Module 3: Analysis of Statically Determinate Structures - Examples

Reactions of Beams and Frames

Example 1: Simply Supported Beam with Point Load

A simply supported beam spans L=10 mL = 10 \text{ m}. It is supported by a pin at AA (left end) and a roller at BB (right end). A vertical point load of P=50 kNP = 50 \text{ kN} is applied at a distance of 4 m4 \text{ m} from support AA. Calculate the reactions at AA and BB.

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Example 2: Cantilever Beam with Uniform Load

A cantilever beam of length L=6 mL = 6 \text{ m} is fixed at the left end (AA) and free at the right end (BB). It carries a uniformly distributed load of w=15 kN/mw = 15 \text{ kN/m} over its entire length. Calculate the reactions at the fixed support AA.

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Example 3: Overhanging Beam with Multiple Loads

A beam is supported by a pin at AA (x=0x=0) and a roller at BB (x=8 mx=8 \text{ m}). The beam overhangs to the right to point CC (x=10 mx=10 \text{ m}). It carries a point load of 40 kN40 \text{ kN} at x=4 mx=4 \text{ m}, and a point load of 20 kN20 \text{ kN} at the overhang tip (x=10 mx=10 \text{ m}). Find the reactions at AA and BB.

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Analysis of Trusses

Example 1: Method of Joints (Basic)

A simple triangular truss consists of three members forming a right triangle. Member ABAB is horizontal (4 m4 \text{ m}), member BCBC is vertical (3 m3 \text{ m}), and member ACAC is the hypotenuse (5 m5 \text{ m}). It is pinned at AA and has a roller at BB. A downward force of 100 kN100 \text{ kN} is applied at joint CC. Find the internal force in member ABAB.

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Example 2: Method of Sections

A Pratt roof truss has 6 horizontal panels of 3 m3 \text{ m} each (18 m18 \text{ m} total span) and a height of 4 m4 \text{ m}. It is supported by a pin at the left and a roller at the right. Downward point loads of 50 kN50 \text{ kN} are applied at the bottom chord joints. Find the force in the top chord member located in the 3rd panel from the left.

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Example 3: Zero-Force Members

Identify zero-force members in a complex truss to simplify analysis.

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Cable Structures

Example 1: Cable with Concentrated Load

A cable is suspended between two supports AA and BB at the same elevation, spanning 20 m20 \text{ m}. A single point load of 100 kN100 \text{ kN} is applied at the midspan, causing a sag of 5 m5 \text{ m}. Find the maximum tension in the cable.

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Example 2: Cable with Uniform Horizontal Load (Parabolic)

A suspension bridge main cable spans 100 m100 \text{ m} horizontally between two towers of equal height. The cable carries a uniform horizontal load of w=20 kN/mw = 20 \text{ kN/m} (representing the deck). The maximum sag in the center is 10 m10 \text{ m}. Find the horizontal tension (HH).

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Example 3: Maximum Tension in Parabolic Cable

Using the same suspension bridge cable from Example 2 (Span 100 m100 \text{ m}, load 20 kN/m20 \text{ kN/m}, sag 10 m10 \text{ m}, H=2500 kNH = 2500 \text{ kN}), calculate the maximum tension in the cable.

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Three-Hinged Arches

Example 1: Symmetrical Three-Hinged Arch

A symmetrical three-hinged parabolic arch has a span of L=40 mL = 40 \text{ m} and a rise (height) of h=10 mh = 10 \text{ m} to the central crown hinge. It is subjected to a vertical point load of 120 kN120 \text{ kN} located 10 m10 \text{ m} horizontally from the left support. Calculate the horizontal thrust (HH) at the supports.

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Example 2: Unsymmetrical Load on Arch

A three-hinged semi-circular arch has a radius of 15 m15 \text{ m}. A uniform horizontal wind load of 5 kN/m5 \text{ kN/m} acts on the left half of the arch. Find the reactions at the base.

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Example 3: Internal Forces in an Arch

Using the symmetrical arch from Example 1, determine the internal shear force, axial force, and bending moment at a section 10 m10 \text{ m} from the left support (directly under the 120 kN120 \text{ kN} load).

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