Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial #1 - CivE. 205 Name: __________________________ I.D:_____________ _______________________________________________________________________________ Exercise 1: For the Beam below:
- Calculate the reactions at the supports and check the equilibrium of point a - Define the points at which there is change in load or beam shape - Draw the Shear Force Diagram (S.F.D.) and Bending Moment Diagram (B.M.D.). Drawing the diagrams to-scale and
show the compression side.
10.5 10.5
0.5 0.5
11.5
+
0
21 22
17
0
22.04
10 KN
a
2 m
3 KN / m
2 m 4 m
S.F.D.
B.M.D.
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Exercise 2: - Draw the (N.F.D.), (S.F.D.), and (B.M.D.) for the following frame.
13
13
3
3 1
1
13
9
1 1 10
10
2 2
The column has zero shear.
38
38
42
2 2
20 18
2
0
2
18
20
C sum of moments = 0
10 KN
4 m
2 m
4 m 1.5 m
a
b
1 m
2 m
1 KN / m 2 KN.m
10 KN
2 KN
2 m
c
N.F.D.
S.F..D.
B.M.D.
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial 3:
7.33
3.33 3.33
6.67
S.F.D.
6.67 4
20
6.66
4
10
B.M.D
1
Beam cross section
0.5 m
0.3 m
2
3
4
5
6
1
2
3
4
5
6
M =10
M =10 V = 6.7
Properties of area: Ix = b.t3/12 = 0.3*0.53/12 = 0.003125 m4 Q6, 4 = A.y’ = 0.125*0.3*0.1875 = 0.00703125 m3 Q3 = A.y’ = 0.25*0.3*0.125 = 0.009375 m3 Q1, 2, 5 = 0 m3
Forces: Mx = 10 KN.m Vy = -6.67 KN
Stresses: σ1, 2 = Mx. y / Ix = -10 * 0.25 / 0.003125 = -800 KN/m2 σ5 = Mx. y / Ix = 10 * 0.25 / 0.003125 = 800 KN/m2 σ6 = Mx. y / Ix = -10 * 0.125 / 0.003125 = -400 KN/m2 σ4 = Mx. y / Ix = 10 * 0.125 / 0.003125 = 400 KN/m2 σ3 = Mx. y / Ix = 10 * 0 / 0.003125 = 0 KN/m2 τ1, 2, 5 = Vy. Q1 / t . Ix = -6.67 * 0 / 0.3*0.003125 = 0 KN/m2
τ6, 4 = Vy. Q6 / t . Ix = -6.67 * 0.00703125 / 0.3*0.003125 = -50 KN/m2
τ3 = Vy. Q3 / t . Ix = -6.67 * 0.009375 / 0.3*0.003125 = -66.67 KN/m2
800 Element1, 2
66.67 Element3
800 Element5
Element6 50 400
Element4
50 400
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
σx = - 800, σy = 0, τxy = 0 tan 2θp = 2τxy/ (σx – σy) θp = 0o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy
2 σ max, min = 0, - 800 σx = 0, σy = 0, τxy = + 66.67 tan 2θp = 2τxy/ (σx – σy) θp = 45o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy
2 σ max, min = +/- 66.67 σx = - 400, σy = 0, τxy = + 50 tan 2θp = 2τxy/ (σx – σy) θp = - 7.02o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy
2 σ max, min = 6.16, - 406.16 σx = 400, σy = 0, τxy = + 50 tan 2θp = 2τxy/ (σx – σy) θp = 7.02o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy
2 σ max, min = 406.16, - 6.16 σx = 800, σy = 0, τxy = 0 tan 2θp = 2τxy/ (σx – σy) θp = 0o σ max, min = (σx + σy) / 2 +/- ((σx – σy) / 2)2 + τxy
2 σ max, min = 800, 0
800 Element 1 & 2
66.67 Element 3
45o
66.67 66.67
Element 6 50 400
7.02o
406.16
6.16
Element 4 50 400
7.02o
406.16
6.16
800 800 Element 5
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Element Mohr’s Circle principle plane max. shear
Tutorial 4 - Ex.1:
5
5 5
5
5 5
45ο
σmax σmax = 10 σmin
= 0
x
σ
τ x
y = (5, -5)
σmax σmin
2θp= 90ο
From point x to σmax
R = 5
τ
σmax
45ο
σmax = 5
σmin = -5
x
σ
y
x
σmax σmin
2θp = 90DFrom point x to σmax
� R = 5
5 5
x θp = 31.7ο θs = 76.7ο x
σmax
τ max
5
5
σ
τ
y
x 2θ from point x to σmax
(σav, τmax)
(σav, -τmax)
(σmax, 0)
2θs from point x to σav & τmax
(σmin, 0)
R = 5.6
τ max = 5.6 σ av = 2.5
x
5
5 σ
τ
x = (0, 5)
y = (5, -5)
2θp from point x to σmax
(σav, -τmax)
(σmax, 0)
2θs from point x to σav & τmax
(σmin, 0)
θp = 59.3ο θs = 13.3ο
x
σmax
τ max
R = 5.6
τ max = 5.6 σ av = 2.5
σ max = 8.1 σ min = -3.1
σ max = 8.1 σ min = -3.1
σ max = 5 σ min = -5
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Element Mohr’s Circle principle plane max. shear
5
5
5
σ
τ
y
x
σmax σmin
2θp = 45ο From point x to σmax
R = 7.07 2θs from point x to σav & τmax
σmax
22.5ο
σmax = 7.07
σmin = -7.07
x
τ max
22.5ο x
τ max = 7.07
5
3
4 σ
τ
y
x
2θp from point x to σmax
(σav, τmax)
(σav, -τmax)
(σmax, 0)
2θs from point x to σav & τmax
(σmin, 0)
R = 5.025
σmax
θ p= 42.1ο
σmax = 8.5
σmin = -1.5
x
τ max
θs = 2.9ο x
τ max = 5.025 σ av = 3.5
5
5 R = 0
τ
σ x = y = (5, 0)
θp = θs = 0 (one point)
5
5
5
5
5
5
τmax
45ο x
σ y x
σmax σmin
R = 5
2θs from point x to σav & τmax
5
5
τ max = 5 σ av = 0
σ max = 7.07 σ min = -7.07
σ max = 8.5 σ min = -1.5
τ
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Exercise 2:
S.F.D.
