(a) x = 10 in.
Equilibrium: Draw a FBD for the interval between A and B where 0 x a, and write the following equilibrium equation:
(750 lb/ft)(1 ft/12 in.)(48 in. ) (2,000 lb) (1,000 lb) 0
Stress: The normal stress at this location can be calculated as follows.
2 2
2
(1.25 in.) 1.227185 in.
4
5,375 lb
4,379.944 psi 4,380 p
1 si
Equilibrium: Draw a FBD for the interval between B and C where a x a b, and write the following equilibrium equation:
(750 lb/ft)(1 ft/12 in.)(48 in. ) (1,000 lb) 0
Stress: The normal stress at this location can be calculated as follows.
P1.16 Two 6 in. wide wooden boards are to be joined by splice plates that will be fully glued on the contact surfaces. The glue to be used can safely provide a shear strength of 120 psi. Determine the smallest allowable length L that can be used for the splice plates for an applied load of P = 10,000 lb. Note that a gap of 0.5 in. is
required between boards (1) and (2). FIGURE P1.16
Solution
Consider a FBD of board (2). The glue on the splice plates provides resistance to the 10,000 lb applied load on both the top and bottom surfaces of board (2). Denoting the shear resistance on a glue surface as V, equilibrium in the horizontal direction requires
0
In other words, each glue surface must be large enough so that 5,000 lb of shear resistance can be provided to board (2). Since the glue has a shear strength of 120 psi, the area of each glue surface on board (2) must be at least
The boards are 6-in. wide; therefore, glue must be spread along board (2) for a length of at least
2
Although we’ve discussed only board (2), the same rationale applies to board (1). For both boards (1) and (2), the glue must be applied along a length of at least 6.9444 in. on both the top and bottom of the boards in order to resist the 10,000 lb applied load.
The glue applied to boards (1) and (2) must be matched by glue applied to the splice plates. Therefore, the splice plates must be at least 6.9444 in. + 6.9444 in. = 13.8889 in. long. However, we are told that a 0.5-in. gap is required between boards (1) and (2); therefore, the splice plates must be 0.5-in. longer.
Altogether, the length of the splice plates must be at least
min 6.9444 in. 6.9444 in. 0.5 in. 14.39 in.
L Ans.
P1.17 For the clevis connection shown in Figure P1.17, determine the maximum applied load P that can be supported by the 10-mm-diameter pin if the average shear stress in the pin must not exceed 95 MPa.
FIGURE P1.17 Solution
Consider a FBD of the bar that is connected by the clevis, including a portion of the pin. If the shear force acting on each exposed surface of the pin is denoted by V, then the shear force on each pin surface is related to the load P by:
0 2
Fx P V V P V
The area of the pin surface exposed by the FBD is simply the cross-sectional area of the pin:
2 2 2
pin pin (10 mm) 78.539816 mm
4 4
A d
If the average shear stress in the pin must be limited to 95 MPa, the maximum shear force V on a single cross-sectional surface must be limited to
2 2
bolt (95 N/mm )(78.539816 mm ) 7, 461.283 N
V A
Therefore, the maximum load P that may be applied to the connection is 2 2(7, 461.283 N) 14,922.565 N 14.92 kN
P V Ans.
P1.18 For the connection shown in Figure P1.18, determine the average shear stress produced in the 3/8-in. diameter bolts if the applied load is P = 2,500 lb.
FIGURE P1.18 Solution
There are four bolts, and it is assumed that each bolt supports an equal portion of the external load P.
Therefore, the shear force carried by each bolt is 2,500 lb
625 lb 4 bolts
V
The bolts in this connection act in single shear. The cross-sectional area of a single bolt is
2 2 2 2
bolt bolt (3 / 8 in.) (0.375 in.) 0.110447 in.
4 4 4
A d
Therefore, the average shear stress in each bolt is
2 bolt
625 lb
5,658.8427 psi
0.110447 in. 5,660 psi
V
A Ans.
