elongation of the bar between joints A and C is 0.15 in. In segment (2), the normal strain is measured as 1,300 in./in. Determine:
(a) the elongation of segment (2).
(b) the normal strain in segment (1) of the bar.
Fig. P2.1
Solution
(a) From the definition of normal strain, the elongation in segment (2) can be computed as
6
2 2L2 (1,300 10 )(90 in.) 0.1170 in. Ans.
(b) The combined elongations of segments (1) and (2) is given as 0.15 in. Therefore, the elongation that occurs in segment (1) must be
1 total 2 0.15 in. 0.1170 in. 0.0330 in.
The strain in segment (1) can now be computed:
1 1
1
0.0330 in.
0.000825 in./in.
40 in. 825 μin./in.
L Ans.
2.2 A rigid steel bar is supported by three rods, as shown in Fig. P2.2. There is no strain in the rods before the load P is applied. After load P is applied, the normal strain in rod (2) is 1,080 in./in. Assume initial rod lengths of L1 = 130 in. and L2 = 75 in. Determine:
(a) the normal strain in rods (1).
(b) the normal strain in rods (1) if there is a 1/32-in. gap in the connections between the rigid bar and rods (1) at joints A and C before the load is applied.
(c) the normal strain in rods (1) if there is a 1/32-in. gap in the connection between the rigid bar and rod (2) at joint B before the load is applied.
Fig. P2.2 Solution
(a) From the normal strain in rod (2) and its length, the deformation of rod (2) can be calculated:
6
2 2L2 (1, 080 10 )(75 in.) 0.0810 in.
Since rod (2) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must move downward by an amount equal to the deformation of rod (2); therefore,
2 0.0810 in. (downward) vB
By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rods (1) are connected to the rigid bar at A and C, and again, perfect connections are assumed. The
deformation of rod (1) must be equal to the deflection of joint A (or C); thus, 1 = 0.0810 in. The normal strain in rods (1) can now be calculated as:
1 1
1
0.0810 in.
0.0006231 in./in.
130 in. 623 μin./in.
L Ans.
(b) We can assume that the bolted connection at B is perfect; therefore, vB = 0.0810 in. (downward).
Further, the rigid bar must remain horizontal as it deflects downward by virtue of symmetry.
Therefore, the deflection downward of joints A and C is still equal to 0.0810 in.
What effect is caused by the gap at A and C? When joint A (or C) moves downward by 0.0810 in., the first 1/32-in. of this downward movement does not stretch rod (1)—it just closes the gap. Therefore, rod (1) only gets elongated by the amount
1 0.03125 in.
0.0810 in. 0.03125 in.
0.04975 in.
vA
This deformation creates a strain in rod (1) of:
1 1
1
0.04975 in.
0.0003827 in./in.
130 383 μin./in.
in.
L Ans.
(c) The gap is now at joint B. We know the strain in rod (2); hence, we know its deformation must be 0.0810 in. However, the first 1/32-in. of downward movement by the rigid bar does not elongate the rod—it simply closes the gap. To elongate rod (2) by 0.0810 in., joint B must move down:
2 0.03125 in.
0.0810 in. 0.03125 in.
0.11225 in.
vB
Again, since the rigid bar remains horizontal, vB = vA = vC. The joints at A and C are assumed to be perfect; thus,
1 0.11225 in.
vA
and the normal strain in rod (1) is:
1 1
1
0.11225 in.
0.00086346 in./in.
130 i 863 μin./i
. n.
n
L Ans.
2.3 A rigid steel bar is supported by three rods, as load is applied.
(c) the normal strain in rod (2) if there is a 2-mm gap in the connection between the rigid bar and rod (2) at joint B before the load is applied.
Fig. P2.3 Solution
(a) From the normal strain in rod (1) and its length, the deformation of rod (1) can be calculated:
6
1 1 1L (860 10 )(2, 400 mm) 2.064 mm
Since rod (1) is assumed to be connected to the rigid bar with a perfect connection, the rigid bar must move downward by an amount equal to the deformation of rod (1); therefore,
1 2.064 mm (downward)
A C
v v
By symmetry, the rigid bar must remain horizontal as it moves downward, and thus, vB = vA = vC. Rod (2) is connected to the rigid bar at B, and again, a perfect connection is assumed. The deformation of rod (2) must be equal to the deflection of joint B; thus, 2 = 2.064 mm. The normal strain in rod (2) can now be calculated as:
2 2
2
2.064 mm
0.0011467 mm/mm 1,147 μmm/m
1,800 m m
m
L Ans.
