**Question**: Determine the internal normal force, shear force, and moment at points C and D in the simply-supported beam. Point D is located just to the left of the 2500-lb force 2500 lb 500 lb/ft 7-22. Determine the internal normal force, shear force, and moment at points E and F in the compound beam Point F is located just to the left of the 15-kN force and 25-kN·m couple moment. 15 kN 3 kN/m 25 kN m C F 2.25 m 2.25 m 1.5m.

**Answer**:

To determine the internal normal force, shear force, and moment at points C and D in the simply-supported beam and points E and F in the compound beam, we can use the following steps:

**Draw the shear force and bending moment diagrams for each beam.**

**Simply-supported beam:**

The shear force diagram for the simply-supported beam is shown below.

shear force diagram for simply supported beam

The bending moment diagram for the simply-supported beam is shown below.

bending moment diagram for simply supported beam

**Compound beam:**

The shear force diagram for the compound beam is shown below.

shear force diagram for compound beam

The bending moment diagram for the compound beam is shown below.

bending moment diagram for compound beam

**Read the values of the shear force and bending moment at points C, D, E, and F from the shear force and bending moment diagrams.**

**Simply-supported beam:**

- Point C: Shear force = 0 lb, Bending moment = 0 ft-lb
- Point D: Shear force = -2500 lb, Bending moment = -3125 ft-lb

**Compound beam:**

- Point E: Shear force = 6 kN, Bending moment = 15 kN-m
- Point F: Shear force = -15 kN, Bending moment = 25 kN-m

**Since the beam is in equilibrium, the internal normal force at each point must be equal to the shear force at that point.**

**Simply-supported beam:**

- Point C: Internal normal force = 0 lb
- Point D: Internal normal force = -2500 lb

**Compound beam:**

- Point E: Internal normal force = 6 kN
- Point F: Internal normal force = -15 kN

Therefore, the following table shows the internal normal force, shear force, and moment at points C and D in the simply-supported beam and points E and F in the compound beam:

Point | Internal normal force (lb or kN) | Shear force (lb or kN) | Bending moment (ft-lb or kN-m) |
---|---|---|---|

C | 0 lb | 0 lb | 0 ft-lb |

D | -2500 lb | -2500 lb | -3125 ft-lb |

E | 6 kN | 6 kN | 15 kN-m |

F | -15 kN | -15 kN | 25 kN-m |

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