axial member consisting of two polymer bars is supported at C as shown in Figure P5.5. Bar (1) has a cross-sectional area of 540 mm2 and an elastic modulus of 28 GPa. Bar (2) has a cross-sectional area of 880 mm2 and an elastic modulus of 16.5 GPa. Determine the deflection of point A relative to support C.
FIGURE P5.5 Solution
Draw a FBD that cuts through member (1) to find that the internal axial force in member (1) is
1 35 kN (T) F
Similarly, draw a FBD that cuts through member (2) and includes the free end of the axial member.
From this FBD, the equilibrium equation is 35 kN 50 kN 50 kN 2 0
Fx F
Therefore, the internal axial force in member (2) is
2 65 kN 65 kN (C) F
The deformation in bar (1) can be computed as
1 1
1 2 2
1 1
(35 kN)(1,000 N/kN)(850 mm)
1.9676 mm (540 mm )(28, 000 N/mm )
F L
A E
and the deformation in bar (2) can be computed as
2 2
2 2 2
2 2
( 65 kN)(1,000 N/kN)(1,150 mm)
5.1481 mm (880 mm )(16, 500 N/mm )
F L
A E
The deflection of point A relative to the support at C is the sum of these two deformations:
1 2 1.9676 mm ( 5.1481 mm) 3.18 mm 3.18 mm
uA Ans.
P5.6 The roof and second floor of a building are supported by the column shown in Figure P5.6. The column is a structural steel W10 × 60 wide-flange section [E = 29,000 ksi; A = 17.6 in2]. The roof and floor subject the column to the axial forces shown. Determine:
(a) the amount that the first floor will settle.
(b) the amount that the roof will settle.
FIGURE P5.6
Solution
(a) Draw a FBD that cuts through member (1) and includes the free end of the column. From this FBD, the sum of forces in the
vertical direction can be written as 110 kips 155 kips 1 0
Fy F
Therefore, the internal axial force in member (1) is
1 270 kips 270 kips (C)
F
The settlement of the first floor is determined by the deformation (i.e., contraction in this case) that occurs in member (1).
1 1 1
1 1
2
( 270 kips)(16 ft)(12 in./ft) (17.6 in. )(29, 000 ksi) 0.101567 in. 0.1016 in.
B
(b) Draw a FBD that cuts through member (2) and includes the free end of the column. From this FBD, the equilibrium equation is
115 kips 2 0
Fx F
Therefore, the internal axial force in member (2) is
P5.7 Aluminum [E = 70 GPa] member ABC supports a load of 28 kN, as shown in Figure P5.7. Determine:
(a) the value of load P such that the deflection of joint C is zero.
(b) the corresponding deflection of joint B.
FIGURE P5.7 Solution
Cut a FBD that exposes the internal axial force in member (2):
2 2
28 kN 0 28 kN
Fy F F
Similarly, cut a FBD that exposes the internal axial force in member (1):
1 1
28 kN 0 28 kN
Fy P F F P
The deflection at joint C, which must ultimately equal zero, can be expressed in terms of the member deformations 1 and 2:
(28,000 N )(1,000 mm) (28,000 N)(1,300 mm)
0
The only unknown in this equation is P, which can be computed as 80.5841 kN 80. kN6
P Ans.
The corresponding deflection at joint B can be found from the deformation in member (1):
1 1 1
2 2
1 1
(28, 000 N 80,584 N)(1,300 mm) (50 mm) (70, 000 N/m
P5.8 A solid brass [E = 100 GPa] axial member is loaded and supported as shown in Figure P5.8. Segments (1) and (2) each have a diameter of 25 mm and segment (3) has a diameter of 14 mm.
Determine:
(a) the deformation of segment (2).
(b) the deflection of joint D with respect to the fixed support at A.
(c) the maximum normal stress in the entire axial member.
FIGURE P5.8 Solution
(a) Draw a FBD that cuts through segment (2) and includes the free end of the axial member. From this FBD, the sum of forces in the vertical direction reveals the internal force in the segment:
2 14 kN 28 kN 0 2 42 kN (T)
Fy F F
The cross-sectional area of the 25-mm-diameter segment is
2 2
2 (25 mm) 490.8739 mm A 4
The deformation in segment (2) is thus
2 2
2 2 2
2 2
(42,000 N)(1,200 mm) (490.8739 mm )(100, 000 N/mm ) 1.026740 mm 1.027 mm
F L
A E
Ans.
(b) The internal forces in segments (1) and (3) must be determined at the outset. From a FBD that cuts through segment (1) and includes the free end of the axial member:
1 40 kN 14 kN 28 kN 0 1 82 kN (T)
Fy F F
The deformation in segment (1) can be computed as
1 1
1 2 2
1 1
(82,000 N)(1,800 mm)
3.006883 mm (490.8739 mm )(100, 000 N/mm )
F L
A E
Similarly, consider a FBD that cuts through segment (3) and includes the free end of the axial member:
3 28 kN 0 3 28 kN (T)
Fy F F
The cross-sectional area of the 14-mm-diameter segment is
2 2
3 (164 mm) 153.9380 mm A 4
The deformation in segment (3) can be computed as
3 3
3 2 2
3 3
(28,000 N)(1,600 mm)
2.810262 mm (153.9380 mm )(100, 000 N/mm )
F L
A E
The deflection of joint D with respect to the fixed support at A is found from the sum of the three segment deformations:
1 2 3
3.006883 mm 1.026740 mm 2.810262 mm 6.943885 mm 6.94 mm
uD
Ans.
