1. The making of participation
1.4. Reconfiguration of objects
P4.2. The ultimate shear strength of the bolts is 300 MPa, and a factor of safety of 4.0 is required with
The allowable shear stress for the bolts is
allow
To support a load of P = 350 kN, the total area subjected to shear stress must equal or exceed:
2
There are six bolts, and each bolt acts in double shear; therefore, the total area subjected to shear stress is
2
(6 bolts)(2 surfaces per bolt) bolt V 4
A d
The minimum bolt diameter can be found be equating the two expressions for AV:
2 2
bolt
bolt
(12 surfaces) 6,333.333 mm 4
4.3 A 14-kip load is supported by two bars as shown in Fig. P4.3. Bar (1) is made of cold-rolled red brass (Y = 60 ksi) and has a cross-sectional area of 0.225 in.2. Bar (2) is made of 6061-T6 aluminum (Y = 40 ksi) and has a cross-sectional area of 0.375 in.2. Determine the factor of safety with respect to yielding for each of the bars.
Fig. P4.3 Solution
Consider a FBD of joint B. Members (1) and (2) are two-force members; therefore, the equilibrium equations can be written as:
2cos50 1cos35 0
Substitute this expression into Eq. (b) to obtain:
cos35 tan 50 sin 35 14 kips 1.549804 14 kips Backsubstitute to obtain F2:
2 1
cos35
(9.033401 kips)(1.274374) 11.511935 kips cos50
Therefore, the factor of safety in bar (1) is
1
4.4 A steel plate is to be attached to a support with three bolts as shown in Fig. P4.4. The cross-sectional area of the plate is 560 mm2 and the yield strength of the steel is 250 MPa. The ultimate shear strength of the bolts is 425 MPa. A factor of safety of 1.67 with respect to yield is required for the plate. A factor of safety of 4.0 with respect to the ultimate shear strength is required for the bolts. Determine the minimum bolt diameter required to develop the full strength of the plate. (Note: consider only the gross cross-sectional area of the plate—not the net area.)
Fig. P4.4 Solution
The allowable normal stress is
allow
The full strength of the plate (based on the gross cross-sectional area) is therefore:
2
max allow (149.7006 MPa)(560 mm ) 83,832.3353 N
P A
The allowable shear stress of the bolts is
allow
The minimum shear area required to support the maximum load P is
max 2
There are three bolts, each acting in single shear, which provide a shear area of
2 2
bolt bolt
(3 bolts)(1 surface per bolt) (3 surfaces)
4 4
AV d d
Equating the two expressions for the shear area, the minimum bolt diameter can be computed:
2 2
bolt
bolt
(3 surfaces) 789.0102 mm 4
18.2994 mm 18.30 mm d
d
Ans.
4.5 In Fig. P4.5, member (1) is a steel bar with a cross-sectional area of 1.35 in.2 and a yield strength of 50 ksi.
Member (2) is a pair of 6061-T6 aluminum bars having a combined cross-sectional area of 3.50 in.2 and a yield strength of 40 ksi. A factor of safety of 1.6 with respect to yield is required for both members. Determine the maximum allowable load P that may be applied to the structure. Report the factors of safety for both members at the allowable load.
Fig. P4.5 Solution
Consider a FBD of joint B and write the following equilibrium equations:
2cos 65 1 0
Substituting Eq. (c) into Eq. (a) gives an expression for P in terms of F1:
2 1
and rearranging Eq. (c) gives an expression for P in terms of F2:
2sin 65
PF (e)
The allowable normal stress in member (1) is
,1
and the allowable force in member (1) is
2
allow,1 allow,1 1 (31.25 ksi)(1.35 in. ) 42.1875 kips
F A (f)
The allowable normal stress in member (2) is
,2
and the allowable force in member (2) is
2
allow,2 allow,2 2 (25.00 ksi)(3.50 in. ) 87.50 kips
F A (g)
Substitute the allowable force in member (1) from Eq. (f) into Eq. (d) to obtain the maximum load P based on the capacity of member (1):
allow,1tan 65 (42.1875 kips) tan 65 90.471386 kips
PF (h)
Repeat with the allowable force in member (2) from Eq. (g) substituted into Eq. (e) to obtain the maximum load P based on the capacity of member (2):
allow,2sin 65 (87.50 kips)sin 65 79.301931 kips
PF (i)
Compare the results in Eqs. (h) and (i) to find that the maximum load P that may be applied is controlled by the capacity of member (2):
max 79.3 kips
P Ans.
