Internal Forces - Examples & Applications

Practical calculation examples and applications for Internal Forces.

Mathematical Theory Examples

Example 1: Basic Internal Forces at a Point

A simply supported beam of length 10 m10\text{ m} carries a central point load of 100 kN100\text{ kN}. Determine the internal normal force, shear force, and bending moment at a point exactly 3 m3\text{ m} from the left support.

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Example 2: Intermediate Shear and Moment Equations

A cantilever beam of length LL is fixed at the left end (x=0x=0) and free at the right end (x=Lx=L). It carries a uniform distributed load ww over its entire length. Derive the equations for shear force V(x)V(x) and bending moment M(x)M(x) as a function of the distance xx from the free end.

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Example 3: Advanced Shear and Moment Diagram Relations

A beam is subjected to a linearly varying distributed load that starts at 00 and reaches w0w_0 at the end. Explain mathematically the shape of the resulting Shear Force Diagram (SFD) and Bending Moment Diagram (BMD).

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Example 4: Overhanging Beam with Point Loads

An overhanging beam is supported at AA (x=0x=0) and BB (x=8.00 mx=8.00\text{ m}), with an overhang extending to CC (x=10.0 mx=10.0\text{ m}). A point load of 40.0 kN40.0\text{ kN} is applied at x=4.00 mx=4.00\text{ m}, and a point load of 20.0 kN20.0\text{ kN} is applied at the free end CC. Determine the internal shear and moment at x=6.00 mx=6.00\text{ m}.

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Example 5: Cantilever with an Applied Moment

A cantilever beam of length 5.00 m5.00\text{ m} is fixed at the left end (x=0x=0). A concentrated clockwise moment of 60.0 kNm60.0\text{ kN}\cdot\text{m} is applied at the free end (x=5.00 mx=5.00\text{ m}). Determine the internal shear and moment at x=2.50 mx=2.50\text{ m}.

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Example 6: Simply Supported Beam with Triangular Load

A simply supported beam of length 6.00 m6.00\text{ m} carries a triangular load that is 00 at the left support AA and increases linearly to 30.0 kN/m30.0\text{ kN/m} at the right support BB. Find the maximum shear force in the beam.

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Example 7: Internal Forces in a Frame

A rigid L-shaped frame consists of a vertical member ABAB (3.00 m3.00\text{ m} tall, AA is the fixed base) and a horizontal member BCBC (4.00 m4.00\text{ m} long). A downward point load of 50.0 kN50.0\text{ kN} acts at the free end CC. Determine the internal forces and moment at the midpoint of member ABAB (height y=1.50 my=1.50\text{ m} from AA).

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Example 8: Shaft with Bending and Torsion

A solid circular shaft is subjected to an internal bending moment of 1.50 kNm1.50\text{ kN}\cdot\text{m} and an internal torsional moment (torque) of 2.00 kNm2.00\text{ kN}\cdot\text{m} at a specific cross-section. Calculate the equivalent internal moment MeM_e and equivalent internal torque TeT_e used in the design of such shafts, using the formulas Me=12(M+M2+T2)M_e = \frac{1}{2}(M + \sqrt{M^2 + T^2}) and Te=M2+T2T_e = \sqrt{M^2 + T^2}.

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Case Studies: Conceptual Theory

Case Study 1: The Importance of Bending Moment Diagrams in Reinforced Concrete

A structural engineer uses Bending Moment Diagrams (BMD) extensively when designing reinforced concrete beams. Explain conceptually why the shape and sign (positive vs. negative) of the BMD dictate where the steel rebar must be placed within the concrete beam.

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Case Study 2: Point Loads and Shear Failure

Consider a heavy piece of machinery resting on a small area of a concrete floor slab. What specific feature on the Shear Force Diagram indicates a high risk of failure, and what kind of failure is this called?

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Case Study 3: Relationship Between Shear Zero-Crossings and Maximum Bending Moments

In beam analysis, locating the points where the shear force diagram crosses the zero axis is a critical step. Explain why these points are significant when determining the design capacity required for a beam.

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Case Study 4: Effect of Internal Hinges in Continuous Structures

A long continuous beam over multiple supports is sometimes designed with an internal hinge inserted at a specific location along its span. Explain how an internal hinge alters the internal forces at that specific point.

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