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यह सामग्री किसी भी मौलिकता का दावा नहीं करती है और इसे निर्धारित पाठ्यपुस्तकों के विकल्प के रूप में उपयोग नहीं किया जा सकता है। मैं उन विभिन्न ओपन स्रोतों और NPTEL/ SWAYAM पाठ्य सामग्री को स्वीकार करना चाहूंगा जिनसे व्याख्यान नोट तैयार किया गया था। जानकारी का स्वामित्व संबंधित लेखकों या संस्थानों के पास है जहां ओपन स्रोत की सामग्री तैयार की गई थी। इसके अलावा, यह दस्तावेज़ व्यावसायिक उद्देश्य के लिए उपयोग करने के लिए नहीं है और ब्लॉगस्पॉट के मालिक किसी भी मुद्दों, कानूनी या अन्यथा, के लिए जिम्मेदार नहीं हैं जो इस दस्तावेज़ के उपयोग से उत्पन्न होते हैं।
Course Name: Structural Mechanics [संरचनात्मक यांत्रिकी]
Course Code: MZSCEA-SM-05
Content Creator: Dr. Mohd. Zameeruddin
Beam Deflections: Calculations of deflection for determinate beams by double integration, Macaulay’s method, moment area method, conjugate beam method, deflection by method of superposition
Energy Principles: Strain energy and strain energy density, strain energy in traction, shear, flexure and torsion - Castigliano's and Engessor's energy theorems, principle of virtual work, application of energy theorems for computing deflections in beams, Maxwell's reciprocal theorem, Williot Mohr diagrams
Method of Consistent Deformation: Different structural systems, concept of analysis, basic assumptions, indeterminacy, choice of unknowns, Castigliano's theorem
Moment Distribution Method: Analysis of continuous beams propped cantilevers, continuous beams - theorem of three moments - analysis of continuous beams settlement effects, thermal effect, Shear Force and Bending Moment diagrams for continuous beams, portal frames with and without sway.
Slope Deflection Method: Analysis of continuous beams, analysis of rigid frames, frames without sway and with sway, settlement effects, introduction to difficulties in frames with sloping legs and gabled frames
Course Outcomes and CO Mapping with BT Levels
|
Course
Id |
Course Outcome
Statements |
Planned Activities |
|
CO1 |
Identify determinate and indeterminate beams and
frames. |
- |
|
CO 2 |
Calculate slopes and deflections at various
locations for different type’s beams and frames. |
Demonstrative
experiments |
|
CO 3 |
Apply energy principles for the analysis of
determinate and indeterminate structures. |
- |
|
CO 4 |
Demonstrates the analysis of both sway and
non-sway frame structures using slope deflection and moment distribution
equations |
Demonstrative
experiments |
|
CO ID |
Action |
POs/PSOs |
Knowledge |
Condition |
Criteria |
|
CO1 |
Identify |
PO1, PO2 and PSO2 |
Factual, Conceptual, Procedural, Meta-cognitive and practical constraints |
Concept of stability
and determinacy, conditions of equilibrium and degree of freedom |
Determinate and
indeterminate structures |
|
CO 2 |
Apply |
PO1,
PO2 and PSO2 |
Conceptual, Procedural, Meta-cognitive and practical constraints |
Internal and
external work done and strain resilience |
Different condition
of loading and deformable capabilities |
|
CO 3 |
Calculate |
PO1,
PO2 and PSO2 |
Conceptual, Procedural, Meta-cognitive and practical constraints |
Relationship
between curvature, slope and deflection |
Various type of loadings
and end conditions |
|
CO 4 |
Demonstrates |
PO1, PO2 and PSO2 |
Factual, Conceptual, Procedural, Meta-cognitive and practical constraints |
Continuous beams
and frames |
End conditions,
distribution factor and rotational factors |
|
CO ID |
PO 1 |
PO 2 |
PO 3 |
PO 4 |
PO 5 |
PO 6 |
PO 7 |
PO 8 |
PO 9 |
PO 10 |
PO 11 |
PO 12 |
|
CO 1 |
2 |
2 |
- |
- |
- |
- |
- |
- |
- |
- |
- |
- |
|
CO 2 |
2 |
2 |
- |
- |
- |
- |
- |
- |
- |
- |
- |
- |
|
CO 3 |
3 |
3 |
- |
- |
- |
- |
- |
- |
- |
- |
- |
- |
|
CO 4 |
3 |
3 |
- |
- |
- |
- |
- |
- |
- |
- |
- |
- |
|
Average |
2.5 |
2.5 |
- |
- |
- |
- |
- |
- |
- |
- |
- |
- |
It is defined as resistance of member to deformation when subjected to
load. Stiffness may be defined
as a ratio of unit force [F] causing a unit displacement [∆] or a ratio of a unit
moment [M] causing a unit deformation [Ɵ].