The factor of safety in member (2) is thus:
FS2 1.6 Ans.
For an applied load of P = 79.3 kips, the force in member (1) can be computed from Eq. (d):
1
79.301931 kips
36.979096 kips tan 65 tan 65
F P
which results in a normal stress of
1
1 2
1
36.979096 kips
27.391924 ksi 1.350 in.
F
A
and thus, its factor of safety is:
,1 1
1
50 ksi
FS Y 27.391924 ksi 1.825
Ans.
4.6 The rigid structure ABD in Fig. P4.6 is supported at B by a 35-mm-diameter tie rod (1) and at A by a 30-mm-diameter pin used in a single shear connection. The tie rod is connected at B and C by 24-mm-diameter pins used in double shear connections. Tie rod (1) has a yield strength of 250 MPa, and each of the pins has an ultimate shear strength of 330 MPa. A concentrated load of P = 50 kN acts as shown at D. Determine:
(a) the normal stress in rod (1).
(b) the shearing stress in the pins at A and B.
(c) the factor of safety with respect to the yield strength for tie rod (1).
(d) the factor of safety with respect to the ultimate strength for the pins at A and B.
Fig. P4.6 Solution
Member (1) is a two-force member that is oriented at with respect to the horizontal axis:
tan 8 m 1.509434
5.3 m
From a FBD of rigid structure ABD, the following equilibrium equations can be written:
cos 60 1cos56.4755 0
x x
(50 kN) 92.7405 kN
(8 m) cos56.4755 (8 m) cos56.4755
F P
Substitute F1 into Eq. (a) to obtain Ax:
1cos56.4755 cos 60 (92.7405 kN) cos56.4755 (50 kN) cos 60 26.2199 kN
Ax F P
and substitute F1 into Eq. (b) to obtain Ay:
The resultant pin force at A is found from Ax and Ay:
2 2 2 2
(26.2199 kN) (120.6144 kN) 123.4314 kN
x y
A A A
(a) The cross-sectional area of the 35-mm-diameter tie rod is:
2 2
1 (35 mm) 962.112750 mm A 4
and thus, the normal stress in rod (1) is:
1
(b) The 30-mm-diameter single shear pin at A has a shear area of
2 2
, (30 mm) 706.858347 mm
V A 4
A
Consequently, the shear stress in pin A is
2
123, 431.4 N
174.6197 MPa
706.858347 m 174. M a
m 6 P
A Ans.
The 24-mm-diameter double shear pins at B and C have a shear area of
2 2
, (2 surfaces) (24 mm) 904.778685 mm
V B 4
904.778685 mm 102.5 MPa
B Ans.
(d) The factor of safety with respect to the ultimate strength for the pin at A is 330 MPa
FSA U 174.6197 MPa 1.890
A
Ans.
and the factor of safety for pin B is 330 MPa
FS 102.5008 MP 3.2
a 2
4.7 Rigid bar ABD in Fig. P4.7 is supported by a pin connection at A and a tension link BC. The 8-mm-diameter pin at A is supported in a double shear connection, and the 12-mm-diameter pins at B and C are both used in single shear connections. Link BC is 30-mm wide and 6-mm thick. The ultimate shear strength of the pins is 330 MPa and the yield strength of link BC is 250 MPa.
(a) Determine the factor of safety in pins A and B with respect to the ultimate shear strength.
(b) Determine the factor of safety in link BC with respect to the yield strength.
Fig. P4.7 Solution
Consider a free-body diagram of the rigid bar. Link BC is a two-force member that is oriented at an angle of with respect to the horizontal axis:
0.6 m
tan 53.1301
0.45 m
The equilibrium equations for the rigid bar can be written as:
cos 53.1301 (8.2 kN)cos70 0 sin 53.1301 (8.2 kN)sin70 0
(0.75 m) cos 53.1301 (0.45 m) sin 53.1301
(1.85 m)(8.2 kN)sin70 (0.75 m)(8.2 kN)cos70 0
x BC x
Solving these three equations simultaneously gives:
20.1958 kN 9.3129 kN 8.4511 kN
BC x y
F A A
The resultant force at pin A is:
2 2 2 2
(9.3129 kN) ( 8.4511 kN) 12.5758 kN
x y
A A A
Before the factors of safety can be determined, we must compute the shear stresses in pins A and B as well as the normal stress in link BC.
