Chapter 10, Problem 7RQ

H.W. Solve problems in (Das) page 249 Problems 9.1, 9.3, and 9.5

3. As the audio frequency of Fig. 11-7 goes down, what components of Fig. 12-4 must be modified for normal operation? OD C₂ 100 HF R₁ 300 Re 300 ww 100A R 8 Voc Rz 10k reset output 3 R7 8 Voc 3 reset output Z discharge VR₁ 5k 2 trigger 2 trigger 7 discharge R 3 1k 5 control voltage threshold 6 5 control voltage 6 threshold GND Rs 2k C. C. 100 GND Uz LM555 1 Ce 0.01 U, LM555 0.01 8.01.4 PRO Fig. 11-7 Audio lutput Pulse width modulator R4 1k ww C7 Re 1k ww R7 100 VR 50k 10μ Ra R10 C₁. R1 3.9k 3.9k 0.14 100k TO w Rs 51 82 3 H 10 Carrier U₁ Ca Input A741 2.2 Us MC1496 PWM signal input R2 0.1100k Uz A741 41 Cs 1 Re 10k VR2 50k VR3 100k 14 12 C3. 3% + Ce 0.1 10μ 5 1A HH C +12V 0.1 O PWM Output C 0.02- R 100k +12 V Demodulated output 6 Ca 0.33 w R 10k R12 100k ww 31 о + 4A741 -12 V Fig. 12-4 PWM demodulator C 1500p

By using the yield line theory, determine the moment (m) for an isotropic reinforced concrete two-way slab shown in figure under a uniformly distributed load (w). m m 2000

Determine the collapse load for the simply supported slab. 3 m 3 m

m A square slab is simply supported along all sides and is to be isotropically reinforced. Determine the ultimate resisting moment (m) per linear meter required just to sustain a uniformly distributed load (w) in kN/m². m

By using the yield line theory, determine the moment (m) for an isotropic reinforced concrete two-way slab (supports on two S.S sides shown in figure under the load (P) (all dimensions are in mm). m m 2000 2000 3000

By using the yield line theory, determine the moment (m) for an isotropic reinforced concrete two-way slab shown in figure under a concentrated force (P) on the free corner. The two line supports of slab is simply supports. m m 2000 2000

1: Determine the load capacity of the one-way uniformly loaded (5 kN/m²) simply supported slab shown in Fig. 2 m 2 m 1.5 m E E

1: Determine the load capacity of the one-way uniformly loaded (5 kN/m²) simply supported slab shown in Fig. Solution: 2 m 2 m هنا الاسناد بسيط، لذلك سيتشكل خط خضوع واحد بالمنتصف ( البلاطة متناظرة) = We [5.0x (2x1.5) 0 = 8/2 :. W;= [m × 8/2 × 1.5] <2 = [1.5m 6] :: We = Wi 15 6 = 1.5 m 6 m = 10 kN.m 8/2] -8=1.0 1.5 m E E L 8/2 δ 28 0 = L/2 L