04.3 Stress and Strain Supplemental Problems
Problem: The frame shown consists of four wooden members: ABC, DEF, BE, & CF. Each member has a 2 in x 4 in cross-section area (oriented as shown) and each connecting pin is ½ in in diameter. Determine the maximum value of the normal stress (a) in member BE, (b) in member CF.
Solution: I need to find the forces acting in members BE & CF. 1st, use the entire frame as a FBD:
∑M
A
= 0 = (40in )D x − (75in )(480lbf )
ccw+
⇒ D x = 900lbf
Next, use member DEF as a FBD: The reaction @ D must be parallel to FBE & FCF: 4 4 D y = D x = (900lbf ) = 1200lbf 3 3 ∑MF = 0 ccw+
4 = −(30in ) FBE − (45in )(1200lbf ) 5 ⇒ FBE = −2250lbf
∑M
E
=0
ccw+
4 = (30in ) FCE − (15in )(1200lbf ) 5 ⇒ FCE = 750lbf Therefore, member BE is in compression and member CE is in tension. Finally, determine the maximum normal stress in each member. Remember to adjust the area of the member for CE since it is in tension:
σ BE =
σ BE
P (− 2250lbf ) = = − 281.3 lbf 2 = 281.3 lbf 2 (Compr) in in (2in )(4in ) A
(750lbf ) = 107.1lbf = 107.1lbf (Tens) P = = in 2 in 2 A (2in )(4 − 0.5)in
Ans
Problem: A 1.5 m concrete post is reinforced with six steel bars, each rod’s diameter is 28 mm. Knowing that Es = 200 GPa and Ec = 25 GPa, determine the normal stresses in the steel and in the concrete when a 1550 kN axial centric force P is applied to the post.
Solution: The rebar and the concrete are well bonded. Therefore, the load is divided (but not equally) and carried the concrete and the steel. Also, the rebar and concrete deformation will be the same. P = Pconc + Pstl
Pconc L δ ⇒ Pconc = Aconc Econc Aconc E conc L P L δ ⇒ Pstl = Astl E stl δ = stl Astl E stl L
δ=
Combining these: P = ( Aconc E conc + Astl E stl )
δ L
From the definition of strain:
ε=
δ L
=
−P ( Aconc E conc + Astl E stl )
Determine the areas: Astl = 6 Aconc =
π 4
π 4
(0.028m )2
= 3.695 × 10 −3 m 2
(0.45m )2 − Astl
(
) (
)
= 159.0 × 10 −3 m 2 − 3.695 × 10 −3 m 2 = 155.3 × 10 −3 m 2
Plugging in the known values:
(
)
− 1550 × 10 3 N 155.3 × 10 −3 m 2 25 × 10 9 N 2 + 3.695 × 10 −3 m 2 200 × 10 9 N 2 m m −6 = −335.3 × 10
ε=
[(
)(
)
(
)(
)]
[Note: the negative sign denotes compression.] Assume that the deformation remains in the elastic region and apply Hooke’s Law:
(
σ conc = E conc ε = 25 × 10 9 N
(
σ stl = E stl ε = 200 × 10 9 N
m2
)(335.3 × 10
)(335.3 × 10 m 2
−6
−6
)
) = 8.383 × 10
6
N
= 8.383MPa m2 = 67.06 × 10 6 N 2 = 67.06 MPa m
These values are well below the yield values for steel and concrete, so the Hooke’s Law assumption is good. Note that even though the steel is a small portion of the cross-section, it carries most of the stress (load).
Ans
Problem: Wooden members A & B are joined by plywood slice plates that are fully glued on the surface of contact. The joint is designed with a ¼ in gap between members A & B (no glue here). Determine the smallest allowed length, L, for the splice plates if the glue shear stress limit is 120 psi.
Solution: Consider the following diagram of the joint:
There are two shear planes, so the load is split between them. There are four contact areas (two on either side) due to the gap. Let each of these glue areas be (l x w) .
τ=
P F = A lw
5.8 × 10 3 lbf 2 lbf ⇒ 120 2 = in l (4in ) ⇒ l = 6.042in
So, the total length is: L = l + gap + l = (60.42in ) + (0.25in ) + (60.42in ) = 12.33in
Ans
Problem: If P = 10 kips in the plate shown, determine the maximum stress when (a) r = 0.50 in, (b) r = 0.625 in.
Solution: I’ll use stress concentration factors to work this problem.