P1.19 The five-bolt connection shown in Figure P1.19 must support an applied load of P = 265 kN. If the average shear stress in the bolts must be limited to 120 MPa, determine the minimum bolt diameter that may be used for this connection.
FIGURE P1.19 Solution
There are five bolts, and it is assumed that each bolt supports an equal portion of the external load P.
Therefore, the shear force carried by each bolt is 265 kN
53 kN 53,000 N 5 bolts
V
Since the average shear stress must be limited to 120 MPa, each bolt must provide a shear area of at least:
2 2
53,000 N
441.6667 mm 120 N/mm
AV
Each bolt in this connection acts in double shear; therefore, two cross-sectional bolt surfaces are available to transmit shear stress in each bolt.
2
2 bolt
441.6667 mm
220.8333 mm per surface 2 surfaces per bolt 2 surfaces
AV
A
The minimum bolt diameter must be
2 2
bolt 220.8333 mm bolt 16.7682 mm 16.77 mm
4d d
Ans.
P1.20 A coupling is used to connect a 2 in. diameter plastic pipe (1) to a 1.5 in. diameter pipe (2), as shown in Figure P1.20. If the average shear stress in the adhesive must be limited to 400 psi, determine the minimum lengths L1 and L2 required for the joint if the applied load P is 5,000 lb.
FIGURE P1.24 Solution
To resist a shear force of 5,000 lb, the area of adhesive required on each pipe is
2
Consider the coupling on pipe (1). The adhesive is applied to the circumference of the pipe, and the circumference C1 of pipe (1) is
1 1 (2.0 in.) 6.2832 in.
C D
The minimum length L1 is therefore
2
Consider the coupling on pipe (2). The circumference C2 of pipe (2) is
2 2 (1.5 in.) 4.7124 in.
C D
The minimum length L2 is therefore
2
1.21 A hydraulic punch press is used to punch a slot in a 0.50-in. thick plate, as illustrated in Fig. P1.21. If the plate shears at a stress of 30 ksi, determine the minimum force P required to punch the slot.
FIGURE P1.21 Solution
The shear stress associated with removal of the slug exists on its perimeter. The perimeter of the slug is given by
perimeter2(3.00 in.) + (0.75 in.) 8.35619 in.
Thus, the area subjected to shear stress is
perimeter plate thickness (8.35619 in.)(0.50 in.) 4.17810 in.2
AV
Given that the plate shears at = 30 ksi, the force required to remove the slug is therefore
2
min V (30 ksi)(4.17810 in. ) 125.343 kips 125.3 kips
P A Ans.
P1.22 The handle shown in Figure P1.22 is attached to a 40-mm-diameter shaft with a square shear key. The forces applied to the lever are P = 1,300 N. If the average shear stress in the key must not exceed 150 MPa, determine the minimum dimension a that must be used if the key is 25 mm long. The overall length of the handle is L = 0.70 m.
FIGURE P1.22 Solution
To determine the shear force V that must be resisted by the shear key, sum moments about the center of the shaft (which will be denoted O):
700 mm 700 mm 40 mm
(1,300 N) (1,300 N) 0
2 2 2
45,500 N
MO V
V
Since the average shear stress in the key must not exceed 150 MPa, the shear area required is
2 2
45,500 N
303.3333 mm 150 N/mm
V
A V
The shear area in the key is given by the product of its length L (i.e., 25 mm) and its width a. Therefore, the minimum key width a is
303.3333 mm2
12.1333 mm
25 m 12.1
m 3 mm
AV
a L Ans.
P1.23 An axial load P is supported by the short steel column shown in Figure P1.23. The column has a cross-sectional area of 14,500 mm2. If the average normal stress in the steel column must not exceed 75 MPa, determine the minimum required dimension a so that the bearing stress between the base plate and the concrete slab does not exceed 8 MPa. Assume b = 420 mm.
FIGURE P1.23