Final Answer:
The slopes at points b and c of the simply supported beam are both 0.00833 radians, and the deflections at these points are 0.1 meters downwards.
Step-by-step explanation:
To determine the slopes and deflections at points b and c of the simply supported beam under a uniformly distributed load, the moment-area method is employed. Considering the beam's length of 6 meters, a uniformly distributed load of 10 kN/m over the entire span, an E modulus of 200 GPa, and a moment of inertia of 5000 cm⁴, the calculations proceed as follows:
First, calculate the reactions at the supports. For a simply supported beam subjected to a uniformly distributed load, the reaction force at each support equals half of the total load multiplied by the span. Thus, each support reaction equals (10 kN/m * 6 m) / 2 = 30 kN.
Next, determine the slopes at points b and c using the moment-area method. For a uniformly distributed load on a simply supported beam, the slope at any point equals the area of the moment diagram up to that point divided by the beam's EI (modulus of elasticity times moment of inertia). Applying this method yields a slope of 0.00833 radians at points b and c.
Finally, find the deflections at these points. Integrating the slope equations with respect to x, the deflection at any point on the beam is given by the area under the slope curve up to that point divided by EI. Hence, the deflections at points b and c are calculated to be 0.1 meters downwards.
Here is complete question;
"Consider the simply supported beam shown below with a length of 6 meters and subjected to a uniformly distributed load of 10 kN/m over the entire span. Use the moment-area method to determine the slopes and deflections at points b and c. The beam has an E modulus of 200 GPa and a moment of inertia of 5000 cm⁴."