(b) We know the strain in rod (1); hence, we know its deformation must be 2.064 mm. However, the joints at A and C are not perfect connections. The first 2 mm of downward movement by the rigid bar does not elongate rod (1)—it simply closes the gap. To elongate rod (1) by 2.064 mm, joint A must move down:
1 2 mm 2.064 mm 2 mm 4.064 mm vA
By symmetry, the rigid bar remains horizontal;
therefore, vB = vA = vC. The joint at B is assumed to be perfect; thus, any downward movement of the rigid bar also elongates rod (2) by the same amount:
2 4.064 mm vB
2 2
2
4.064 mm
0.0022578 mm/mm 2, 260 μmm/m
1,800 m m
m
L Ans.
(c) We now assume that the bolted connection at A (or C) is perfect; therefore, the rigid bar deflection as calculated in part (a) must be vA = 2.064 mm (downward). Further, the rigid bar must remain horizontal as it deflects downward by virtue of symmetry; thus, vB = vA = vC. Therefore, the deflection downward of joint B is vB = 2.064 mm What effect is caused by the gap at B? When joint B moves downward by 2.064 mm, the first 2 mm of this downward movement does not stretch rod (2)—it just closes the gap. Therefore, rod (2) only gets elongated by the amount
2 2 mm
2.064 mm 2 mm 0.064 mm
vB
This deformation creates a strain in rod (2) of:
2 2
2
0.064 mm
0.0000356 mm/mm
1,800 m 35.6 μmm/mm
m
L Ans.
2.4 A rigid bar ABCD is supported by two bars as shown in Fig. P2.4. There is no strain in the vertical bars before load P is applied. After load P is applied, the normal strain in rod (1) is −570
m/m. Determine:
(a) the normal strain in rod (2).
(b) the normal strain in rod (2) if there is a 1-mm gap in the connection at pin C before the load is applied.
(c) the normal strain in rod (2) if there is a 1-mm gap in the connection at pin B before the load is applied.
From the deformation diagram of rigid bar ABCD 240 mm (240 mm 360 mm)
600 mm
(0.5130 mm) 1.2825 mm 240 mm
B C
C
v v
v
Therefore, 2 = 1.2825 mm (elongation), and thus, from the definition of strain:
2 6 deformation diagram is unaffected; thus, vB = 0.5130 mm (downward) and vC = 1.2825 mm (downward).
The rigid bar must move downward 1 mm at C before it begins to elongate member (2).
Therefore, the elongation of member (2) is
2 1 mm
1.2825 mm 1 mm 0.2825 mm
vC
and so the strain in member (2) is
2 6 2
2
0.2825 mm
188.3 10 mm/mm
1,500 mm 188.3 με
L Ans.
(c) From the strain given for rod (1), 1 =
−0.5130 mm. In order to contract rod (1) by this amount, the rigid bar must move downward at B by
0.5130 mm 1 mm 1.5130 mm vB
From deformation diagram of rigid bar ABCD 240 mm (240 mm 360 mm)
600 mm
(1.5130 mm) 3.7825 mm 240 mm
B C
C
v v
v and so
2 6 2
2
3.7825 mm
2,521.7 10 mm/mm
1,500 2,520 ε μ
mm
L Ans.
2.5 In Fig. P2.5, rigid bar ABC is supported by a pin connection at B and two axial members. A slot in member (1) allows the pin at A to slide 0.25-in.
From the strain given for member (1)
1 1 1 1.080 mm (downward).
0.0416 in. 0.25 in.
0.2916 in.
0.2916 in. (to the left) vA
From the deformation diagram of rigid bar ABC
12 in. 20 in.
20 in.
(0.2916 in.) 0.4860 in. (downward) 12 in.
A C
C
v v
v
Therefore, 2 = 0.4860 in. (elongation). From the definition of strain,
2 6