Pin shear stresses:
The 8-mm-diameter pin at A is supported in a double shear connection; therefore, the shear force acting on one shear plane (which is simply equal to the cross-sectional area of the pin) is half of the resultant force at pin A: VA = 6.2879 kN. The cross-sectional area of the pin at A is:
2 2
(8 mm) 50.265 mm
pin A 4
A
and therefore, the shear stress in pin A is:
(6.2879 kN)(1000 N/kN) 2
125.0940 N/mm 125.0940 MPa VA
The 12-mm-diameter pin at B and C is supported in a single shear connection and so the shear force acting on one shear plane is the entire force in link BC: VB = 20.1958 kN. The cross-sectional area of the pin at B is:
2 2
(12 mm) 113.097 mm
pin B 4
A
and therefore, the shear stress in pin B is:
2 2
(20.1958 kN)(1000 N/kN)
178.5697 N/mm 178.5697 MPa 113.097 mm
Normal stress in link:
The normal stress in link BC is:
(20.1958 kN)(1000 N/kN) 2
112.1986 N/mm 112.1986 MPa (30 mm)(6 mm)
Factors of safety:
For pin A, the factor of safety is:
330 MPa
FS 125.0940 MP 2.6
a 4
FSB U 178.5697 MPa 1.848
B
Ans.
The factor of safety in link BC is:
250 MPa
FS 112.1986 M 2.23
Pa
4.8 In Fig. P4.8, davit ABD is supported at A by a single shear pin connection and at B by a tie rod (1). The pin at A has a diameter of 1.25 in. and the pins at B and C are each 0.75-in.-diameter pins. Tie rod (1) has an area of 1.50 in.2. The ultimate shear strength in each pin is 80 ksi, and the yield strength of the tie rod is 36 ksi. A concentrated load of 25 kips is applied as shown to the davit structure at D. Determine:
(a) the normal stress in rod (1). respect to the horizontal axis:
tan 9 ft 0.75 36.870
12 ft
From a FBD of rigid structure ABD, the following equilibrium equations can be written:
(25 kips) cos 60 1cos 36.870 0
(25 kips) 40.1465 kips
(9 ft) cos 36.870
F
Substitute F1 into Eq. (a) to obtain Ax:
1cos 36.870 (25 kips) cos 60 (40.1465 kips) cos 36.870 (25 kips) cos 60 19.6172 kips
Ax F
and substitute F1 into Eq. (b) to obtain Ay:
1sin 36.870 (25 kips) sin 60 (40.1465 kips) sin 36.870 (25 kips) sin 60 45.7386 kips
Ay F
The resultant pin force at A is found from Ax and Ay:
2 2 2 2
(19.6172 kips) (45.7386 kips) 49.7680 kips
x y
(b) The 1.25-in.-diameter single shear pin at A has a shear area of
2 2
, (1.25 in.) 1.2272 in.
V A 4
A
Consequently, the shear stress in pin A is
2
The 0.75-in.-diameter double shear pins at B and C have a shear area of
2 2
, (2 surfaces) (0.75 in.) 0.8836 in.
V B 4
(d) The factor of safety with respect to the ultimate strength for the pin at A is 80 ksi
4.9 The pin-connected structure is subjected to a load P as shown in Fig. P4.9. Inclined member (1) has a cross-sectional area of 250 mm2 and has a yield strength of 255 MPa. It is connected to rigid member ABC with a 16-mm-diameter pin in a double shear connection at B.
The ultimate shear strength of the pin material is 300 MPa. For inclined member (1), the minimum factor of safety with respect to the yield strength is FSmin = 1.5.
For the pin connections, the minimum factor of safety with respect to the ultimate strength is FSmin = 3.0.
(a) Based on the capacity of member (1) and pin B, determine the maximum allowable load P that may be applied to the structure.
