# rectangular load

## Problem 847 | Continuous Beams with Fixed Ends

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## Problem 846 | Continuous Beams with Fixed Ends

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## Problem 845 | Continuous Beams with Fixed Ends

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## Problem 827 | Continuous Beam by Three-Moment Equation

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## Problem 822 | Continuous Beam by Three-Moment Equation

**Problem 822**

Solve Prob. 821 if the concentrated load is replaced by a uniformly distributed load of intensity w_{o} over the middle span.

Answers:

$M_2 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\beta}{4(\alpha + 1)(1 + \beta) - 1}$

$M_3 = -\dfrac{w_o L^2}{4} \cdot \dfrac{1 + 2\alpha}{4(1 + \alpha)(1 + \beta) - 1}$

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## Problem 820 | Continuous Beam by Three-Moment Equation

**Problem 820**

Solve Prob. 819 if the concentrated load is replaced by a uniformly distributed load of intensity w_{o} over the first span.

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## Problem 334 | Equilibrium of Parallel Force System

**Problem 334**

Determine the reactions for the beam loaded as shown in Fig. P-334.

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## Solution to Problem 691 | Beam Deflection by Method of Superposition

## Solution to Problem 689 | Beam Deflection by Method of Superposition

**Problem 689**

The beam shown in Fig. P-689 has a rectangular cross section 4 inches wide by 8 inches deep. Compute the value of P that will limit the midspan deflection to 0.5 inch. Use E = 1.5 × 10^{6} psi.