# rectangular load

# Problem 846 | Continuous Beams with Fixed Ends

# Problem 845 | Continuous Beams with Fixed Ends

# Problem 827 | Continuous Beam by Three-Moment Equation

# 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:

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

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

# 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.

# Problem 334 | Equilibrium of Parallel Force System

**Problem 334**

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

# 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.

# Solution to Problem 687 | Beam Deflection by Method of Superposition

**Problem 687**

Determine the midspan deflection of the beam shown in Fig. P-687 if E = 10 GPa and I = 20 × 10^{6} mm^{4}.