(b) Rigid member ABC is supported by a double shear pin connection at A. Using FSmin = 3.0, determine the minimum pin diameter that may be used at support A.
Fig. P4.9 Solution
The allowable normal stress for inclined member (1) is
allow
and the allowable shear stress for the pins is
allow
(a) The allowable axial force in member (1) based on normal stress is
2
allow allow 1 (170.00 MPa)(250 mm ) 42,500 N
F A (a)
Member (1) is connected with a 16-mm-diameter double shear pin. The shear area of this pin is:
2 2
(2 surfaces) (16 mm) 402.1239 mm
V 4
A
Consequently, the shear force V (which is applied by the inclined member) that can be applied to the pin is limited to
2
allow allow V (100.0 MPa)(402.1239 mm ) 40, 212.4 N
V A (b)
Comparing the two values given in Eq. (a) and Eq. (b), the maximum allowable force in member (1) is
1 40, 212.4 N
F (c)
Member (1) is a two-force member that is oriented at with respect to the horizontal axis:
1.4 m
tan 0.6087 31.3287
2.3 m
From a FBD of rigid structure ABC, the following equilibrium equations can be written:
1cos 31.3287 0 the maximum load P:
1 15,027.8 N 15.03 kN
P F
Ans.
Substitute F1 and P into Eq. (d) to obtain Ax:
1cos31.3287 (15,027.8 N) (40, 212.4 N) cos31.3287 19,321.5 N
Ax P F
and substitute F1 into Eq. (e) to obtain Ay:
1sin 31.3287 (40, 212.4 N)sin 31.3287 20,908.3 N
Ay F
The total shear area required to support the resultant pin force is
2
Since the pin at A is in a double shear connection, the shear area provided by the pin is
2
(2 surfaces) pin V 4
A d
Equate these two expressions and solve for the required minimum pin diameter:
2 2
pin
pin
(2 surfaces) 284.689 mm 4
4.10 After load P is applied to the pin-connected structure shown in Fig. P4.10, a normal strain of
= +550 is measured in the longitudinal direction of member (1). The cross-sectional area of member (1) is A1 = 0.60 in.2, its elastic modulus is E1 = 29,000 ksi, and its yield strength is 36 ksi.
(a) Determine the axial force in member (1), the applied load P, and the resultant force at pin B.
(b) The ultimate shear strength of the steel pins is 54 ksi. Determine the minimum diameter for the pin at B if a factor of safety of 2.5 with respect to the ultimate shear strength is required.
(c) Compute the factor of safety for member (1) with respect to its yield strength.
Fig. P4.10 Solution
(a) Given the strain in member (1), its stress can be computed from Hooke’s law:
1 E1 1 (29,000 ksi)(0.000550 in./in.) 15.9500 ksi
The force in member (1) is the product of the normal stress and the cross-sectional area:
2
Equilibrium equations for horizontal member ABC can be written as:
Substituting the value F1 = 14.79 kips into these equations gives:
7.7044 kips 9.5611 kips 3.8842 kips 3.88 kips
x y
B B P Ans.
The resultant force at pin B is:
2 2 2 2
( 7.7044 kips) (9.5611 kips) 12.2789 kips 12.28 kips
x y
B B B Ans.
(b) The pin at B is supported in a double shear connection. Therefore, the shear force acting on one shear plane of the pin is half of the resultant force: VB = 6.1395 kips. The allowable shear stress for the pin is computed from the ultimate shear strength and the factor of safety:
54 ksi
Next, the cross-sectional area required for a single shear plane can be determined:
6.1395 kips 2
0.2842 in.
21.6 ksi
B B
a B
B a
V V
A A
To provide AB, the pin at B must have a diameter of at least:
2 2
0.2842 in. 0.602 in.
4dB dB
Ans.
(c) The factor of safety for member (1) with respect to its yield strength is:
1 1
36 ksi
FS Y 15.9500 ksi 2.26
Ans.
4.11 The simple pin-connected structure carries a factor of safety of 2.0 is required in both the strut and the pin at C, determine the maximum load P that can be supported by the structure.
