Aquí tenemos unos ejercicios de mecánica estatica

1. UNIVERSIDAD FERMIN TORO
VICERRECTORADO ACADEMICO
FACULTAD DE INGENIERIA
Mecanica Estatica
Integrantes
AnthonyRojas c.i: 23537385

2. Ejercicios:
1)
∑𝐹𝑦 = 0: 𝑅∆𝑦 + 𝑅 𝐵 = 5𝑘𝑛 (1)
∑𝑀∆ = 0: − 20(5) 𝑘𝑛 + 25𝑅 𝐵 = 0 => 𝑅 𝐵 =
100𝑘.𝑚
25𝑚
= 4𝑘 Así 𝑅∆𝑦 = 5𝑘𝑛 − 4𝑘𝑛 = 1𝑘𝑛
∑𝐹𝑦 = 0 1𝑘𝑛 − 𝑉1 = 0 => 𝑉1 = 1𝑘𝑛
∑𝑀1 = 0: − 1𝑘𝑛( 𝑋) + 𝑀1 = 0 => 𝑀1 = 𝑋 ( 𝑘𝑛. 𝑚)
X=0 => 𝑀1 = 0
X=0 =>𝑀1 = 20
∑𝐹𝑦 = 0 1𝑘𝑛 − 𝑉2 − 5𝑘𝑛 = 0 => 𝑉2 = 4𝑘𝑛
∑𝑀2 = 0: − 1𝑘𝑛𝑋 + 5( 𝑋 − 20) + 𝑀2 = 0 => 𝑀2 = 𝑋 − 5𝑋 + 100
𝑀2 = −4𝑘𝑛𝑋 + 100

3. 2)
𝐹𝑇 = 𝑎𝑟𝑒𝑎 𝑑𝑒 𝑡𝑟𝑖𝑎𝑛𝑔𝑢𝑙𝑜 =
1
2
𝑏ℎ =
1
2
(3𝑝)1200, 𝐹𝑇 = 1800𝑙𝑏
∑𝐹𝑦 = 0: 𝑅 𝐵𝑦 + 𝑅 𝐷𝑦 = 1800 + 4650 = 6450 => 𝑅 𝐵𝑦 + 𝑅 𝐷𝑦 = 6450𝑙𝑏

4. ∑𝑀𝐵 = 0 1(1800) − 3(4650) + 6𝑅 𝐷𝑦 = 0
𝑅 𝐷𝑦 =
12150
6
𝑙𝑏 = 2025𝑙𝑏
𝑅 𝐵𝑦 = 6450𝑙𝑏 − 2025𝑙𝑏 = 4425𝑙𝑏
𝐹1 =
1
2
𝑏ℎ =
1
2
(400𝑥) 𝑥 , 𝐹1 = 200𝑥2
∑𝐹𝑦 = 0
−𝑉1 − 200𝑥2 = 0 , 𝑉1 = −200𝑥2 {
𝑥 = 0 => 𝑉1 = 0
𝑥 = 3 => 𝑉1 = −1800𝑙𝑏
∑𝑀1 = 0:
1
2
𝑥𝐹1 + 𝑀1 = 0 => 𝑀1 = −
1
3
𝑥(200𝑥2 = −
200
3
𝑥3 𝑠𝑖 𝑥 = 0 => 𝑀1 = 0
𝑠𝑖 𝑥 = 3 => 𝑀1 = −1800𝑙𝑏𝑝
3<x<6
∑𝐹𝑦 = 0: −1800 + 4425 − 𝑉2 = 0; 𝑉2 = 2625𝑙𝑏
∑𝑀2 = 0 , ( 𝑥 − 2) 𝐹𝑇 − 4425( 𝑥 − 3) + 𝑀2 = 0
=> 𝑀2 = −1800( 𝑥 − 2) + 4425( 𝑥 − 3) = 2625𝑥 − 9675 , {
𝑥 = 3 => 𝑀2 = −1800𝑙𝑏
𝑥 = 6 => 𝑀2 = 6075𝑙𝑏

5. −1800𝑙𝑏 + 4425𝑙𝑏 − 4650 − 𝑉3 = 0 => 𝑉3 = −2025𝑙𝑏
∑𝑀3 = 0
1800( 𝑥 − 2) − 4425( 𝑥 − 3) + 4650( 𝑥 − 6) + 𝑀3 = 0
𝑀3 = −2025𝑥 + 18225(𝑙𝑏𝑝)
X=6 => 𝑀3 = 6075𝑙𝑏
X=9 => 𝑀3 = 0