Tugevusõpetus I kodutöö 4 / Stength of materials I HW 4 (0)

Homework nr 4 in STRENGTH of MATERIALS I (MHE0011)
Variant
Title
A
B
Strength Analysis of a Beam
8
9
Student
Student Code
Delivery Date
Teacher
18.12. 2017
Priit Põdra
Hot-rolled INP section , manufactured of steel S235 must be used for a beam. The beam will be loaded by the concentrated load F and line distributed load p.
Beam dimensions obey the relationship :
b = a/2.
The value of concentrated load is F = 10 kN and the line distributed load is uniform with the intensity: p = F/b.
The value of design factor is: [S] = 4.
The beam load case is to be chosen according to the student code last digit A. The dimensions are to be chosen according to the last but one digit of the student code B.
Required milestones:

Compose a scaled structural model of the beam (according to particular A and B);

Calculate the supports’ reactions;

Compose the scaled diagrams of bending moment M and shear force Q;

Determine the beam’ critical sections (or critical section), compose the strength condition for bending and determine the appropriate INP section with minimum material usage ;

Compose a scaled sketch of the chosen INP section with diagrams of normal stress  and shear stress , acting at the critical section(s);

Calculate the values of safety factor for both bending and shear and check the adequate strength of the beam;

Formulate the answer .
Beam load case according to the last digit of the student code A
1
2
3
4
5
6
7
8
9
0
Beam dimensions according to the last but one digit of the student code B
1
2
3
4
5
a = 6,0 m; c = 2,4 m
a = 5,5 m; c = 2,5 m
a = 5,0 m; c = 2,6 m
a = 4,5 m; c = 2,5 m
a = 4,0 m; c = 1,8 m
6
7
8
9
0
a = 3,5 m; c = 1,5 m
a = 3,0 m; c = 1,6 m
a = 2,5 m; c = 1,1 m
a = 2,0 m; c = 1,1 m
a = 1,5 m; c = 0,7 m

Table of Contents

Table of Contents 2
1.Problem Statement 3
2. Solution 4
2.1.Calculating the supports’ reactions 4
2.2.Composing the scaled diagrams of bending moment M and shear force Q 5
2.3.Determining the beam’ critical sections (or critical section), composing the strength condition for bending and determining the appropriate INP section with minimum material usage 6
2.4.Composing a scaled sketch of the chosen INP section with diagrams of normal stress σ and shear stress τ, acting at the critical section(s)’ 7
3. Final Results 12

Problem Statement

Sketch of the problem
Initial data:
F is concentrated load
p is line distributed load
[S] is design factor
[S] = 4
F = 10 kN
a =2000 mm
c = 1100 mm
b = a/2 = 2000/2 = 1000 mm
p = F/b = 10/1 =10 kN*m-1

Solution

Calculating the supports’ reactions

Equation of equilibrium for point A
p = 10*1 = 10 kN
L is length of distributed load
Equation of equilibrium for point B
Equation of equilibrium for F
This means that support reactions’ values and directions are correct

Composing the scaled diagrams of bending moment M and shear force Q

The diagram of shear force Q is done by step method . From that, I calculated the bending moment M:
= FC * BC = 1,1*10 = 11 kNm
= FB * BD = 10,5 * 0,5 = 5,25 kNm
= FC * BC - FB * BD= 1,1*10 - 0,5 * 0.5 = 10,75 kNm

Determining the beam’ critical sections (or critical section), composing the strength condition for bending and determining the appropriate INP section with minimum material usage

Critical sections:
Section AE:
Q = 10,5 kN
M = 0 kNm
Section BC:
Q = 10 kN
M = 11 kNm
Minimum safe moment of resistance:
When bending is acting about the axis y
Uniform beam is safe enough, if its cross -section moment of resistance value is not smaller than 18,7 cm3
Above is a part of a table of hot-rolled I-profiles.
It shows the profile that corresponds to Wx≥[W]=18,7 cm3, is I 80.
Maximum bending stress
Check for adequate bending strength
-> The profile is not strong enough
Let’s choose I 200
Check for adequate bending strength
-> The profile is strong enough

Composing a scaled sketch of the chosen INP section with diagrams of normal stress σ and shear stress τ, acting at the critical section(s)’

Maximum shear stress at points O1
Q is cross-section shear force
SO1 is statical moment of half -section about axis y
I is cross-section axial moment of inertia
s is cross-section width at O1
Statical moment of half-section about axis y
I = 2140 cm4
Q = QB
Steel yield strength for shear
Check for adequate bending strength
Safety of cross-section BC for shear is adequate
Diagram of normal stress σ and shear stress τ of cross section BC
Shear stress at the cross-section A
Maximum shear stress at points O2
Q is cross-section shear force
SO2 is statical moment of half-section about axis y
I is cross-section axial moment of inertia
s is cross-section width at O2
Statical moment of half-section about axis y
I = 3060 cm4
Q = QA
Steel yield strength for shear
Check for adequate bending strength
Safety of cross-section A for shear is adequate
Diagram of normal stress σ and shear stress τ of cross section A
Because the bending moment MA = 0, then also the bending stress σ = 0

Final Results

Minimum safe profile: I-profile DIN 1025-1 – I 200
Critical points of a beam are A and B
Evaluation table (to be filled by teacher)
Correctness of the answer
Explanations of methodics
Quality of illustrations
Symbols ’ clarifications
Overall tidiness
Total