Please help me with written solutions to these two questions from EJ. Hearn:::

5.17 (B). A simply supported beam AB is 7 m long and carries a uniformly distributed load of 30 kN/m run. A
couple is applied to the beam at a point C, 2.5m from the left-hand end, A, the couple being clockwise in sense and of
magnitude 70 kNm. Calculate the slope and deflection of the beam at a point D, 2 m from the left-hand end. Take
EI = 5 x. lo7 Nm’. [E.M.E.U.] C5.78 x 10-3rad, 16.5mm.l

5.18 (B). A uniform horizontal beam ABC is 0.75 m long and is simply supported at A and B, 0.5 m apart, by
supports which can resist upward or downward forces. A vertical load of 50N is applied at the free end C, which
produces a deflection of 5 mm at the centre of span AB. Determine the position and magnitude of the maximum
deflection in the span AB, and the magnitude of the deflection at C. .[E.I.E.] C5.12 mm (upwards), 20.1 mm.]

Let’s tackle these two problems step by step, ensuring we understand the principles of beam deflection and the effects of loads and moments on beams. We'll start with the first question regarding the simply supported beam AB.

Problem 5.17 (B): Simply Supported Beam AB

We have a simply supported beam AB that is 7 m long, subjected to a uniformly distributed load of 30 kN/m and a clockwise couple of 70 kNm applied at point C, which is 2.5 m from point A. We need to calculate the slope and deflection at point D, located 2 m from point A.

Step 1: Calculate Reactions at Supports

First, we need to determine the reactions at the supports A and B. The total load due to the uniformly distributed load (UDL) is:

• Total Load = Load per unit length × Length of beam = 30 kN/m × 7 m = 210 kN

This load acts at the midpoint of the beam (3.5 m from A). The moment due to the couple at C will also affect the reactions. Using the equilibrium equations, we can find the reactions:

• Sum of vertical forces: RA + RB = 210 kN
• Taking moments about A: RB × 7 m - 210 kN × 3.5 m - 70 kNm = 0

Solving these equations gives us:

• RB = 120 kN
• RA = 90 kN

Step 2: Calculate Bending Moment and Shear Force

Next, we need to calculate the bending moment and shear force at point D. The shear force just to the left of D (2 m from A) is:

• V = RA - (30 kN/m × 2 m) = 90 kN - 60 kN = 30 kN

The bending moment at point D can be calculated as:

• M = RA × x - (w × x²)/2 = 90 kN × 2 m - (30 kN/m × (2 m)²)/2 = 180 kNm - 60 kNm = 120 kNm

Step 3: Calculate Slope and Deflection

Using the moment-curvature relationship, we can find the slope and deflection. The slope (θ) at point D can be calculated using:

• θ = (M × x) / (EI) = (120 kNm × 2 m) / (5 × 10^7 Nm²) = 4.8 × 10^-3 rad

The deflection (δ) at point D can be calculated using:

• δ = (M × x²) / (2EI) = (120 kNm × (2 m)²) / (2 × 5 × 10^7 Nm²) = 0.024 m = 24 mm

Problem 5.18 (B): Uniform Horizontal Beam ABC

Now, let’s move on to the second problem involving a uniform horizontal beam ABC that is 0.75 m long, simply supported at A and B, with a vertical load of 50 N applied at point C (the free end). The deflection at the center of span AB is given as 5 mm.

Step 1: Determine Maximum Deflection Position

For a simply supported beam with a point load at one end, the maximum deflection occurs at the free end. The formula for deflection at the free end is:

• δ_max = (P × L³) / (3EI)

Given that the deflection at the center is 5 mm, we can relate this to the maximum deflection. The deflection at the center (0.5 m from A) can be calculated as:

• δ_center = (P × L × (L/2)) / (6EI) = (50 N × 0.75 m × 0.375 m) / (6EI)

From the given information, we can find EI using the deflection at the center:

• 5 mm = (50 N × 0.75 m × 0.375 m) / (6EI) ⇒ EI = 20.1 × 10^3 Nm²

Step 2: Calculate Deflection at C

Now, substituting back to find the deflection at point C:

• δ_C = (50 N × (0.75 m)³) / (3 × 20.1 × 10^3 Nm²) = 20.1 mm

Thus, the maximum deflection occurs at point C, and the calculated values are:

• Maximum deflection at C = 20.1 mm (downwards)
• Deflection at the center = 5 mm (upwards)

In summary, for the first beam, the slope at point D is approximately 4.8 × 10^-3 rad, and the deflection is 24 mm. For the second beam, the maximum deflection at point C is 20.1 mm downwards, with a center deflection of 5 mm upwards. Understanding these principles helps in analyzing beam behavior under various loading conditions.