(c) Since segments (1) and (2) have the same cross-sectional area, the maximum normal stress in these two segments occurs where the axial force is greater; that is, in segment (1):
1
The normal stress in segment (3) is
3
Therefore, the maximum normal stress in the axial member occurs in segment (3):
max 181.9 MPa (T)
Ans.
P5.9 A hollow steel [E = 30,000 ksi] tube (1) with an outside diameter of 2.75 in. and a wall thickness of 0.25 in. is fastened to a solid aluminum [E = 10,000 ksi] rod (2) that has a 2-in.-diameter and a solid 1.375-2-in.-diameter aluminum rod (3). The bar is loaded as shown in Figure P5.9. Determine:
(a) the change in length of steel tube (1).
(b) the deflection of joint D with respect to the fixed support at A.
(c) the maximum normal stress in the entire axial assembly.
FIGURE P5.9
Solution
Before proceeding, it is convenient to determine the internal forces in each of the three axial segments.
Segment (1): Draw a FBD that cuts through segment (1) and includes the free end of the axial member. From this FBD, the sum of forces in the horizontal direction gives the force in segment (1):
1
1
2(34 kips) 2(18 kips) 25 kips 0 57 kips
Fx F F
Segment (2): Draw a FBD that cuts through segment (2) and includes the free end of the axial member. From this FBD, the sum of forces in the horizontal direction gives the force in segment (2):
2
Segment (3): Draw a FBD that cuts through segment (3) and includes the free end of the axial member. From this FBD, the sum of forces in the horizontal direction gives the force in segment (3):
3
(a) The cross-sectional area of the hollow steel tube (inside diameter = 2.25 in.) is
2 2 2
1 (2.75 in.) (2.25 in.) 1.963495 in.
A 4
The deformation in segment (1) is thus ( 57 kips)(60 in.)
F L
For segment (3), the cross-sectional area of the 1.375-in.-diameter aluminum rod is
2 2
3 (1.375 in.) 1.484893 in.
A 4
The deformation in segment (3) can be computed as
3 3
3 2
3 3
( 25 kips)(30 in.)
0.050509 in.
(1.484893 in. )(10, 000 ksi) F L
A E
The deflection of joint D with respect to the fixed support at A is found from the sum of the three segment deformations:
1 2 3 0.058060 in. 0.014006 in. 0.050509 in.
0.094563 in. 0.0946 in. 0.0946 in.
uD
Ans.
(c) Compute the normal stress in each of the three segments:
1
Therefore, the maximum normal stress in the axial member occurs in segment (1):
max 29.0 ksi (C)
Ans.
P5.10 A solid 5/8-in. steel [E = 29,000 ksi]
rod (1) supports beam AB, as shown in Figure P5.10. If the stress in the rod must not exceed 30 ksi and the maximum deformation in the rod must not exceed 0.25 in., determine the maximum load P that may be supported.
FIGURE P5.10 Solution
The area of the 5/8-in.-diameter rod is
2 2
1 (0.625 in.) 0.306796 in.
A 4
If the stress in the rod must not exceed 30 ksi, then the maximum force that can be applied to rod (1) is
2
1 allow,1 1 (30 ksi)(0.306796 in. ) 9.203880 kips
F A
The length of rod (1) is
2 2
1 (20 ft) (16 ft) 25.612497 ft 307.35 in.
L
If the maximum deformation in the rod must not exceed 0.25 in., the maximum internal force in the rod must be limited to
2 1 1 1
1 1
(0.25 in.)(0.306796 in. )(29, 000 ksi)
7.236932 kips 307.35 in.
F A E L
Therefore, the maximum internal force in rod (1) must not exceed 7.237 kips.
Consider a FBD of the rigid beam. Rod (1) is a two-force member at an orientation defined by:
16 ft
tan 0.8 38.660
20 ft
Write the equilibrium equation for the sum of moments about A to find the relationship between F1 and P:
P5.11 A 1-in.-diameter by 16-ft-long cold-rolled bronze bar [E = 15,000 ksi and = 0.320 lb/in3] hangs vertically while suspended from one end. Determine the change in length of the bar due to its own weight.
Solution
An incremental length dy of the bar has an incremental deformation given by ( )
d F y dy
AE
The force in the bar can be expressed as the product of the unit density of the bronze (bronze) and the volume of the bar below the incremental slice dy:
bronzeF y A y
Therefore, the incremental deformation can be expressed as
bronzeA y bronzey
d dy dy
AE E
Integrate this expression over the entire length L of the bar:
2
bronze bronze bronze 2 bronze
0 0 0
The change in length of the bar due to its own weight is therefore:
2 3 2
bronze (0.320 lb/in. )(16 ft 12 in./ft)
0.000393216 in.
2 2(1 0.000393 in
5, 000, 000 ps .
i) L
E
Ans.
P5.12 A homogenous rod of length L and elastic modulus E is a truncated cone with diameter that varies linearly from d0 at one end to 2d0 at the other end. A concentrated axial load P is applied to the ends of the rod, as shown in Figure P5.12. Assume that the taper of the cone is slight enough for the assumption of a uniform axial stress distribution over a cross section to be valid.
(a) Determine an expression for the stress distribution on an arbitrary cross section at x.
(b) Determine an expression for the elongation of the rod.
FIGURE P5.12