B.M.D
41 kips
12.2 kips
7.8 kips
43 kips
16 kips
239.4 kip.ft
40 kip.ft
I
I
1
Beam cross section
24 in
12 in
2
3
Properties of area: Ix = b.t3/12 = 1*23/12 = 0 .667 ft4 Q2 = A.y’ = 1*1*0.5 = 0.5 ft3 Q1 =Q3 = 0
Forces: Mx = 239.4 kip.ft Vy = 12.2 kips
Stresses: σx1 = Mx. y / Ix = -239.4 * 1 / 0.667 = -359.1 kip/ft2 σx2 = Mx. y / Ix = -239.4 * 0 / 0.667 = 0 kip/ft2 σx3 = Mx. y / Ix = 239.4 * 1 / 0.667 = 359.1 kip/ft2 τy1 = Vy. Q1 / t . Ix = 12.2 * 0 / 1*0.667 = 0 kip/ft2 τy2 = Vy. Q2 / t . Ix = 12.2 * 0.5/ 1*0.667 = 9.15 kip/ft2 τy3 = Vy. Q3 / t . Ix = 12.2 * 0 / 1* 0.667 = 0 kip/ft2
9.15
359.1
359.1
Element1
Element2
Element3
Element1, R = 179.55 Element3, R = 179.55 Element2, R = 9.15
2θp from point x to σmax x
y = (0, -9.15)
σ max = 9.15 σ min = -9.15
9.15
-9.15
45ο x
σmax
σ min = -359.1 σ max = 0 σ max = 359.1 σ min = 0
x y y x
τ τ
τ
σ σ σ
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial #5
The state of strain at a point on a wrench has components εx = 150 (10-6), εy = 200 (10-6), and γxy = -700 (10-6).
- Use Mohr’s circle to determine the equivalent in-plane strains on an element oriented at an angle of θ = 30o clockwise from the original position.
- Sketch the deformed elements at the original and the new orientation. - Sketch the elements at the principal plane and the maximum shear plane. - Determine the absolute maximum shear strain.
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II
Tutorial # 6
1. From the given strains, calculate the state of plain stress (draw the element).
Strain Values εx = + 350 (10-6), εy = + 600 (10-6), γxy = - 400 (10-6)
τxy = G.γxy = 11.28*106 * (-400)* 10-6 = - 4512 psi
E. εx = σx - υ. σy
29*350 = σx - .285 σy (1)
29*600 = σy - .285 σx (2)
By solving Eq. (1) and (2) σx = 16444.3 psi & σy = 22085.2 psi 2. Draw Mohr’s circles for stress and for strain. 3. Calculate the required yield stress σyield for the material so that to prevent failure with respect to both
Von Mises and TRESCA criteria, with a factor of safety of 2.5. Von Mises: For 2-D: σ1
2 - σ1. σ2 + σ2
2 < σyield
2 σ12
- σ1. σ2 + σ22
< (σyield / F.S)2
24585.752 -24585.75*13943.75 + 13943.752 = 456069715.5625 = (σyield / F.S)
2 σyield = 53389.5 psi TRESCA: τmax (3-D) = σmax / 2 = 12292.875 psi τmax = σyield / 2 τmax = σyield / 2 F.S σyield = 2*2.5*12292.875 σyield = 61464.375 psi Select a material with σyield = 61464.375 psi answer
22085.2
4512
16444.3
R= 5321 x
y y σmax = 24585.75
σmin = 13943.75
σmax = 24585.75
Dr. Tarek Hegazy CIV E 205 - Mechanics of Materials II