Mathematically,
K = F/∆
K = M/Ɵ
There are two types of
stiffness
Absolute Stiffness
It is expressed in terms
of the material's Young's modulus (E), the moment of inertia of the
cross-section (I), and the length of the member (L).
For a beam with one end
hinged and other end fixed, absolute stiffness is 4EI/L
Relative Stiffness
The relative stiffness of
a member is the ratio of a member's stiffness to the total stiffness of all
members connected to a joint.
Mathematically,
Kr = K1/K2
Where Kr is
relative stiffness, K1 and K2 are stiffness of two different
components or members respectively.
Relative stiffness is
expressed in terms of I and L, omitting the constant E.
Relative stiffness may be defined as the ratio of moment of inertia to the length of that member K = I/L.
Note: Video Lecture Link - Stiffness
Distribution factor:
When a moment is applied
at a joint to create rotation without causing the members to translate, the
moment is distributed among all of the members joined at that joint in
proportion to their stiffness. The ratio of summative stiffness at the joint to
the member stiffness is referred as Distribution Factor (DF).
Mathematically,
DFi = Ki/∑K
DFi is the distribution factor of ith joint
Ki is the stiffness of ith joint
∑K = ∑K1+K2+K3+K4
is summation of stiffness of all member meeting at a joint.
Carry Over Factor [COF]
It may be defined as the ratio
of moment developed or induced at one end due to a moment applied at another
end.
For a beam with one end
hinged and other end fixed, the carry over moment at fixed end is half the applied
moment at the hinge end, hence COF is 0.50
For a beam with both ends
hinged, the carry over moment will be of equal magnitude, hence COF is 1.0
Flexural Rigidity
The product of Young’s Modulus (E) and Moment of Inertia (I) is called Flexural Rigidity (EI) of Beams. The unit is N / mm2.
Sway
Sway is the lateral
movement of the joints in a portal frame due to the unsymmetric in dimensions, unsymmetric
loads, unequal moments of inertia, and different end conditions.
Video Lecture - Carry over factor
- Enlist different methods to find deflection in the beam and explain any one method. [DBATU, Summer 2025]
- A cantilever beam of span 5m is subjected to a concentrated load of 100 kN applied at the free end. Assume EI, the flexural rigidity of the beam is constant. Calculate the maximum deflection at the free end of the beam in terms of EI [DBATU, Summer 2025]
- Explain the concept of static and kinematic indeterminacy with examples [DBATU, Summer 2025]
- Determine the deflection at free end of a cantilever beam of length 5m subjected to a point load of 100 kN at the free end. Use the strain energy method. Assume uniform flexural rigidity as "EI". [DBATU, Summer 2025]
- Determine the reactions at the propped end for a propped cantilever beam of span 10m subjected to a uniformly distributed load of 5 kN/m. Also find the bending moment at the fixed end. [DBATU, Summer 2025]
- A fixed beam of span "L" is subjected to a point load "W" kN at mid span. Draw shear force diagram and bending moment diagram. Locate the point of contraflexure. Assume flexural rigidity, EI is constant. Use consistent deformation method. [DBATU, Summer 2025]
- A fixed beam of span 10m is subjected to uniformly distributed load of magnitude 50 kN/m. throughout the span. Construct the shear force and bending moment diagram. Locate the point of contraflexure. Flexural rigidity EI is constant. Use consistent deformation method. [DBATU, Summer 2025]
- State and explain theorem of three moments. [DBATU, Summer 2025]
- Explain the terms: a) Carry over moment, b) Carry over factor and c) stiffness factor [DBATU, Summer 2025]
- State and explain the assumptions made in the slope deflection method while developing this method. [DBATU, Summer 2025]
- Develop slope deflection equations for two span continuous beam ABC having each span (AB & BC) of length 4m and subjected to a uniformly distributed load of 10 kN/m. Support A is hinged, and B and C are rollers. Assume constant EI for both the span. [DBATU, Summer 2025]
- Analyze the two span continuous beam ABC by the slope deflection method, having both ends as fixed ones. Span AB is having 4m length, moment of inertia as I and subjected to central point load of 40 kN. Span BC is having 6m length and moment of inertia as 2I and is subjected to a uniformly distributed load of 20 kN/m over the entire span BC. Draw the bending moment diagram and elastic curve/deflected profile. [DBATU, Summer 2025]
- Analyze the two span continues beam ABC as shown in figure below by the Moment Distribution method and draw the bending moment diagram and deflection profile. [DBATU, Summer 2025]
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