σ max = K
P A
Using the figure in the Appendix: D 5in r 0.5in = =2 = = 0.20 ⇒ K = 1.94 d 2.5in d 2.5in P 10 × 10 3 lbf σ max = K = (1.94 ) = 10.35 × 10 3 lbf 2 = 10.35ksi in Amin (2.5in )(0.75in )
(
)
D 5in r 0.625in = =2 = = 0.25 ⇒ K = 1.82 d 2.5in d 2.5in P 10 × 10 3 lbf σ max = K = (1.82 ) = 9.707 × 10 3 lbf 2 = 9.71ksi in Amin (2.5in )(0.75in )
(
Ans (a)
)
Ans (b)
Solution: I need to find the forces acting in members BE & CF. 1st, use the entire frame as a FBD:
∑M
A
= 0 = (40in )D x − (75in )(480lbf )
ccw+
⇒ D x = 900lbf
Next, use member DEF as a FBD: The reaction @ D must be parallel to FBE & FCF: 4 4 D y = D x = (900lbf ) = 1200lbf 3 3 ∑MF = 0 ccw+
4 = −(30in ) FBE − (45in )(1200lbf ) 5 ⇒ FBE = −2250lbf
∑M
E
=0
ccw+
4 = (30in ) FCE − (15in )(1200lbf ) 5 ⇒ FCE = 750lbf Therefore, member BE is in compression and member CE is in tension. Finally, determine the maximum normal stress in each member. Remember to adjust the area of the member for CE since it is in tension:
σ BE =
σ BE
P (− 2250lbf ) = = − 281.3 lbf 2 = 281.3 lbf 2 (Compr) in in (2in )(4in ) A
(750lbf ) = 107.1lbf = 107.1lbf (Tens) P = = in 2 in 2 A (2in )(4 − 0.5)in
Ans
Problem: A 1.5 m concrete post is reinforced with six steel bars, each rod’s diameter is 28 mm. Knowing that Es = 200 GPa and Ec = 25 GPa, determine the normal stresses in the steel and in the concrete when a 1550 kN axial centric force P is applied to the post.
Solution: The rebar and the concrete are well bonded. Therefore, the load is divided (but not equally) and carried the concrete and the steel. Also, the rebar and concrete deformation will be the same. P = Pconc + Pstl
Pconc L δ ⇒ Pconc = Aconc Econc Aconc E conc L P L δ ⇒ Pstl = Astl E stl δ = stl Astl E stl L
δ=
Combining these: P = ( Aconc E conc + Astl E stl )
δ L
From the definition of strain:
ε=
δ L
=
−P ( Aconc E conc + Astl E stl )
Determine the areas: Astl = 6 Aconc =
π 4
π 4
(0.028m )2
= 3.695 × 10 −3 m 2
(0.45m )2 − Astl
(
) (
)
= 159.0 × 10 −3 m 2 − 3.695 × 10 −3 m 2 = 155.3 × 10 −3 m 2
Plugging in the known values:
(
)
− 1550 × 10 3 N 155.3 × 10 −3 m 2 25 × 10 9 N 2 + 3.695 × 10 −3 m 2 200 × 10 9 N 2 m m −6 = −335.3 × 10
ε=
[(
)(
)
(
)(
)]
[Note: the negative sign denotes compression.] Assume that the deformation remains in the elastic region and apply Hooke’s Law:
(
σ conc = E conc ε = 25 × 10 9 N
(
σ stl = E stl ε = 200 × 10 9 N
m2
)(335.3 × 10
)(335.3 × 10 m 2
−6
−6
)
) = 8.383 × 10
6
N
= 8.383MPa m2 = 67.06 × 10 6 N 2 = 67.06 MPa m
These values are well below the yield values for steel and concrete, so the Hooke’s Law assumption is good. Note that even though the steel is a small portion of the cross-section, it carries most of the stress (load).
Ans
Problem: Wooden members A & B are joined by plywood slice plates that are fully glued on the surface of contact. The joint is designed with a ¼ in gap between members A & B (no glue here). Determine the smallest allowed length, L, for the splice plates if the glue shear stress limit is 120 psi.
Solution: Consider the following diagram of the joint:
There are two shear planes, so the load is split between them. There are four contact areas (two on either side) due to the gap. Let each of these glue areas be (l x w) .
τ=
P F = A lw
5.8 × 10 3 lbf 2 lbf ⇒ 120 2 = in l (4in ) ⇒ l = 6.042in
So, the total length is: L = l + gap + l = (60.42in ) + (0.25in ) + (60.42in ) = 12.33in
Ans
Problem: If P = 10 kips in the plate shown, determine the maximum stress when (a) r = 0.50 in, (b) r = 0.625 in.
Solution: I’ll use stress concentration factors to work this problem.
σ max = K
P A
Using the figure in the Appendix: D 5in r 0.5in = =2 = = 0.20 ⇒ K = 1.94 d 2.5in d 2.5in P 10 × 10 3 lbf σ max = K = (1.94 ) = 10.35 × 10 3 lbf 2 = 10.35ksi in Amin (2.5in )(0.75in )
(
)
D 5in r 0.625in = =2 = = 0.25 ⇒ K = 1.82 d 2.5in d 2.5in P 10 × 10 3 lbf σ max = K = (1.82 ) = 9.707 × 10 3 lbf 2 = 9.71ksi in Amin (2.5in )(0.75in )
(
Ans (a)
)
Ans (b)