Fig. P4.11 Solution
From a FBD of the rigid bar, the following equilibrium equations can be written:
1 0
Substitute this result into Eq. (a) to express Cx in terms of P:
1 2.2
Cx F P
The resultant reaction force at pin C can now be expressed as a function of P:
2 2 2 2
(2.2 ) ( ) 2.41661
x y
C C C P P P (e)
Strut AB:
The allowable normal stress for strut AB is
allow
Therefore, the allowable axial force for strut AB is
2
allow,1 allow 1 (30 ksi)(0.25 in. ) 7.50 kips
F A
From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations on the strut normal stress is
allow,1
Pin C:
The allowable shear stress for the pin at C is
allow
54 ksi
27 ksi FS 2.0
U
The 0.375-in.-diameter double shear pin at C has a shear area of
2 2
(2 surfaces) (0.375 in.) 0.2209 in.
V 4
A
thus, the allowable reaction force at C is
2 allow
allow V (27 ksi)(0.2209 in. ) 5.9641 kips
C A
From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations on the pin shear stress is
allow max
5.9641 kips
2.4680 kips 2.41661 2.41661
P C (g)
Maximum load P:
Comparing the results in Eqs. (f) and (g), the maximum load P that may be applied to the rigid bar is
max 2.4680 kips 2.47 ipsk
P Ans.
4.12 In Fig. P4.12, rigid tee-beam ABC is supported at A by a single shear pin connection and at B by a strut, which consists of two 30 mm wide by 8 mm thick steel bars. The pins at A, B, and D are each 12 mm in diameter. The yield strength of the steel bars in strut (1) is 250 MPa, and the ultimate shear strength of each pin is 500 MPa. Determine the allowable load P that may equilibrium equations can be written:
x x 0
Substitute this result into Eq. (b) to express Ay in terms of P:
1 ( 2.181818 ) 1.181818 Ay P F P P P
The resultant reaction force at pin A can now be expressed as a function of P:
2 2 2 2
(0) ( 1.181818 ) 1.181818
x y
A A A P P (e)
The allowable normal stress for strut (1) is
allow
Therefore, the allowable axial force for strut (1) is
allow,1 allow 1 (83.333333 MPa)(2 30 mm 8 mm) 40, 000 N
F A
From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress limitation is
The allowable shear stress for the pins at A, B, and D is
allow
The cross-sectional area of a 12-mm-diameter pin is
2 2
pin (12 mm) 113.097336 mm A 4
The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is
2 allow ,
allow V A (166.666667 MPa)(113.097336 mm ) 18,849.5560 N
A A
From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is
allow max
18,849.5560 N
15,949.627 N 1.181818 1.181818
P A (g)
The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut (1) based on the capacity of the pins at B and D is
2
allow,1 allow V B, (166.666667 MPa)(2)(113.097336 mm ) 37,699.1121 N
F A
From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins at B and D is
allow,1 max
37,699.1121 N
17, 278.7611 N 2.181818 2.181818
P F (h)
Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar is
max 15,949.627 N 15.95 kN
P Ans.
4.13 In Fig. P4.13, rigid bar ABC is supported at A by a single shear pin connection and at B by a strut, which consists of two 2 in. wide by 0.25 in. thick steel bars.
The pins at A, B, and D each have a diameter of 0.5 in.
The yield strength of the steel bars in strut (1) is 36 ksi, and the ultimate shear strength of each pin is 72 ksi.
Determine the allowable load P that may be applied to the rigid bar at C if an overall factor of safety of 3.0 is required. Use L1 = 36 in. and L2 = 24 in.
Fig. P4.13 Solution
From a FBD of the rigid beam ABC, the following equilibrium equations can be written:
x x 0
Substitute this result into Eq. (b) to express Ay in terms of P:
1 ( 1.666667 ) 0.666667 Ay P F P P P
The resultant reaction force at pin A can now be expressed as a function of P:
2 2 2 2
(0) ( 0.666667 ) 0.666667
x y
A A A P P (e)
Strut (1):
The allowable normal stress for strut (1) is
allow
Therefore, the allowable axial force for strut (1) is
allow,1 allow 1 (12.0 ksi)(2 2 in. 0.25 in.) 12.0 kips
F A
From Eq. (d), the maximum load that may be applied to the rigid bar based on the strut normal stress limitation is
Pins:
The allowable shear stress for the pins at A, B, and D is
allow
The cross-sectional area of a 0.5-in.-diameter pin is
2 2
pin (0.5 in.) 0.196350 in.
