1. Strength of Materials University of Technology Second Class
Building and Construction Engineering Department
Strength of Materials
Second Class
2014 – 2015
prepared by
Professor Dr. Nabeel Al-Bayati
Structural Engineering Branch

2. Bending Stresses in Beams
Lecture -7
Bending Stresses in Beams
Lecture -7, Part 2
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

3. Lecture -7-part-2 (Bending Stresses in Beams)
7-11 Bending of Members Made of Several Materials:
Consider a composite beam formed from two materials with E1 and E2.
Normal strain varies linearly.
Piecewise linear normal stress variation.
Neutral axis does not pass through section centroid of composite section.
Elemental forces on the section are
Define a transformed section such that
(Ratio of the moduli
of elasticity).
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

4. Lecture -7-part-2 (Bending Stresses in Beams)
Solved Example
Bar is made from bonded pieces of steel (Es = 29x106 psi) and brass (Eb = 15x106 psi). Determine the maximum stress in the steel and brass when a moment of 40 kip*in is applied.
Solution Steps:
Step 1: Transform the bar to an equivalent cross section made entirely of brass
Step 2: Evaluate the cross sectional properties of the transformed section
Step 3: Calculate the maximum stress in the transformed section. This is the correct maximum stress for the brass pieces of the bar.
Step 4: Determine the maximum stress in the steel portion of the bar by multiplying the maximum stress for the transformed section by the ratio of the moduli of elasticity.
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

5. Lecture -7-part-2 (Bending Stresses in Beams)
7-12 Reinforced Concrete Beams:
Concrete beams subjected to bending moments are reinforced by steel rods.
The steel rods carry the entire tensile load below the neutral surface. The upper part of the concrete beam carries the compressive load.
In the transformed section, the cross sectional area of the steel, As, is replaced by the equivalent area nAs where n = Es/Ec.
To determine the location of the neutral axis,
The normal stress in the concrete and steel
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

6. Lecture -7-part-2 (Bending Stresses in Beams)
Solved Example
A concrete floor slab is reinforced with 5/8-in-diameter steel rods. The modulus of elasticity is 29x106psi for steel and 3.6x106psi for concrete. With an applied bending moment of 40 kip*in for 1-ft width of the slab, determine the maximum stress in the concrete and steel.
Solution Steps:
Step 1: Transform to a section made entirely of concrete.
Step 2: Evaluate the geometric properties of the transformed section.
Step 3: Calculate the maximum stresses.
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

7. Lecture -7-part-2 (Bending Stresses in Beams)
7-13: Solved Examples:
Example 7-21:Knowing that a beam of the cross section shown of two materials (Aluminum and steel) is bent about a horizontal axis and that the bending moment is 60 kip.ft, EAL= 10*106 psi and Es = 30*106 psi. determine the maximum stress in each material.
Aluminum
steel
4in
8in
3in
bAL=3in
Solution :
Transform the section to an equivalent cross section made entirely of steel
y’A, in3
y’, in
Area, in2
96 40 4 10
8*3 =24
1*4 = 4 1 2
∑y’A=136
∑A = 28
steel
4in
8in
1in
N.A. y’
3in
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

8. Lecture -7-part-2 (Bending Stresses in Beams)
Example 7-22:Knowing that a beam of the cross section shown of two materials (Wood and steel) is bent about a horizontal axis and that the (all) wood = 1200psi, (all) steel= 20000psi, EWood= 1.2*106 psi and Es = 30*106 psi. Find the allowable moment.
b = 8in
6in
3in
1/2in
12in
3in
6in
N.A.
8/25in
=0.32in y’ 2 1 3
steel
Solution : transform wood section to steel
steel
1/2in
Wood
y’A, in3
y’, in
Area, in2
12.75
24.96
0.75
6.5
0.25
2*0.5 =1
12*0.32 =3.84
6*0.5=3 1 2 3
∑y’A=38.46
∑A = 7.84
steel
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

9. Lecture -7-part-2 (Bending Stresses in Beams)
N.A.
=0.32in
y’=4.9 2 1 3
13-4.9
=8.1 13
0.5in
=7.6
=4.4
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

