HASAN AFTAB SAEED, PhD, PE
COMBINED LOADING
Mechanics of Materials II
HASAN AFTAB SAEED, PhD, PE
TAKING STOCK
We have considered five types of loads and the stresses they cause
axial force 𝑁
We now know how to determine the
stress in a member subjected
separately to an internal axial
force 𝑁 , a shear force 𝑉 , a
bending moment 𝑀 , a torsional
moment 𝑇, or an internal pressure 𝑝
shear force 𝑉
bending moment 𝑀
torsional moment 𝑇
internal pressure 𝑝
HASAN AFTAB SAEED, PhD, PE
COMBINED LOADING
In most mechanical/structural design and analysis however, the cross section of a
member will be subjected to several of these loadings simultaneously (in
combination)
When this happens, the principle of superposition comes in handy to determine the
resultant stress caused by the combination of loads
HASAN AFTAB SAEED, PhD, PE
PRINCIPLE OF SUPERPOSITION
The resultant stress at a point can be determined by algebraically summing the
stress caused by each separate load component provided linear structural response
Linear structural response
▪
Stress components are linearly related to their corresponding loads
→ For linear elastic behaviour, this is satisfied
▪
The loading must not significantly change the configuration of the load vis a vis
member geometry
→ For small deformations, this is satisfied
Linear structural response is expected from all problems in this course
HASAN AFTAB SAEED, PhD, PE
COMBINED LOADING: GENERAL PROCEDURE
Stress on any point of interest 𝑃 is computed in four steps:
1. Cut the member on the cross-section containing 𝑃
2. Determine 𝑁 and 𝑉 (at the centroid 𝑂 of the cross-section), and 𝑀 and 𝑇 about the
principal axes of inertia for the cross section
3. Compute 𝜎𝑁 , 𝜏𝑉 , 𝜎𝑀 and 𝜏 𝑇 at 𝑃 separately
4. Use superposition to algebraically add 𝜎𝑁 to 𝜎𝑀 , and 𝜏 𝑇 to 𝜏𝑉 to get the resultant
normal stress 𝜎 and the resultant shear stress 𝜏 (respectively) at 𝑃
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 1
The 500 kg engine is suspended from the jib crane at the position shown. On a
properly oriented stress element, show the state of stress at point 𝐵 on the
cross section of the boom at section 𝑎– 𝑎. Point 𝐵 is just above the bottom
flange.
Identification of 𝐷𝐸 as a two-force member
is crucial for analysis
Why?
0.21 MPa
𝐵
𝑦
𝑥 0.21 MPa
3.26 MPa
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 2
The rib-joint pliers are used to grip the smooth pipe 𝐶. If the force of 100 N is
applied to the handles, determine the state of stress at points 𝐴 and 𝐵 on the
cross section of the jaw at section 𝑎 − 𝑎 . Indicate the results on a stress
element for each point.
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 3
Determine the magnitude and the location of the maximum normal stress
developed on the section 𝑎 − 𝑎.
𝜎max = 24 MPa (C)
at point 𝐵
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 4
The drill is in contact with the wall and is subjected to the torque and force
shown. Determine and show on a properly oriented stress element the state of
stress at point 𝐴 on the cross section of drill bit at section 𝑎– 𝑎.
𝑦
𝑧
𝑥
𝐴
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 5
The solid rod shown has a radius of 7.5 mm. If it is subjected to the loading
shown, determine and show on a stress element the state of stress at point 𝐴.
Vector analysis
214.10 MPa
175.05 MPa
HASAN AFTAB SAEED, PhD, PE
EXAMPLE 5 (SUMMARY OF X-SECTION STRESSES)
Point 𝐴