A 4
Pin A:
The pin at A is a single shear connection; therefore, AV,A = Apin. The allowable reaction force at A is
2 allow ,
allow V A (24.0 ksi)(0.196350 in. ) 4.7124 kips
A A
From Eq. (e), the maximum load that may be applied to the rigid bar based on the limitations of pin A is
allow
The pins at B and D are in double shear connections; therefore, AV,B = 2Apin. The allowable force in strut (1) based on the capacity of the pins at B and D is
2
allow,1 allow V B, (24.0 ksi)(2)(0.196350 in. ) 9.4248 kips
F A
From Eq. (d), the maximum load that may be applied to the rigid bar based on the limitations of the pins at B and D is
Maximum load P:
Comparing the results in Eqs. (f), (g), and (h), the maximum load P that may be applied to the rigid bar is
max 5.6549 kips 5.65 ipsk
P Ans.
4.14 A concentrated load of P = 70 kips is applied to beam AB as shown in Fig. P4.14. Rod (1) has a diameter of 1.50 in., and its yield strength is 60 ksi. Pin A is supported in a double shear connection, and the ultimate shear strength of pin A is 80 ksi.
(a) Determine the normal stress in rod (1).
(b) Determine the factor of safety with respect to the yield strength for rod (1).
(c) If a factor of safety of 3.0 with respect to the ultimate strength is specified for pin A, determine the minimum required pin diameter.
Fig. P4.14 Solution
Tie rod (1) is a two-force member that is oriented at with respect to the horizontal axis:
8.5 ft
tan 0.7083 35.3112
12 ft
From a FBD of beam AB, the following equilibrium equations can be written:
1cos35.3112 0 (cos35.3112 ) 0.833333 ft (sin 35.3112 )(12 ft)
F
Substitute F1 into Eq. (a) to obtain Ax:
1cos35.3112 (73.5272 kips) cos35.3112 60.0 kips
Ax F
and substitute F1 into Eq. (b) to obtain Ay:
1sin 35.3112 70 kips (73.5272 kips)sin 35.3112 27.50 kips
Ay P F
The resultant pin force at A is found from Ax and Ay:
2 2 2 2
(60.0 kips) (27.5 kips) 66.0019 kips
x y
A A A
(a) The cross-sectional area of the 1.5-in.-diameter tie rod (1) is
2 2
1 (1.5 in.) 1.7671 in.
A
The normal stress in tie rod (1) is:
(c) The allowable shear stress is:
allow
The shear area required for pin A is thus
2
Since pin A is in a double shear connection, the minimum required pin area is
2
2 pin
2.4751 in.
1.2375 in.
2 shear surfaces 2 AV
A
Thus, the diameter of pin A must be
2 2
4.15 Beam AB is supported as shown in Fig. P4.15. Tie rod (1) is attached at B and C with double shear pin connections while the pin at A is attached with a single shear connection. The pins at A, B, and C each have an ultimate shear strength of 54 ksi, and tie rod (1) has a yield strength of 36 ksi. A concentrated load of P = 16 kips is applied to the beam as shown. A factor of safety of 3.0 is required for all components. Determine:
(a) the minimum required diameter for tie rod (1).
(b) the minimum required diameter for the double shear pins at B and C.
(c) the minimum required diameter for the single shear pin at A.
Fig. P4.15 Solution
Tie rod (1) is a two-force member that is oriented at with respect to the horizontal axis:
8.5 ft
tan 0.7083 35.3112
12 ft
From a FBD of beam AB, the following equilibrium equations can be written:
1cos35.3112 0 (cos35.3112 ) 0.833333 ft (sin 35.3112 )(12 ft)
F
Substitute F1 into Eq. (a) to obtain Ax:
1cos35.3112 (16.8062 kips) cos35.3112 13.7143 kips
Ax F
and substitute F1 into Eq. (b) to obtain Ay:
1sin 35.3112 16 kips (16.8062 kips)sin 35.3112 6.2857 kips
1sin 35.3112 16 kips (16.8062 kips)sin 35.3112 6.2857 kips