2. Unit 2- Stresses in Beams
Topics Covered
Lecture -1 – Review of shear force and bending
moment diagram
Lecture -2 – Bending stresses in beams
Lecture -3 – Shear stresses in beams
Lecture -4- Deflection in beams
Lecture -5 – Torsion in solid and hollow shafts.
3. Beam Deflection
Recall: THE ENGINEERING BEAM THEORY
σ M E
= =
y I R
Moment-Curvature Equation
v (Deflection)
y
€ NA
A B
x
A’ B’
If deformation is small (i.e. slope is “flat”):
4. 1 dθ
∴ ≈
R dx
R
δy
and δθ ≈ (slope is “flat”)
δx
B’
δy 1 d2y €
A’ ⇒ ≈ 2
R dx
€
€
Alternatively: from Newton’s Curvature Equation
y € ⎛ d 2 y ⎞ ⎛ dy ⎞ 2
R ⎜ 2 ⎟ if ⎜ ⎟ <<<< 1
1 ⎝ dx ⎠ ⎝ dx ⎠
= 3
y = f (x) R ⎛ 2 ⎞ 2
⎛ dy ⎞
⎜1+ ⎜ ⎟ ⎟ 1 d2y
x ⎝ ⎝ dx ⎠ ⎠ ⇒ ≈ 2
€ R dx
€
€
5. From the Engineering Beam Theory:
M E 1 M d2y
= = = 2
I R R EI dx
d2y
⇒ ( EI ) 2 = M
dx
€Flexural € € Bending
Stiffness Moment
Curvature
€
6. Relationship
A C B Deflection = y
dy
Slope =
dx
d2y
A C B Bending moment = EI 2
y dx
d3y
Shearing force = EI 3
dx
d4 y
Rate of loading = EI 4
dx
€
7. Methods to find slope
and deflection
Double integration method
Moment area method
Macaulay’s method
8. Double integration method
d 2 y ⎛ 1 ⎞
⎟ M Curvature
Since, 2 = ⎜
dx ⎝ EI ⎠
dy ⎛ 1 ⎞
⇒ = ⎜ ⎟ ∫ M ⋅ dx + C1 Slope
dx ⎝ EI ⎠
€
⎛ 1 ⎞
⇒ y = ⎜ ⎟ ∫ ∫ M ⋅ dx⋅ dx + ∫ C ⋅ dx + C
1 2 Deflection
⎝ EI ⎠
€
Where C1 and C2 are found using the boundary conditions.
€ Curvature Slope Deflection
y
R dy
dx
9. Double integration method
Simple supported
W
Slope Deflection
L/2 L/2 dy
A C B Slope = Deflection = y c
dx
yc 2 WL3
WL =−
= θA = θB = − 48EI
16EI
L
Uniform distributed load
x
€ Slope
€
Deflection
w/Unit length
A C dy
B Slope = Deflection = y c
dx
yc 2 5 WL3
WL =−
= θA = θB = − 384 EI
24 EI
L
€
€
10. Macaulay’s method
The procedure of finding slope and deflection for
simply supported beam with an eccentric load is very
laborious.
Macaulay’s method helps to simplify the calculations
to find the deflection of beams subjected to point
loads.
11. Moment-Area Theorems
• Consider a beam subjected to arbitrary
loading,
dθ d 2 y M
= =
dx dx 2 EI
θD xD
M
∫ dθ = ∫ EI dx
θC xC
xD
M
θ D − θC = ∫ EI
dx
xC
dx
CD = Rdθ = dx
€
€ R dθ
• First Moment-Area Theorem:
€
area under BM diagram between
€ C and D.
9 - 11
12. Moment-Area Theorems
• Tangents to the elastic curve at P and P’
intercept a segment of length dt on the vertical
through C.
M
dt = xdθ = x dx
EI
xD x
M 1 D 1 −
tC D = ∫ x
EI
dx = ∫ xMdx = EI A x
EI xC
xC
− A= total area of BM diagram between C & D
€ x = Distance of CG of BM diagram from C
• Second Moment-Area Theorem:
The tangential deviation of C with respect to
€ D is equal to the first moment with respect to
a vertical axis through C of the area under
the BM diagram between C and D.
9 - 12
15. An Exercise- Moment of Inertia – Comparison
1
Load
Maximum distance of
4 inch to the centroid I2
2 x 8 beam
Load 2
I1
Maximum distance of 1 inch to 2 x 8 beam
the centroid
I2 > I1 , orientation 2 deflects less
Ken Youssefi Engineering 10, SJSU 15