10. Lecture -7-part-2 (Bending Stresses in Beams)
Home Work : Solve the previous example (7-22) by transform steel section to wood
Example 7-23: For the beam shown:
Draw bending moment diagram along beam
Draw stress distribution due to Max Positive Moment only.
where EWood= 14kN/mm2 and ESteel = 210 kN/mm2.
steel
Wood
400mm
200mm
10mm
360mm
Section y-y
w=10kN/m 4m A B
1.5m
hinge
M=65kN.m C D y
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

11. Lecture -7-part-2 (Bending Stresses in Beams)
Solution :
w=10kN/m
Member CD: RC= RD = 10*4/2 = 20 kN
MC = 0 , MD = 0, M(mid-span CD) = 10*42/8 = 20kN.m C D RC RD 4m
w=10kN/m
40kN.m C A B
Member ABC: RA
M=65kN.m
RC=20kN
 MA = , MA *3* *3 = 0, MA = 40kN.m
1.5m
1.5m
 MC = , *3 * 1.5 – RA *3 = 0, RA = 50kN x
RA=50kN
RD=20kN
40kN.m
RC=20kN
Shear
VA = RA = 50kN
VB = 50 – 10 *1.5 = 35
VC = 50 – 10 *3 = 20
VD = 50 – 10 * 7 = -20 = RD
VA=50kN
VD=20kN
VC=20kN
VB=35kN
S.F.D
Moment
x = 0 to 1.5m
MA = - 40kN.m
MBA = - 40 – 10*1.5*1.5/2 + 50* 1.5 = 23.75kN.m
x = 1.5m to 3m
MBC = – 65 = kN.m
MC = 0
Mmax (Positive) = kN.m
+23.75kN.m
+20kN.m A B C D
-40kN.m
kN.m
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

12. Lecture -7-part-2 (Bending Stresses in Beams)
Wood
6000mm
3000mm
10mm
360mm y’
N.A.
100mm 2 1 3 4
200mm
Transform steel sections to wood
y’A, mm3
y’, mm
Area, mm2
150000 5
=375
360/2+10=190
2/3*360+10=250
3000*10 =30000
6000*10 = 60000
200*360 = 72000
2*(100*360/2)=36000 1 2 3 4
∑y’A=
∑A =
3000mm
Wood
6000mm
10mm
360mm
y’=228.94
N.A.
100mm 2 1 3 4
200mm
151.06
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

13. Lecture -7-part-2 (Bending Stresses in Beams)
400mm
A=13.634MPa
B=12.73
C=0.848
D=1.317
E=19.76
F=20.66
N.A.
Stress distribution at point B
due to max. positive moment A
10mm
Steel B C
c2=151.06
141.06
N.A.
Wood
360mm
218.94
c1=228.94 D
10mm
Steel E F
200mm
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

14. Lecture -7-part-2 (Bending Stresses in Beams)
Example 7-24: Circular sections of aluminum and steel having maximum bending moment = 80kN.m., where EAluminum= 70*103MPa and ESteel = 21*104 MPa. Find max. bending stress for steel and aluminum and draw stress distribution of the section.
Steel
Aluminum
150mm
300mm
Solution :
Transform steel sections to aluminum
Steel
Aluminum
150mm
300mm
3*150=450mm
Steel to Aluminum
N.A.
3*150=450mm
Steel to Aluminum
150mm
Aluminum
300mm
R1=150
R2=75
a=75mm
b=225mm
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

15. Lecture -7-part-2 (Bending Stresses in Beams)
Steel
Aluminum
150mm
300mm A B C D E F
N.A.
75mm
A=26.827MPa
B=13.413
C=40.24
D=40.24
E=13.413
F=26.827
N.A.
Stress distribution of the composite section
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)

16. Prof. Dr. Nabeel Al-Bayati
Thank you for listening
Prof. Dr. Nabeel Al-Bayati
Prof. Dr. Nabeel Al-Bayati
Lecture -7-part-2 (Bending Stresses in Beams)